Elementary Astronomical Calculations.pdf

For learning Elementary Astronomical Calculations

1. 1
Elementary Astronomical Calculations:
Gravity and motion ( Lecture –1)
•By Sukalyan Bachhar
•Senior Curator
•National Museum of Science & Technology
•Ministry of Science & Technology
•Agargaon, Sher-E-Bangla Nagar, Dhaka-1207, Bangladesh
•Tel:[phone](Off), Contact: 01923522660;
•Websitw: www.nmst.gov.bd; Facebook: Buet Tutor
&
•Member of Bangladesh Astronomical Association
•Short Bio-Data:
First Class BUET Graduate In Mechanical Engineering [1993].
Master Of Science In Mechanical Engineering From BUET [1998].
Field Of Specialization Fluid Mechanics.
Field Of Personal Interest Astrophysics.
Field Of Real Life Activity Popularization Of Science & Technology From1995.
An Experienced TeacherOf Mathematics, Physics & Chemistry for O-,A-, IB-&
Undergraduate Level.
Habituated as Science Speaker for Science Popularization.
Experienced In Supervising For Multiple Scientific Or Research Projects.
17 th BCS Qualified In Cooperative Cadre (Stood First On That Cadre).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

2. 2
Basic facts: Measurements of angles:
•Therearemainlythreemeasurementsystemsforangles.
•(1)Degreesystem;(2)Gradesystem&(3)Radiansystem.
•#(1)Degreesystem:Onerightangle=90
0
.
–Meaningofsuperscript:‘0’Degree;‘’Minute&‘’Seconds.
–Thissystemofmeasurementispopularoneinengineeringfield.
•#(2)Gradesystem:Onerightangle=100
g
.
–Meaningofsuperscript:‘g’Grade;‘’Minute&‘’Seconds.
–Thissystemofmeasurementissuitableforcalculatingasitisbasedon100
insteadofdegreesystemwhichisbasedon60.
•#(3)Radiansystem:Onerightangle=
c
/2
Meaningofsuperscript:‘c’Radian[‘c’comesfromcircle].
–Thissystemofmeasurementisthemostclosetonature(relatedtocircle)
andwidelyusedinpurescience.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

3. 3
Basic facts: Concept about radian
•Definitionofoneradian:Oneradianisananglesubtendedatcentrebythearcofacirclewhoselengthis
equaltotheradiusofthecircle.
•Definitionof(pi):Itisexperimentallyestablishedthattheratioofcircumference(orperimeter)ofa
circletoitsdiameterisalwaysconstant.
Thisconstantisknownas.
•Actualvalueofisstillundiscovered,itisusedas:
3.1415926535897932384626433832795;butasroughestimate,22/7.
•Circumferenceofacircle:
Fromtheabovedefinition:Circumference/Diameter=C/D=
But,diameter=2Radius.So,C/2r=C=2r.
•i.e,C=2r
[N.B.:Where‘C’‘D’and‘r’standforcircumference,diameterandradius
ofacirclerespectively.]
•Itisclearthat:incaseofsubtendinganangleatcentrebythearcofany
circle,theanglesubtendedatcentreisdirectlyproportionaltoitsarclength..
•Now,fromdefinitionofoneradian:Angle(inradian)=arclength/radius.
=s/rs=r.
[N.B.:Where‘’‘s’and‘r’standforangle,arclengthandradiusofacirclerespectively.]
•Thatis:onecompleteangle=Circumference/radius=2r/r=2.So,2
c
=360
0
90
0
=
c
/2
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

4. 4
Basic facts: Terminology of a right-angled triangle
•Let us consider a triangle, ABC.
Of which one angle, ABC= 90
0
or right angle.
•The angle, ACB = is introduced
as the Angle of consideration.
•Opposite arm of the right angle is known as Hypotenuse.
•The arm opposite the is introduced to be Opposite.
•The arm adjacent to (but not Hypotenuse ) is introduced to be Adjacent
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

5. 5
Basic facts: Trigonometric ratios
•Name of trigonometric ratios are:
•Sine sin ; Cosine cos ; Tangent tan; Cosecant cosec or csc ; Secant
sec ; Cotangent cot.
•Definition of trigonometric ratios:
• ; ;
• ; ;Hypotenuse
Opposite
sin Hypotenuse
Adjacent
cos Adjacent
Opposite
tan Opposite
Hypotenuse
eccos Adjacent
Hypotenuse
sec Opposite
Adjacent
cot
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

6. 6
Basic Facts: Relationship among the trigonometric ratios
•From above definition, one can easily find:
• 
• 
• 
•Also,
• 

eccos
1
Opposite
Hypotenuse
1
Hypotenuse
Opposite
sin === 

eccos
1
sin= 

sec
1
1
cos 









Adjacent
HypotenuseHypotenuse
Adjacent 

sec
1
cos 

cot
1
1
tan 









Opposite
AdjacentAdjacent
Opposite 

cot
1
tan 


tan
cos
sin


















Adjacent
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite 


tan
cos
sin

Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

7. 7
Basic Facts: Trigonometric inverse functions
•In general:
•For Trigonometric Function:  (y) fx xfy
-1
  
 
 xarctan or (y)tanx xtany
xarccos or (y)cosx xcosy
xarcsin or (y)sinx xsiny
1-
1-
-1


   
010
560.8290sin 0.829056sin :Example 

Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

8. 8
Basic Facts: Circle and sphere
•Circle: Equation: (x -h)
2
+ (y -k)
2
= r
2
Centre ( h, k) and radius = r
•Sphere: Equation: (x -h)
2
+ (y -k)
2
+ (Z -l)
2
= r
2
Centre( h, k, l) and radius = r
Surface area of sphere = 4r
2
Volume of sphere= (4/3)r
3
•Ellipse: Equation: (x/a)
2
+ (y/b)
2
= 1
a semi major axis, bsemi minor axis,
e eccentricity of the ellipse & e = (1 –(b/a)
2
)
Focus : ( ae, 0) & : ( -ae, 0) ; Centre (0, 0)
Area of the ellipse = ab
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

9. 9
Basic Facts: Equations of motions
(for constant acceleration or force)2asv
at
2
1
s
atv
:0)(u zero isvelocity initial withequations The
2asuv
at
2
1
uts
atuv
:Equations The
2
2
22
2







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

10. 10
Basic Facts: Equations of motions
(for constant acceleration or force)2gh-uv
/2guh 0,v :height maximum For ; gt
2
1
uth
u/gt 0,v :height maximum For ; gtuv

onaccelerati nalgravitatio g- a
reached height hs
:uvelocity initial withup thrownbody of equations The
2ghv
gt
2
1
h
gtv

onaccelerati nalgravitatio g a
released isbody whichfrom height hs
:0)(u zero isvelocity initial body with falling of equations The
22
22
2
2











Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

11. 11
Basic Facts: Newton’s law of gravitation
•Newton’s law of gravitation:In universe, any two bodies attract each other. This
force of attraction is directly proportional to product of their masses and inversely
proportional to square of distance apart between them. This force acts along the
line joining between them.
•Mathematically,
•Where: F is gravitational force; m
1and m
2are masses of two bodies;
r is the distance apart between them and
G is universal gravitational constant.
And G = 6.7 10
-11
N-m
2
/kg
2
.2
21
2
21
221
r
mm
GF
r
mm
F
r
1
F & mmF 
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

12. 12
Basic Facts: Newton’s law of gravitation
Gravitational acceleration on any object:
A body of mass ‘m’ is placed on the surface of the planet
of radius ‘r’ & mass ‘M’.
Force acting on mass ‘m’ is due to gravity22
2
r
GM
g
r
GMm
mgF
gonaccelerati nalgravitatio onaccelerati and
onacceleratimass force But,
r
GMm
Fgravity to due Force The





13. 13
Basic Facts: Angular motion
•Angular displacement is designated by .
•Angular velocity is designated with by .
•Angular acceleration is designated with by .
•The moment of inertia, I, of a body about an axis is defined by
where r
i is the perpendicular distance from the axis of a particle of mass m
i, and
the summation is taken over the whole of the body.
•Torque is designated by .on. so and 3, 2, 1,i Where; rmI i.e.
...rmrmI
2
ii
2
22
2
11



Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

14. 14
Basic Facts: Angular motion
(for constant angular acceleration or torque)distance pendicularr velocity, linearv mass,m
mvr
r
v
mr particles of Momentum Angular
Iω Momentum Angular
Momentum Angular
distance larperpendicur Force,F rFτ
(ττ Torque
distancer on,accelerati Lineara αra
r
v
a
t
ω
α
on(αnaccelerati Angular
distancer velocity, Linearv
r
v
ω
T
2π
ω
t
θ
ω
)velocity(ω Angular
2
2






Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

15. 15
Basic Facts: Angular motion
(for constant angular acceleration or torque) velocity angular Finalω velocity angular Initialω
2ααωω 3.
αt
2
1
tωθ 2.
αtωω 1.
Motion Angularof Equation
0
2
0
2
2
0
0
2




Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

16. 16
Basic Facts: Light
•Speed of Light in Vacuum = 2.99792458 10
8
m/s.
•Computational value of Speed of Light = 2.99792458 10
8
m/s.
•Speed of Light = c = Frequency () wave length ()
•Energy = E = Plank Constant (h) Frequency ()
•Plank Constant (h) = 6.63 10
34
J-s.
•Inverse Square Law:Intensity of light is inversely proportional to the square of
the distance of the source.2
1/rI
distance of square
1
Intensity 
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

17. 17
Basic Facts: Lightm 10 cm 10 Angstrom1
wavelenthof Unit
/T103)(λ h wavelengtingCorrespond
law s Wien':energy maximum
emitsbody Black whichat Wavelength
eTemperatur AbsoluteT
AreaSurfaceA
.degJ/m105.67Constant BoltzmanStefan)Constant(σ
TAConstantE
Body Black fromEnergy of emissiom of Rate
10-8-
6
m
428
4







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

18. 18
Basic Facts: Light
Color Frequency (THz) Wavelength(nm)
Violet 668-789 380-450
Blue 606-668 450-495
Green 526-606 495-570
Yellow 508-526 570-590
Orange 484-508 590-620
Red 400-484 620-750
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

19. 19
Basic Facts: Light
The Electromagnetic Spectrum
Wave Type Wavelength (m) Frequency (Hz) Energy (J)
Radio waves > 0.1 < 3 x 10
9
< 2 x 10
-24
Microwaves 10
-3
-0.1 3 x 10
9
-3 x 10
11
2 x 10
-24
-2 x 10
-22
Terahertz waves 10
-3
-10
-4
3 x 10
11
-3 x 10
12
2 x 10
-22
-2 x 10
-21
Infrared 7 x 10
-7
-10
-3
3 x 10
11
-4 x 10
14
2 x 10
-22
-3 x 10
-19
Optical (visible light)4 x 10
-7
-7 x 10
-7
4 x 10
14
-7.5 x 10
14
3 x 10
-19
-5 x 10
-19
Ultraviolet 10
-8
-4 x 10
-7
7.5 x 10
14
-3 x 10
16
5 x 10
-19
-2 x 10
-17
X-rays 10
-11
-10
-8
3 x 10
16
-3 x 10
19
2 x 10
-17
-2 x 10
-14
Gamma rays < 10
-11
> 3 x 10
19
> 2 x 10
-14
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

20. 20
Basic Facts: Light
•Table of astronomical constants
Quantity Symbol Value
Relative
uncertainty
Astronomical Unit AU 1.496 ×10
11
m -
Speed of light c 299 792 458 m s
−1
defined
Constant of gravitation G 6.674 28×10
−11
m
3
kg
−1
s
−2
1.0×10
−4
Parsec=A/tan(1") pc 3.085 677 581 28×10
16
m 4.0×10
−11
Light-year= 365.25cD ly 9.460 730 472 5808×10
15
m defined
Hubble constant H
0
70.1km s
−1
Mpc
−1
0.019
Solar luminosity L
☉
3.939×10
26
W
= 2.107×10
−15
S D
−1
variable,
±0.1%
Electron volt eV 1.602 ×10
−19
J defined
PI  3.1416 defined
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

21. 21
Basic Facts: Units and conversion
•Prefix Symbol Factor Numerically Name
•Tera T 10
12
10000000000 thousand billion
•Giga G 10
9
1 000 000 000 billion**
•Mega M 10
6
1 000 000 million
•Kilo k 10
3
1 000 thousand
•Centi c 10
-2
0.01 hundredth
•Milli m 10
-3
0.001 thousandth
•Micro μ 10
-6
0.000 001 millionth
•Nano n 10
-9
0.000 000 001 billionth**
•Pico p 10
-12
0.000000000001 thousand billionth
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

22. 22
Basic Facts: Greek Alphabets
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

23. 23
Gravitational force exerted by Sun And Earth on Moon
2.2FF 2.2
101.50
103.85
106
102

R
R
M
M

F
F
have, weequations above theof ratio theTaking
Constant onalGraviotati UniversalG Moon theof Massm
km 103.85Moon fromEarth of DistanceR kg 106 Earth theof MassM
km 101.50Moon fromSun of DistanceR kg 102 Sun theof MassM
Where,
R
mGM
F Earth theMoon toon Force lGravitiona
R
mGM
F Sun theMoon toon Force lGravitiona
R
mGM
F n,Gravitatio of law sNewton' From
ES
2
8
5
24
30
2
S
E
E
S
E
S
5
E
24
E
8
S
30
S
2
E
E
E
2
S
S
S
2
s





























Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

24. 24
Gravitational force exerted by Sun And Earth on Moon
•ThissuggeststhatSunexertsalmosttwiceasgreataforceonMoon
astheEarthdoes.
•ThusitisnotpropertosaythatMoonorbitstheEarth.Moonactually
orbittheSun,withtheEarthcausingthecurvatureofMoon’sorbitto
change.
•Moon’spathisalwaysconcavetowardsthesun,becausetheNet
forceonMoonisalwaysinward,evenwhenitisbetweentheEarth
andSun.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

25. 25
Gravitational acceleration on the Earth
*Theweightofanobjectofmassmatthe
surfaceoftheEarthisobtainedbymultiplying
themassmbytheaccelerationdueto
gravity,g,atthesurfaceoftheEarth.The
accelerationduetogravityisapproximately
theproductoftheuniversalgravitational
constantGandthemassoftheEarthM,
dividedbytheradiusoftheEarth,r,squared.
(WeassumetheEarthtobesphericaland
neglecttheradiusoftheobjectrelativetothe
radiusoftheEarthinthisdiscussion.)
*Themeasuredgravitationalaccelerationatthe
Earth'ssurfaceisfoundtobeabout9.80m/s
2
.22
r
GM
g
r
GMm
mgF


Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

26. 26
Gravitational acceleration on the Earth
•*Theaccelerationthatanobjectexperiencesbecauseofgravitywhenitfalls
freelyclosetothesurfaceofamassivebody,suchasaplanet.Alsoknownas
theaccelerationoffreefall,itsvaluecanbecalculatedfromtheformula
•g=GM/(R+h)
2
•whereMisthemassofthegravitatingbody(suchastheEarth),Risthe
radiusofthebody,histheheightabovethesurface,andGisthegravitational
constant(=6.6742×10-11N·m2/kg2).Ifthefallingobjectisat,orverynearly
at,thesurfaceofthegravitatingbody,thentheaboveequationreducesto
•g=GM/R
2
•InthecaseoftheEarth,gcomesouttobeapproximately9.8m/s2(32ft/s
2
),
thoughtheexactvaluedependsonlocationbecauseoftwomainfactors:the
Earth'srotationandtheEarth'sequatorialbulge.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

27. 27
Gravitational acceleration on the planets
m/s 3.66g
m/s 2.60.05539.8gMercuryon on Accelerati nalGravitatio
9.8g and 2.6
R
R
0.0553,
M
M

Mercury.consider usLet
R
R
M
M
g g
R
R
M
M

g
g

earthfor
R
MG
g
planetor body other any for
R
MG
g
:be willfollowing theplanet;or body other any on on accelerati nalgravitatio the
found becan it Earth, theof surface on theon accelerati lgravitiona theKnowing
2
22
22
2
2






























Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

28. 28
Gravitational acceleration on the planets
Acceleration Due to Gravity Comparison
Body Mass RatioRadius Ratio
g / g-Earth
Acceleration Due
to Gravity, "g" [m/s²]
Acceleratio
n Due
to Gravity,
"g" [m/s²]
Sun 332776
0
109.091 27.95 274.13
Mercury 0.0553 0.383 0.37 3.59
Venus 0.815 0.949 0.90 8.87
Earth 1.000 1.000 1.00 9.81
Moon 0.0123 0.273 0.17 1.62
Mars 0.107 0.533 0.38 3.77
Jupiter 317.8 11.200 2.65 25.95
Saturn 95.2 9.450 1.13 11.08
Uranus 14.5 4.010 1.09 10.67
Neptune 17.1 3.880 1.43 14.07
Pluto 0.0021 0.187 0.04 0.42
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

29. 29
Masses of Sun and Earth
•Mass of Sun: 
 

kg 101.98
360024365.25
4π
106.67
101.5
M Sun The of Mass
units. mks 106.67Constant nalGravitatio G
s 360024365.25Earth of Revolution of Time T
m 101.5Earth and Sun the between DistanceR
2ππ/earth the ofvelocity Angularω Earth, Massofm Sun, of MassM
Where,

T
4π
.
G
R

G
ωR
M Sun The of Mass
R
GMm
mRω
Sun of n AttractionalGravitatio - Force lCentripeta
have, WeEarth. on Sun the of attraction nalgravitatio
theby provided is force lcentripeta neccessary The Sun. the around revolves Earth The
30
2
2
11-
3
11
11-
11
2323
2
2












Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

30. 30
Masses of Sun and Earth
•Mass of Earth: 
 
kg 106
36002427.3
4π
106.67
103.84
M Earth The of Mass
s 36002427.3Moon of Revolution of Time T
m 103.84Earth and Moon the between DistanceR
Where,

T
4π
G
R
M Earth The of Mass
R
mGM
ωmR
have, WeMoon. of motion orbital
the for force lcentripeta neccessary the provides Earth the of attraction nalgravitatio The
24
2
2
11-
3
8
8
M
23
M
E
2
M
E2
MM










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

31. 31
Escape velocity  
 
km/s 11.2m/s 101.12 m
106.376
106106.672
VEarth the of Velocity Escape
m 106.376 Earth of RadiusR
kg 106 Earth of MassM
units mks 106.67Constant nalGravitatioG
Earth For
Earth of Velocity Escape the find us Let
R
2GM
V VelocityEscape
get, wegrearrangin and m Cancelling
0
R
GMm
-mV
2
1
Hence,
R
GMm
-Energy Potential nalGravitatio mV
2
1
EnergyKinetic
have, wezero. be shouldbody the ofenergy potential pluskinetic wordsother In body. the of
energy total zero to correspondvelocity minimum The planet. the ofvelocity escape as known
ismvelocity The planet. the to back returns never it that sovelocity minimum
certain a with withthrown be shouldbody the planet, a of clutch nalgravitatio this from
escape to order In cotect. its inbody a on attraction nalgravitatio exerts planetEvery
4
6
2411-
e
6
24
11-
e
2
2










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

32. 32
Escape velocitykm/s 60km/s
11.2
318
11.2VJupiter of Velocity Escape
11.2
1
R
R
and 318
M
M
whereJupiter, of Velocity Escape the find us Let

R
R
M
M
VVPlanet of Velocity Escape

R
M2G
VPlanet of Velocity Escape
R
2GM
VEarth of Velocity Escape
have, We.comparisonby planets the all ofvelocity escape calculate can weEarth
the withplanets the of ratios radius and mass and Earth the of Velocity Escape the Knowing
e
ee
ee


















IsaacNewton'sanalysisof
escapevelocity.ProjectilesA
andBfallbacktoearth.
ProjectileCachievesa
circularorbit,Danelliptical
one.ProjectileEescapes.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

33. 33
Escape velocity
Following table gives the Escape Velocity of All planets (Including the Sun and Moon):
Body Mass Ratio Radius Ratio Escape velocity
km/s
Sun 3327760 109.091 620
Mercury 0.0553 0.383 4.3
Venus 0.815 0.949 10.4
Earth 1.000 1.000 11.2
Moon 0.0123 0.273 2.4
Mars 0.107 0.533 5.0
Jupiter 317.8 11.200 60.0
Saturn 95.2 9.450 36.0
Uranus 14.5 4.010 21.3
Neptune 17.1 3.880 23.5
Pluto 0.0021 0.187 2.4
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

34. 34
Escape velocity from solar system  
 
:planets of distances at Sun to reference with
velocity escape the find can weunits alastronomic in Planets the of distance the Knowing
km/s 42m/s 104.2
m/s
101.5
102106.672
Distance sEarth' at Sun From Velocity Escape
Earth) of case (In m 10 1.5Sun from DistanceR
kg 102Sun of MassM units mks 106.67GravityConstantofG
where
R
2GM
velocity Escape
have, Wesystem,. solar the from escapey permanantlbody the that so Earth
of distance a atvelocity escape the find us Let Sun. from distance the on on depends which
required, isvelocity minimum certain a system solar from escape should it that so launced
be to is spacecraft a If gravity. sSun' from escapely neccessari notmay body the but imparted.
isvelocity y necccessar if planet froma ecsapemay body a though Even orbit. their in planets
the keep to force lcentripeta neccessary provides Sun the of attraction nalGravitatio
4
11
3011
11
3011









Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

35. 35
Escape velocity from solar system
Location with respect to V
e
[2]
Location with respect to V
e
[2]
on theSun, the Sun's gravity: 617.5km/s
onMercury, Mercury's gravity: 4.3km/s at Mercury, the Sun's gravity: 67.7km/s
onVenus, Venus' gravity: 10.3km/s at Venus, the Sun's gravity: 49.5km/s
onEarth, the Earth's gravity: 11.2km/s at the Earth/Moon,the Sun's gravity: 42.1km/s
on theMoon,
the Moon's
gravity:
2.4km/s at the Moon, the Earth's gravity: 1.4km/s
onMars, Mars' gravity: 5.0km/s at Mars, the Sun's gravity: 34.1km/s
onJupiter, Jupiter's gravity: 59.5km/s at Jupiter, the Sun's gravity: 18.5km/s
onSaturn, Saturn's gravity: 35.6km/s at Saturn, the Sun's gravity: 13.6km/s
onUranus, Uranus' gravity: 21.2km/s at Uranus, the Sun's gravity: 9.6km/s
onNeptune, Neptune's gravity: 23.6km/s at Neptune, the Sun's gravity: 7.7km/s
in thesolar
system,
theMilky Way's
gravity:
≥ 525km/s
[3]
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

36. 36
Kepler’s laws
•JohanesKepler(1571-1630)wasfirsttoconceivethelawsof
planetarymotion.AstronomicalobservationsofMarsled
Keplertotheellipticalorbits.
•Kepler’slawsarestatedasfollows:
•FirstLaw:Everyplanetmovesroundthesunonaelliptic
pathkeepingthesunoneofthefocusoftheellipse.
•Secondlaw:ThelineconnectingtheplanetandtheSun
sweepsoutequalareainequaltime.
•Thirdlaw:Thesquareoforbitalperiodictimeofaplanetis
directlyproportionaltothecubeofsemi-majoraxisofthe
ellipse.3
a
GM
4π2
τ ,accurately More
3
a
2
τ ally,Mathematic
2


Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

37. 37
Proof of Kepler’s second law
*
Kepler’s Second Law
A planet in its path around the sun sweeps out equal areas in equal times.
SupposeatagiveninstantoftimetheplanetisatpointPinitsorbit,
movingwithavelocitymeterspersecondinthedirectionalongthe
tangentatP(seefigure).Inthenextseconditwillmovevmeters,
essentiallyalongthisline(thedistanceisofcoursegreatly
exaggeratedinthefigure)sotheareasweptoutinthatsecond
isthatofthetriangleSPQ,whereSisthecenterofthesun.
TheareaoftriangleSPQisjust½basexheight.ThebasePQisvmeterslong,theheightisthe
perpendiculardistancefromthevertexofthetriangleatthesunStothebaselinePQ,whichis
justthetangentialvelocityvector.
Hence
ComparingthiswiththeangularmomentumLoftheplanetasitmovesaroundthesun,
itbecomesapparentthatKepler’sSecondLaw,theconstancyoftheareasweepingrate,istelling
usthattheangularmomentumoftheplanetaroundthesunisconstant.
Infact,
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

38. 38
Proof of Kepler’s third lawlaw. third sKepler' the proves table following The units. alAstronomic
in R distance and yearsin T period the select us letnumbers, large avoid to oder In
Constant
R
T
R T
RConstantT constant is
GM
4π
equation, this From
R
GM
4π
T
get, weg,rearrangin and sides both from m'' Cancelling
Revol of Time T
Sun of Mass M
revolution of TimePeriodic T 2ππ/velocity Angularω
Sun from Planet of DistanceR etMassofplanm
Where,

R
GMm
mRω
Sun of n AttractionalGravitatio Force lCentripeta
have, Wesun. the of attraction
onalgraviatatiby provided whichplanet a on act should force lcentripeta motion orbital For
3
2
32
32
2
3
2
2
2
2









Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

39. 39
Proof of Kepler’s third law
Planet Period T
(years)
Distance R
( Au)
Mercury 0.241 0.387 1.002
Venus 0.615 0.723 1.000
Earth 1.000 1.000 1.000
Mars 1.880 1.524 0.998
Jupiter 11.90 5.204 1.005
Saturn 29.50 9.582 0.989
Uranus 84.00 19.201 0.996
Neptune 165.0 30.047 1.003
Pluto 248.0 39.236 1.018 
32
32
/Auy
/RT
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

40. 40
Elliptical motion
•Inthesolarsystem,theEllipticalmotionisquitecommon.Alltheplanetsmove
aroundtheSunandallsatellitesoftheplanetsmovearoundtherespective
planetsintheEllipticalorbits.Henceitsinstructivetolearnmoreaboutthe
Ellipticalmotion.
•IntheEllipticalmotion
PPerihelionNearestpointtotheSun
AAphelionFarthestpointtotheSun
CCentreoftheellipse
CP=CA=aSemimajoraxis
CQ=CR=bSemiminoraxis
FFocusPositionoftheSun
F’EmptyFocus
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

41. 41
Elliptical motion

 
 
VectorPosition r
neglected) be (can Planet the of Massm
Sun the of MassM
units mks 106.67 Contant alGrvitationG Where
a
1
r
2
mMG V
ellipse the on pointany at Planet the ofvelocity the for expression the accept shall We
e1-a b axis minor semi The
e1ab eaba eabr
FCQ triangle angle right In
ar 2a2r
2aconstant rr Q, at is planet the If
2aconstantrr Ellipse,any For
F from VectorPositionrJF
F from VectorPositionrFJ
e 1a aeaCFCA FA Distance Aphelion
e1-a ae-aCF-CPFP Distance Perihelion
aeeCpCF
CP
CF
eEllipse the ofy Ecentricit The
11-
2
2
22222222222







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


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







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

42. 42
Elliptical motion
•Velocity at Perihelion
•Velocity at Aphelion
 

 


 
e1
e1
a
mMG
V
e1a
e12
mMG
a
1
e1a
2
mMG V
e1-ar perihelion At
2
P
2
P





















 
 

 


 
e1
e1
a
mMG
V
e1a
e12
mMG
a
1
e1a
2
mMG V
e1ar aphelion At
2
P
2
P






















Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

43. 43
10. Elliptical motion
•Magnitude of the position vector ‘r’  
 
   
 
   
 
plotted. be can ellipse the and found
be can r'' vector position the x'' of values different for known are e and a of values the If
ecosx1
e1a
r
ecosx14are14a
4arecosxe4a4ar4a
4raecosxe4arrar24a
r-2ar 2arr
cosx2ae2r2aerr2ar
cosx2ae2r2aerr
x-180cos2ae2r2aerr
JFF triangle the to cosines oflaw the Applying
2
22
222
22222
2222
222
222











Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

44. 44
Elliptical motion applied to Hale-Bopp comet
•TheHale-Boppcometwasthebrightest
cometsobservedinpastseveraltears.The
cometasnearesttheSunon1
st
April1997.Its
periodictimeistoolong,about2500years.
Naturallyitsorbitisquiteelongated.Asan
applicationoftheEllipticalmotion,Hale-Bopp
cometisveryidealcelestialobject.Withthe
knowledgeofeccentricityofitsorbitseveral
aspectsofthemotionofthecometcanbe
studied.
•Eccentricityoftheorbitofthehale-Boppcomet
=e=0.995074405
•Sinceeccentricityisnearlyonetheorbitis
almostparabolic.
•Periheliondistance=0.914091158AU(given)
•Thesemimajoraxis‘a’
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

45. 45
Elliptical motion applied to Hale-Bopp comet
*  

 
  
 
Sun to compared neglected be can comet Bopp-Hale of Mass m
kg 10 2 Sun of Mass M
units mks 10 6.67 constant nalGravitatio G

e1
e1
a
mMG
V
Perihelion at Speed
km 10555.3678 AU370.2452 AU450.665074401185.57965e1a distance Aphelion
orbit. sEarth' the insideslightly arrives comet Bopp-Hale the means Wkich
km 101.580.94109115 AU80.91409115 distance Perihelion
AU18.396681 AU0.99507441-185.5795 e1-ab have, We
b'' axis minor Semi
AU39.236only is Pluto of axos major Semi
e1-a Distance Perihelion have, We
AU185.57965 AU
50.995074401-
80.91409115
e1-
Distance Perihelion
a
30
11-
P
8
8
22















Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

46. 46
Elliptical motion applied to Hale-Bopp comet
* 
 
 yrs2528.1074
AU1
AU185.57965
yr1
R
R
TT
R
R

T
T

comet. Bopp-Hale and Earth of timePeriodic compare shall We
RT :Law 3rd sKepler' use We
T'' TimePeriodic
km/hr 391.5792 km/s 0.108772 m/s 108.772
e1
e1
a
mMG
V
Aphelionat Speed
km/hr 158400 km/s 44 m/s 104.4097506
m/s
0.995071
0.995071
101.5185.57965
010 210 6.67
V
3/2
2
3
E
H
HH3
E
3
H
2
E
2
H
32
a
4
11
3011-
P
































Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

47. 47
Barycenter
•The barycenter is the point between two objects where they balance each
other. It is the center of mass where two or more celestial bodies orbit each
other.
•When a moon orbits a planet, or a planet orbits a star, both bodies are actually
orbiting around a point that lies outside the center of the primary (the larger body).
•For example, the moon does not orbit the exact center of the Earth, but a point on a
line between the Earth and the Moon approximately 1,710 km below the surface of
the Earth, where their respective masses balance. This is the point about which the
Earth and Moon orbit as they travel around the Sun.
•
The barycenter is one of the foci of the elliptical orbit of each body. This is an
important concept in the fields of astronomy and astrophysics.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

48. 48
Barycenter

 

Earth. the of surface the
inside km 1746 km 4654-6400 is, System Moon Earth the of barycenter The
km 6400 Earth the of radius The
km 4654 km 384000
0.01227MM
0.01227M
r
R and M ,M of values the ngsubstituti equation this In
have, We
R
MM
M
r
M
MM
r
R
1
r
R
M
M
rRMrM
B, barycenter the at balances system the Since
M 0.01227 Moon the of Mass M
Earth the of Mass M
Earth the of center the from barycenter the of Distance r
km 384000 Moon and Earth between distance Average R
System Moon-Earth 1 *
ee
e
me
me
m
m
me
m
e
me
em
e














Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

49. 49
Barycenter
 


km. 48000 km 696000-744000 , distance
a at sun the of surface the outside is System Jupiter-Sun
the of barycenter The km. 696000 is Sun the of radius The
km 744000
km 107.8
0.000955MM
0.000955M
r barycenter Jupiter-Sun
equation, earlier the Using
0.000955MM
1047
1
Jupiter the of Mass M
Sun the of Mass M
km 107.8 R system this For
System Jupiter-Sun 2 *
8
ss
s
ssj
s
8








Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

50. 50
Barycenter
 


surface. sPluto' the outside
km 900 km 1200-2110 , distance a at is System Charon-Pluto
the of barycenter the Hence km. 1200only is Pluto the of radius The
km 2110
km 19600
0.12MM
0.12M
r barycenter Charon-Pluto
equation, earlier the Using
0.12M Charon the of Mass M
Pluto the of Mass M
km 19600 R system this For
System Charon-Pluto 3 *
pp
p
pc
p








Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

51. 51
Barycenter
•Examples
Larger
body
m
1
(m
E
=1)
Smaller
body
m
2
(m
E
=1)
a
(km)
r
1
(km)
R
1
(km)
r
1
/R
1
Remarks
Earth 1 Moon 0.0123 384,000 4,670 6,380 0.732
The Earth has a perceptible "wobble"; seetides.
Pluto 0.0021 Charon
0.000254
(0.121m
Pluto
)
19,600 2,110 1,150 1.83
Both bodies have distinct orbits around the barycenter, and as such Pluto and Charon were considered as adouble planetby many before the redefinition
ofplanetin August 2006.
Sun 333,000 Earth 1
150,000,000
(1AU)
449 696,000 0.000646
The Sun's wobble is barely perceptible.
Sun 333,000 Jupiter
318
(0.000955m
Sun
)
778,000,000
(5.20 AU)
742,000 696,000 1.07
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

52. 52
Barycenter
*
Two bodies similar mass Pluto –Charon System Earth –Moon System
Sun -Earth System Binary Star System
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

53. 53
Conversion of The Coordinates
•InAstronomycoordinatesofstarsorplanets
aregivenwithreferencetoeithereclipticor
celestialequator.LongitudeandLatitudesare
theecliptic,whileRightascensionand
declinationareequatorialcoordinates.Many
timesitbecomesnecessarytoconvertthe
coordinatesfromoneformtoanother.
•1)EquatorialtoEcliptic
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

54. 54
Conversion of the Coordinates
•1) Equatorial to Ecliptic
•2) Ecliptic to Equatorial 
  

     RAsinisinndeclinatiocosicosndeclinatiosinlatitudesin
RAcos
isinndeclinatiotanicosRAsin
longitudetan


 
   
 
      longitudesinisinlatitudecosicoslatitudesinndeclinatiosin
longitudecos
isinlatitudetanicoslongitudesin
RAtan



Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

55. 55
Elementary Astronomical Calculations:
Gravity and motion ( Lecture –1)
Thank You
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

56. 56
Elementary Astronomical Calculations:
Earth and Motion ( Lecture –2)
•By Sukalyan Bachhar
•Senior Curator
•National Museum of Science & Technology
•Ministry of Science & Technology
•Agargaon, Sher-E-Bangla Nagar, Dhaka-1207, Bangladesh
•Tel:[phone](Off), Contact: 01923522660;
•Websitw: www.nmst.gov.bd; Facebook: Buet Tutor
&
•Member of Bangladesh Astronomical Association
•Short Bio-Data:
First Class BUET Graduate In Mechanical Engineering [1993].
Master Of Science In Mechanical Engineering From BUET [1998].
Field Of Specialization Fluid Mechanics.
Field Of Personal Interest Astrophysics.
Field Of Real Life Activity Popularization Of Science & Technology From1995.
An Experienced TeacherOf Mathematics, Physics & Chemistry for O-,A-, IB-&
Undergraduate Level.
Habituated as Science Speaker for Science Popularization.
Experienced In Supervising For Multiple Scientific Or Research Projects.
17 th BCS Qualified In Cooperative Cadre (Stood First On That Cadre).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

57. 57
Energy released by an asteroid on the collision course
with the Earth
*
*Mainasteroidbelt
Mainarticle:Asteroidbelt
ThemajorityofknownasteroidsorbitwithinthemainasteroidbeltbetweentheorbitsofMarsandJupiter,
generallyinrelativelylow-eccentricity(i.e.,notveryelongated)orbits.Thisbeltisnowestimatedtocontain
between1.1and1.9millionasteroidslargerthan1km(0.6mi)indiameter,[29]andmillionsofsmaller
ones.[30]Theseasteroidsmayberemnantsoftheprotoplanetarydisk,andinthisregion
theaccretionofplanetesimalsintoplanetsduringtheformativeperiodofthesolarsystemwaspreventedby
largegravitationalperturbationsbyJupiter.
[edit]Trojans
Mainarticle:Trojanasteroids
Trojanasteroidsareapopulationthatshareanorbitwithalargerplanetormoon,butdonotcollidewithit
becausetheyorbitinoneofthetwoLagrangianpointsofstability,L4andL5,whichlie60°aheadofand
behindthelargerbody.
ThemostsignificantpopulationofTrojanasteroidsaretheJupiterTrojans.AlthoughfewerJupiterTrojans
havebeendiscoveredasof2010,itisthoughtthatthereareasmanyasthereareasteroidsinthemain
belt.
AcoupletrojanshavealsobeenfoundorbitingwithMars.[note2]
[edit]Near-Earthasteroids
Mainarticle:Near-Earthasteroids
Near-Earthasteroids,orNEA's,areasteroidsthathaveorbitsthatpassclosetothatofEarth.Asteroidsthat
actuallycrosstheEarth'sorbitalpathareknownasEarth-crossers.AsofMay2010,7,075near-Earth
asteroidsareknownandthenumberoveronekilometreindiameterisestimatedtobe500-1,000.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

58. 58
Energy released by an asteroid on the collision course
with the Earth
*
  


J 104.8m/s 15000kg 104.2
2
1
mv
2
1
energyKinetic
kg 104.2 kg 300015π
3
4
asteroid of Mass
m/s 15000 km/s 15vVelocity
kg/m 3000 Density
m 15rasteroid of Radius
Densityπr
3
4
Density Volume Mass
have We
m/s in Velocity V
kg in Asteroidof Mass m Where
mv
2
1
object moving ofenergy Kinetic
15272
7
3
3
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

59. 59
Energy released by an asteroid on the collision course
with the Earth
*bomb.atomic Hiroshima 76 about to
equivalentenergy releases km/s 15 spped withtravelling m 30
of diameter of asteroid anby releasedenergy the means Which
76
106.3
104.8
energy bomb Hiroshima
asteroid ofenergy Kinetic
Thus,
J 10 6.3 J 10 4.2 15energy bomb Hiroshima
J 10 4.2 about is kilotone 1
energy TNT of kilotones 15 about had bomb Hiroshima
energy. bomb Hiroshima withot compare us Let
energy. of amount highvery a
is this Still . J 10 2 be asteroid the withleftenergy the of amount Let
explodes. it before ionfragmentat
and slowing noise, in lost isenergy kinetic this of half that assume us Let
13
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60. 60
Linear velocity at any point on the surface of Earth
i.AngularvelocityisaratioofthetotalangularThemeasurement
throughwhichaparticlerotatesinagivenunitoftime.Ifweuseto
standforangularvelocity,wehave .
ReviewingthemotionoftheEarth,recallthatthe
1.Earthhasanangularvelocityof radiansperhour.
2.linearvelocityofapointontheEarth'ssurfacewascalculatedby
multiplyingthisangularvelocitybytheradiusoftheEarth=6400km.
.
ii.Usingthisasaguide,wedefinelinearvelocity,v,to
be whereisangularvelocityinradiansandristheradius.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

61. 61
Linear velocity at any point on the surface of Earth 
 
   
 
   
   km/hr. 0 km/hr 90cos1675 velocity the pole, At
km/s 1536 m/s 23.5cos1675
23.5φ Dhaka, At; km/hr
360024
2π
23.5cos106.4
T
2π
φRcosωφRcos
Earth the of speed AngularlatitudecosEarth the of Radius
pointany at seed Linear
0
0
003
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

62. 62
To Find inclination and eccentricity of Earth’s orbit0.01671
40.0188227270.00004203-70.01670861
7T0.00004203-70.01670861 ty Eccentrici
deg 23.4390453 deg 0.018227240.013-23.43929 nInclinatio
40.01882272T and 2452232.5 JD
35) Chapter (Ref.
2001 November 19 for have wedate that for T and JD the calculatedalready have We
2001 November 19 on orbit sEarth' ofty Eccentrici and nInclinatio find us Let
7T0.0000420370.01670861tyEccentrici
0.013T-23.43929 nInclinatio
by given are orbit sEarth' the ofty eccentrici and ninclinatio the
of values eapproximat the T, of powers higher Neglecting values. these in variation
slow a is There constant. not are orbit sEarth' the ofty eccentrici and ninclinatio The
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

63. 63
Precessional motion of Earth 
  (nearly) 950 years 13.3371.66 in sign Asterism1
through shiftsequinox vernal hence degrees, 13.33 Asterism1
(nearly) 2150 years3071.66 in sign 1
through shiftsequinox vernal hence degrees, 30 sign 1
year1 year
71.66
6060
requiresec arc 50.2
years71.66 years
360
25800
require degrees 1
years25800 require degrees 360

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Vernal Equinox Asterism Years of Degrees
Entry Exit Occupation Covered
3520 BC 2570 BC Rohini 950 13.33
2570 BC 1620 BC Krittika 950 13.33
1620 BC 670 BC Bharni 950 13.33
670 BC 280 AD Ashwan 950 13.33
280 AD 1230 AD Revati 950 13.33
1230 AD 2008 AD Uttara Bhadrapada 768 10.67
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

64. 64
Extent of Earth’s orbit from the angular diameter of Sun’s disk 
km 410000 Aphelionof Distance
km 358000 Perihelion of Distance
384000km a axis major Semi
minarc 29.36 Diameter AngularMinx
minarc 33.65 Diameter AngularMax
orbit sMoon' of distances Aphelionand Perihelion Find :Example

0.0167 0.98331-
a
0.9833a-a

PC
SP-PC
PC
SC
e ty Eccentrici The
km 152096185 km 147099615-1495979002 SP-2aSA Distance Aphelion
km 147099615 km 1495979000.9833 SP Distance Perihelion
km 149597900 a accept shall We
0.9833a
64.13
31.532a
SP Distance Perihelion
64.13
31.53
2a
SP

axis major Semi a where; 2aSPSA But
64.13
31.53
31.5332.60
31.53
SPSA
SP

32.60
31.53
SA
SP
Earth. the and
Sun the between distance the to to alproportion
inversely is disk solar of diameter Angular
distance AphelionSA Aphelion,A
distance Perihelion SP ,Perohelion P
figure the In
July) (4 minarc 32.53 Diameter AngularMinx
January) (4 minarc 32.60 Diameter AngularMax
m
m
m
m
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

65. 65
Angular diameters of the outer planets at opposition
•AtoppositionouterplanetisclosetotheEarthanditsbrightnessismaximumasviewedfromthe
Earth.Theformulafortheangulardiameterwhenplanetatoppositioncanbeeasilyderived.
•Inthefigure
aAngularradiusofaplanet
rActualradiusoftheplanet,
RDistancebetweenthePlanetandtheEarth
•We shall apply this formula to the planet Mars. The distance
of Mars at opposition varies because its orbit is elliptical.
Minimum distance of Mars at opposition = 0.382 AU
Maximum distance of Mars at opposition = 0.666 AU
Radius of Mars = 3400 km
•At opposition when Mars is at Minimum distance from the Earth
its angular.
•At opposition when Mars is at Minimum distance from the Earth its angular.
• 
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R
r
R
r
R
r
1-1-
2sin2aplanet the of Diameter Angular sina asin  
 
sec.arc 24.5 Mars of diameter angular maimum Apparent
secarc 24.48secarc 3600
deg 2sin
opposition at mars of diameter angular Maximum
1-
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3
3
11
104.32
104.32
105.1382.0
3400000
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

66. 66
Angular diameters of the outer planets at opposition
Following table gives,
Maximum apparent angular diameters of outer planets. 
secarc14Apparent
1095.12
105.1666.0
3400000
3
11

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diameter angular minimum
deg 2sin
opposition at Mars the of diameter angular Maximun
1-
Planet Minimum distance Radius Apparent Angular
From the Earth (AU) (km) Diameter (arc sec)
Mars 0.382 3400 24.50
Jupiter 3.950 71000 49.40
Saturn 8.040 60000 20.50
Urenus 17.000 25500 4.05
Neptune 28.709 24850 2.40
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

67. 67
Angular diameter of Mercury and Venus at their maximum
elongation
•ElongationisananglebetweentheSun,theearthand
thePlanet.WhenelongationismaximumMercuryand
Venuscanbeseenintheskybeforesunriseorafter
sunset.TofindtheangulardiametersofMercuryand
Venusattheirmaximumelongationitisnecessaryto
findtheirdistanceatthatevent.
•Tofindthemaximumelongationsofthesetwoplanets
wehaveconstructedtherightangletriangles.The
adjacentsideoftheserightangletriangleswillgiveus
thedistancesoftheseplanetsfromtheEarthwhenthey
areatmaximumelongations.   
 
sec.arc 24.4
101.50.682
6050000
2sin elongation maximum at Venusof Diameter Angular
0.682AU 47cos AU1 47EScos EV Venusof Distance
AU1 ES
deg 47elongation maximun isSEV angle EVS triangle angle right In
sec.arc 7.6
101.50.8829
2450000
2sin elongation maximum atMercury of Diameter Angular
0.8829AU 28cos AU1 28EScoselongation maxEScos EM Mercury of Distance
AU1 ES
deg 28elongation maximun is SEM angle EMS triangle angle right In

11
1-
00
11
1-
00m
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Venus2)
Mercury 1)
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

68. 68
The Earth viewed from the outer planets
•It is very interesting to find how the Earth will be seen from the outer planets. The maximum
elongation of the Earth from the outer planets will give us an idea of how it will be seen in the sky
of these planets and the angular diameters will indicate the size of its disk. For the sake of
simplicity we shall consider the average distances of the outer planets from the Sun, we have
following formulae. 
 
 
     
 
 
 
  
c15.30arcse
101.51.15
6400000
2sin Earth of diameter Angular
AU1.15 41cos 1.524 elongatonMScosEM
elongaton maximum at distance Planet - Earth EM
41
1.524
1
sin Mars from Elongation Max
AU1.524 distance Sun - Mars
AU1 distance Sun - Earth
Mars. consider shall weeaxmple, an As
elongationcos distance SunPlanetelongation maximum at Distance Planet - Earth
elongation maximum at Distance Planet - Earth
Earth of Radius
2sin Earth of Diameter Angular2)
Distance Sun - Planet
Distance Sun - Earth
sinElongation Maximun 1)
11
1-
0
01-m
1-
1
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

69. 69
The Earth viewed from the outer planets
•IntheskyofMars,theEarthwillbeseenbeforesunriseoraftersunsetupyo41degreesinthe
Sky.Atmaximumelongationitsangulardiameterwillbeabout15.3arcseconds.Similar
calculationcanbedoneforotherouterplanets.Theresultsarecompliedinthefollowingtable.
•.
Planet Dist. From (Earth-Planet) dist. Max Elongation Angular
Sun (AU) at max elongation (AU) (Degree) Diameter (arc sec)
Mars 1.524 1.15 41 15.30
Jupiter 5.204 5.12 11 3.44
Saturn 9.582 9.53 6 1.85
Uranus 19.201 19.17 3 0.92
Neptune 30.047 30.03 2 0.58
Pluto 39.236 39.22 1.5 0.45
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

70. 70
View of the Earth and the Moon from Mercury and Venus
•The Earth is an outer planet for the Mercury and the Venus. The Earth will be seen from night side
of these planets at oppositions.
•At oppositions we shall assume following distances
•Mercury (1 –0.4) AU =0.6 AU
•Venus (1 –0.7) AU =0.3 AU
•Diameter of the Earth = 12800 km
•Distance between the Earth and the Moon –38400 km only. degree a half about is it Venusthe from and minutesarc 15 aboutonly is
Moon the and Earth the of separation angular maximumMercury the from viewed As*
min.arc 29.3
101.50.3
384000
sinMoon and Earth between separation Angular
sec.arc 56.66
101.50.3
12800
sin Earth the of diameter Angular
Venusthe From Earth the of Size*
min.arc 14.66
101.50.6
384000
sinMoon and Earth between separation Angular
sec.arc 29.3
101.50.6
12800
sin Earth the of diameter Angular
Mercury the From Earth the of Size*
8
1-
8
1-
8
1-
8
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

71. 71
Tidal forces of the planets on the Earth
•Tidal Force:Tidal force is defined as differential gravitation force (across the position) on a body by
any another body.  
earth. the from Planet the of Distance R & planet the of Mass M Where;
R
M
Force Tidal
:r'' of insted R'' Using
dr
r
2GMm
Force Tidal
dr
r
2GMm
dr(-2)r GMm- dr
r
GMm
-
r
dr
r
F
dF
:Force Tidal of Definition From
r
GMm
- F
:nGravitatio ofLaw sNewton' From
:Traetment alMathematic
3
3
3
1-2-
2
2
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

72. 72
Tidal forces of the planets on the Earth 
 
0.00011
0.28
102.445
Sun the of Force Tidal
Venusthe of Force Tidal
4.2r 0.28AUAU0.721- Earth the from Venusthe of distance Minimum
102.445m M102.445 Venusof Mass
Earth? onThe Venusthe of force tidal maximum the is What:2 Example*
0.0000131012.87
4.2
10953.688
Sun the of Force Tidal
Jupiter the of Force Tidal
4.2r 4.2AUAU1-5.2 Earth the from Jupiter the of distance Minimum
10953.688m M10953.688 Jupiter of Mass
Earth? onThe Jupiter the of effect tidal maximum the is What1: Example*
r
m
Sun the of Force Tidal
Planet the of Force Tidal
have, wegSimplifyin
AUin distance of multiple r )/MMm ( ratio mass of multiple m
Where,
AU1r
AU1
M
Mm
Sun the of Force Tidal
Planet the of Force Tidal
have, We AU,in distance the and Sun the of mass of terms in expressed is planet of mass If
R
R
M
M
Earth the from Sun the of Distance
Earth the from Sun the of Distance
Sun the of Mass
Planet the of Mass
Sun the of Force Tidal
Planet the of Force Tidal
3
6-
6-
S
6-
6
3
6-
6-
S
6-
3
SP
3
S
S
3
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

73. 73
Tidal forces of the planets on the Earth
•Following table gives maximum tidal forces of the Sun, The Moon and the Planet on the Earth. It is
clear that tidal force of the Moon is about 2 times the tidal force of the Sun.
Planet Actual
Distance (AU)
Minimum
Distance (AU)
Mass 10
-6
(Sun = 1)
Tidal Force
(Sun = 1)
Mercury 0.387 0.613 0.1659 0.0000007
Venus 0.72 0.28 2.445 0.00011
Mars 1.52 0.52 2.445 0.000002
Jupiter 5.2 4.2 953.688 0.000013
Saturn 9.5 8.5 285.55 0.0000005
Uranus 19.2 18.2 143.6 0.000000001
Neptune 30 29 51.45 0.000000002
Pluto 39.5 38.5 0.0000000000001
Moon 356000 km 2.1
Sun 1 1 1 1
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

74. 74
Fall of the Moon towards the Earth
•TheMoonrevolvesaroundtheEarth.Thenecessarycentripetalforceisprovidedbythe
Earth’sgravitationalattractionontheMoon.Ifcentripetalforcesuddenlyvanishes,the
Moonwouldmpvealongatangenttoitsorbit.BecauseofcentripetalforceMoon
continuouslyfallstowardstheEarth.FromtheNewton’slawofGravitationfalloftheMoon
towardstheEarthcanbeestimated.
•Inthefigure
•E=CenteroftheEarth
•EA=r=radiusoftheEarth=6400km
•EB=R=RadiusoftheMoon’sOrbit=384400km
•BCindicatesdirectionoftheMoonwouldmove,
•Intheabsenceofcentripetalforce.
•CD=falloftheMoontowardstheEarthin1sec.
•BEC=spaceanglethroughwhichtheMoonmovesin1sec.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

75. 75
Fall of the Moon towards the Earth


 


m. 4.9 about through Earth the towars falls Moon the hour 1 in Thus
mm. 1.3616 through Earth the towards falls Moon the sec. 1 in means Which
m 101.3616 CD
m
36002427.332
2π
2
384400000
CD
36002427.332
2π
a
be willsec. 1 in covered a'' angle The
days. 27.332 is Moon the of period orbital the Since
2
a
R1
2
a
1R CD
2
a
1
2
a
1
2
a
1
1

acos
1
and
2
a
1acos small;very is a'' angle Since
1
acos
1
R R -
acos
R
CD
acos
R
CDR
CDR
R
acos
R ED EB But
DCED
EB
EC
EB
acos
3-
2
22
2
1
2
2
2




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


















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






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
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




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


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







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

76. 76
Acceleration of Moon towards he Earth
•ThecentripetalForceprovidedbytheEarth’sgravitykeepstheMooninitsorbit.
WhichmeansateveryinstantMoonisacceleratedtowardstheEarth.Fromthe
knoledgeoftheorbitalvelocityandperiodictimeofthewecanfindtheacceleration
ofMoontowardsthecenteroftheEarth.
•OrbitalvelocityoftheMooncanbefoundfromthefollowingformula 
22
8
2
32
3
8
m/s 0.00272m/s
103.844
101.024
r
v
a
Earth the of center the towards Moon the of onaccelerati
distance
velocity of sq.mass
onaccelerati massF
Moon. the on acting force lCentripeta The
m/s. 101.024 m/s
36002427.3
103.8442π
v
days 27.3 timePeriodic T
km 384400 Earth the of center the from Moon the of distance Average r ere, Wh
T
2π
v

timeperiodic
orbit the of ncecircumfere
velocity Orbital






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







r
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

77. 77
Acceleration of Moon towards he Earth
* 

Earth. the towards mm. 1.36 through
s fall it time same the at 1sec. in km 1 about through moves It
km/s 1.024 Moon the ofvelocity Since
mm 1.36 m 0.00136 m 10.00270
2
1
h
h Earth the towards falls Moon the whichthrough distance s
sec. 1 time t
m/s 0.00272 Moon the of onaccelerati a
0 velocity initial u
at
2
1
uts
formula. sNewton' the Use
Earth. the towards falls Moon the rate what At:2 Example

m/s 0.006 m/s
101.5
1030
r
v
a
Sun the towards onaccelerati sEarth'
m 101.5 Sun the from Earth the of distance average
m/s 1030 km/s 30 Earth the ofvelocity Orbital
Sun. the towards Earth the of onaccelerati the is What1: Example
2
2
2
22
8
2
32
8
3













Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

78. 78
Curious Coincidence between the Sun, Earth and Moon
•Thenumber108
FollowingcoincidencesareobservedwithreferencetoNumber
•DiameteroftheEarth108=DiameteroftheSun(nearly)
12800km108=1382000km(ActualDiameteroftheSun=1392000km.
•DiameteroftheSun108=DistancebetweenTheEarthandtheSun(nearly)
1392000km108=150336000km(Actualaveragedistancebetweenthem=150000000km.)
•DiameteroftheMoon108=DistancebetweenTheEarthandtheMoon(nearly)
3500km108=378000km(Actualaveragedistancebetweenthem=384000km.)
•Thenumber400
FollowingcoincidencesareobservedwithreferencetoNumber
•DistanceoftheSunandtheMoonfromtheEartharenotconstant,duetoellipticalmotionthese
distancescontinuouslychangeandtheratiobecomesnearly400.Duetothesetwocoincidences
thetotalSolareclipsehasbecomepossible.400)(nearly 390
km 384000
km 150000000
Moon the of distance Average
Sun the of distance Average
400)(nearly 398
km 3500
km 1392000
Moon the of Diameter
Sun the of Diameter


Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

79. 79
Curious Coincidence between the Sun, Earth and Moon
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

80. 80
Zero gravity point between the Earth and the Moon 
 

 
km 345946 km
0.111
384000
/MM1
R
h
M
M
1
h
R

M
M
h
h - R

h - R
M
h
M

h - R
GM
h
GM

h-R
GM
h
GM


h
GM
h
GM
have, wepointgravity zero At
2
1
EarthMoon
2
1
Earth
Moon
Earth
Moon
2
2
2
Moon
2
Earth
2
Moon
2
Earth
2
point zero to center EarthMoon & Earth between Distance
Moon
2
Earth
2
Moon
2
Earth
point zero to c enter Earth
point zero to c enter Earthpoint zero to c enter Earth




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

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


Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

81. 81
Synodic time
•Synodictimeisanintervalbetweentwosuccessiveheliocentricconjunctionsofthe
planetsinlongitudes.
•Heliocentricconjunctionindicatestheclosestapproachoftheplanets.Wehaveto
determinethesyndonictimesoftheplanetswithreferencetotheEarth.
•ForInnerplanetstheyareclosetotheEarthattheirInferiorconjunctionthatiswhen
theyarebetweentheEarthandtheSun.
•ForouterplanetstheyareclosetotheEarthattheiroppositionthatiswhentheyare
beyondtheEarth.
•Ingeneralthesyndonictimecanbecalculatedfromtheangularvelocitiesofthe
planets.Relativeangularvelocity=Differencebetweentheangularvelocitiesofthe
planets.
•Angularvelocityisgivenbythefollowingformula
=2/T
Let,Trelativeperiodictime;Tperiodictimeoffirstplanet;Tperiodic
ofsecondplanet.
Thenwehave,
•2/T=2/T-2/T 1/T=1/T-1/T
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

82. 82
Synodic time
•Synodic Time of Inner Planets
•In this case Tperiodic time of inner planets which is smaller than the Earth
• T periodic time of the Earth which is one year.
•Then we have
•1/T =1/T-1/1
•For example
•For Mercury: T= 0.241 years.
•1/T =1/0.241-1/1 = 4.1493 –1 = 3.1413
•Synodic time of Mercury = T =1/3.1413 yr. = 0.3175 yr. = 115.89 days =116 days
(nearly)
•Synodic Time of Outer Planets
•In this case Tperiodic time of the Earth which is one year
• T periodic time of the planets which is outer than the Earth
•Thus we have
•1/T =1/1 -1/T 
•For example
•For Mars: T= 1.88 years.
•1/T =1/1-1/1.88 = 1 –0.5319=0.468
•Synodic time of Mars = T =1/0.468 yr. = 2.1367 yr. = 779.9 days =780 days (nearly).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

83. 83
Synodic time
•Following table gives the orbital period and synodic time of all the planets:
Planets Orbital
Period
(Years)
Synodic
Years
Synodic
days
Mercury 0.241 0.317 116
Venus 0.616 1.641 585
Mars 1.880 2.136 780
Jupiter 11.900 1.090 399
Saturn 29.500 1.035 378
Uranus 84.000 1.013 369.65
Neptune 165.000 1.006 367.477
Pluto 248.000 1.004 366.72
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

84. 84
Maximum elongation of Mercury and Venus
•MercuryandVenusaretheinnerplanets.Theycanonlybeobserved,
fewhoursbeforesunriseorfewhoursaftersunset.IngeneralElongation
isdefinedastheanglebetweenisdefinedastheanglebetweentheSun,
theEarthandthePlanet.
•Ifanelongationofaplanetislessthan10degrees,itcannotbeobserved
intheglareoftheSun.AsviewedfromtheEarththemaximumelongation
oftheMercurydoesnotdoesnotexceedmorethan28degreesandthat
theVenusmorethan47degrees.SincetheEarthrotatesthrough15
degreesinonehour,theMercurycanbeobservedabout2hoursandthe
Venusaboutthreehoursbeforesunriseoraftersunset;iftheirelongations
aremaximum.AtaphelionthedistanceoftheplanetfromtheSunis
maximum.ThuswheneithertheMercuryortheVenusisattheAphelionits
elongationismaximum.IfweknowtheApheliondistanceoftheplanets
theirmaximumelongationcaneasilybefound.SSun,EEarth,M
 Mercury & V  Venus
LettheMercuryandtheVenusbeataphelion,thenMaximumElongationof
theMercuryistheangleSEM,andMaximumelongationoftheVenusis
theangleSEV,SMEandSVEaretherightangletriangles.  
 
(nearly) deg 28 deg. 27.8 Mercury the of Elongation Maximum
0.4666989sin Elongation
0.4666989
AU1
AU0.4666989
SE
SM
Elongation of Sine
AU0.4666989 AU0.20562910.3871 e1a Mercury the of distance Aphelion
0.205629 e Mercury the of orbit the ofty Eccentrici
AU0.3871 a Mercury the of axis major Semi
Mercury the of Elongation Maximum*
1-






Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

85. 85
Maximum elongation of Mercury and Venus
•Elongation of a planet can be East or West.
•Ifthelongitudeoftheplanetisgreaterthanlongitudeofthe
Sun.Elongationis‘East’.Inthiscaseplanetcanbeseenafter
sunset.
•Ifthelongitudeoftheplanetissmallerthanlongitudeofthe
Sun.Elongationis‘West’.Inthiscaseplanetcanbeseen
beforesunrise.
•Iftheelongationoftheplanetislessthan10degreestheplanet
cannotbeseen.  
 
(nearly) deg 48 deg. 46.7 Venusthe of Elongation Maximum
0.7282243sin Elongation
0.7282243
AU1
AU0.7282243
SE
SV
Elongation of Sine
AU0.7282243 AU0.00677210.723326 e1a Venusthe of distance Aphelion
0.006772 e Venusthe of orbit the ofty Eccentrici
AU0.723326 a Venusthe of axis major Semi
Venusthe of Elongation Maximum*
1-






Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

86. 86
Maximum elongation of Mercury and Venus
SIGNS OF ZODIAC
Approximate date of entry of the Sun
Sign (বাাংলা) Western Sign Indian Sign
Aries (মেষ) 21 March 14 April
Taurus(বৃষ) 21 April 14 May
Gemini(মেথুন) 21 May 14 June
Cancer(কককট) 21 June 14 July
Leo(ম িংহ) 21 July 14 August
Virgo(কনযা) 21 August 14 September
Libra(তু লা) 21 September 14 October
Scorpios(বৃশ্চিক) 21 October 14 November
Sagittarius(ধনু) 21 November 14 December
Capricornus(েকর) 21 December 14 January
Aquarius(কু ম্ভ) 21 January 14 February
Pisces(েীন) 21 February 14 March
INDIAN ASTERISMS
1. Ashwani 10. Megha 19. Mula
2. Bharani 11. Purva Phalguni 20. Purv-ashadha
3. Krittika 12. Uttar Phaguni 21. Uttara-ashadha
4. Rohani 13. Hasta 22. Shravan
5. Mriga 14. Chitra 23. Dhanishta
6. Ardra 15. Swati 24. Shatataraka
7. Punarvasu 16. Vishakha 25. Purva-bhadrapada
8. Pushya 17. Anuradha 26. Uttara-bhadrapada
9. Ashlesha 18. jyeshtha 27. Revati
* 1 Sign = 30 degrees 1 Asterism = 13.33 degrees
* Ayanansha= Vernal Equinox –Indian First Point of Aries = 24 degrees (nearly)
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

87. 87
Elementary Astronomical Calculations:
Earth and Motion ( Lecture –2)
Thank You
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

88. 88
Elementary Astronomical Calculations:
Telescope and Calendar ( Lecture –3)
•By Sukalyan Bachhar
•Senior Curator
•National Museum of Science & Technology
•Ministry of Science & Technology
•Agargaon, Sher-E-Bangla Nagar, Dhaka-1207, Bangladesh
•Tel:[phone](Off), Contact: 01923522660;
•Websitw: www.nmst.gov.bd; Facebook: Buet Tutor
&
•Member of Bangladesh Astronomical Association
•Short Bio-Data:
First Class BUET Graduate In Mechanical Engineering [1993].
Master Of Science In Mechanical Engineering From BUET [1998].
Field Of Specialization Fluid Mechanics.
Field Of Personal Interest Astrophysics.
Field Of Real Life Activity Popularization Of Science & Technology From1995.
An Experienced TeacherOf Mathematics, Physics & Chemistry for O-,A-, IB-&
Undergraduate Level.
Habituated as Science Speaker for Science Popularization.
Experienced In Supervising For Multiple Scientific Or Research Projects.
17 th BCS Qualified In Cooperative Cadre (Stood First On That Cadre).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

89. 89
Magnification power of a telescope
•Letusconsidersimplerefractingtelescope.Theobjectiveformsanimageofadistantobjectonits
focalplane.Ifthefocalplaneofeyepiecejustcoincideswiththefocalplaneoftheobjective,the
finalimagewillbeformedatinfinity.Buteyepieceissoarrangedthattheimageisobserved.Let
usconsideridealsituation. 
 
140
5
700
power magnifying cm 5 f & cm 700 f If :Example
changed. be can power magnifying eyepiece the of length focal changingby that clear is It
eyepiece of length focal
objective of length focal
power magnifying by, given is power magnifying its or teesope a of ionmagnificat hence

f
f
x/f
x/f
a
b
magnifying Angular a'' than greater is b''generally
f
x
b b btan smallvery is b'' if
f
x
btan
eyepiece at x object theby made angle b eyepiece; of length focal f eyepiece; for objective of length x
infinity. at formed is image
situation ideal in and eyepiece for object an as behaves objectiveby formed image The
f
x
a a atan smallvery is a'' if
f
x
atan
objective the at image theby made angle a objective; of length focal f ; objectiveby formed image of length x Let,
e0
e
o
0
e
ee
e
00
0







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

90. 90
Resolving power of a telescope
•Resolvingpowerisabilityofatelescopetoseparatetheimagesofstarswhichare
veryclosetoeachother.Italsoallowstodiscerndetailsinanextendedobject.
•Theminimumresolvableangledependsonthediameterofthetelescopeobjective
andwavelengthofthelightbeingobserved.Ingeneral, 
sec.arc 101.2258 sec.arc
10
10500
2062651.221.22θ
nm. 500 of h wavelengtat resolved be can that angle smallest whatm. 10 diameter has telescope Keck :Example
seen. be can stars the of images separate
two i.e. resolved be can yhey apart, sec.arc 1.5 than more are stars if means This
sec.arc 1.5
0.1
106
2062651.221.22θ Then,
m. 106 observed h Wavelengtm. 0.1 objective telescope of Diameter
1.22.by multiplied is angle resolvable minimum account into ndiffractio take To
image. the spreds that pattern ndiffractio produces telescope a of aperture circular The
units. same in measured are objective of diameter and length waveThe
secarc 206265 sec arc 360057.29577 deg. 57.29577 radian 1
radian 1 in secondsarc of number indicates 206265 Where
objective of diameter
length wave
206265θ angle resolvable minimum
2-
9
min
7
min
7-
min








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
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 






:Example
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

91. 91
Resolving power of a radio telescope
•Sun,Stars,Nebulae,Galaxiesandmanyotherobjectsemitradiowaves.Hence
RadioAstronomyhasbecomeanessentialtoolofmodernastronomy.
•ResolvingpowerformulaofopticaltelescopecanbeappliedtoRadioTelescopealso.
•ResolvingPowerofRadioTelescope=206265sec.arc 412.53
1000
0.2206265
Power Resolving
by, given power resolving have will WaveRadio cm 20 with
workingm 1000 aperture an withTelescope Radio A possible. not is thisy Technicall
across km. 41.253 m
1
0.2206256
D
D
0.2206256
sec.arc 1
Telescope. Radio a requires h wavelengtcm 20 at sec.arc 1 resolve to example, For telescope.
optical than larger times 400000 be to has Telescope Radio the angle same resolve to want we
If wave.vsible nm 500 than longer times 400000 is WaveRadio cm 20 means Which
400000
m 10 500
m 0.2

be, willh wavelengttwo these of ratio the then
m), (0.2 cm 20 is
waveRadio of that and nm 500 is
light visible of h wavelengtSuppose
example, For less. quite is diameter
given a of Telescope Radio a of Power
Resolving Hence light. visible of that than
greater far are wavesRadio of ngth Wavele
Aperture D h wavelengtRadio λ Where,
D
λ
206262Telescope Radio of Power Resolving
9-






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
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

92. 92
Scale of an image
•Apparent sizes of the objects like Stars, Sun, Moon, Planets are generally expressed in
angular units. But it is convenient to express the scale of an image in linear units. For
example, apparent size of moon is 0.5 degrees. Suppose a telescope produces the
image of Moon 1 cm across. The scale image is 0.5 deg/cm or 2 cm/deg.
•The scale image only depends on focal length of mirror or lens. Numerically linear length
‘S’ of an image corresponding to 1 degree in the Sky is given by
across cm. 14.5nearly is telescope Haleby produced Moon the of Image
cm 14.5mm 1.45 mm 1
12.4
1800
to scorrespond sec.arc 1800
milimeter 1 to scorrespond sec.arc 12.4 since,
sec.arc 1800 deg. 0.5 Moon the of size Apparent
telescope? thisby produced Moon the of image scale the is What
sec.arc 12.4 sec.arc
10000.29
60 60
scorrespond milimeter 1 Hence
sec.arc 60 60 to scorrespond milimeters 1000 0.29 Thus,
(nearly) m/deg. 0.29S
m/deg. 0.29316 m/deg. 16.80.01745S
telecope? theby produce image scale the
is Whatmeters. 5 is diameter its and meters 16.8 is Telescope Hale of length Focal :Example
0.01745f S
degrees 57.3 radian 1 since
deg. 57.3
f
S










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

93. 93
Brightness of an image
•Anopticaltelescopeisanextensionofoureye.Itcangathermorelightsothatfaintobjects
canbeseen.Lightconsistsofphotons.Inordertomakeafaintobjectbrighter,more
numberofphotonsshouldenteroureye.Hencefarawayfaintobjectscanbeonly
observedwithbigtelescopes.
•Thebrightnessofanimageisameasureoftheamountoflightenergythatisconcentrated
intoaunitareaoftheimage,suchasasquaremillimeter.Thebrightnessofanimage
determineshowlong,timewouldberequiredtorecordtheimagephotographically.
Brightnesscanbedeterminedfromlightgatheringpowerofatelescopeanditsfocal
length. 
 
 
eye. an
of pupil mm 5 than light more times 40000 collect can objective meter 1 means Which
m

d1
d2

have, We
telescope of diameterd2 and pupil of diameterd1 If
meter. 1 diameter of telescope
a and mm 5 diameter pupil a witheye. naked at power gathering Light Compare 1: Example
diameter power gathering Light or
diameter
4
π
power gathering Light
diameter
4
π
circle a of Area
Hence, circular. are objectives the of Most collect. can it light muchhow determines telescope a of objective an
of area The power. gathering its called is light more collect to telescope a of Ability
:Power gathering Light The*
2
2
2
4
2
3
2
104
105
m1


















Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

94. 94
Brightness of an image
•Anopticaltelescopeisanextensionofoureye.Itcangathermorelightsothatfaintobjectscan
beseen.Lightconsistsofphotons.Inordertomakeafaintobjectbrighter,morenumberof
photonsshouldenteroureye.Hencefarawayfaintobjectscanbeonlyobservedwithbig
telescopes.
•Thebrightnessofanimageisameasureoftheamountoflightenergythatisconcentratedinto
aunitareaoftheimage,suchasasquaremillimeter.Thebrightnessofanimagedetermines
howlong,timewouldberequiredtorecordtheimagephotographically.Brightnesscanbe
determinedfromlightgatheringpowerofatelescopeanditsfocallength. 
 
 
eye. an
of pupil mm 5 than light more times 40000 collect can objective meter 1 means Which
104
m 105
m 1
d1
d2

have, We
telescope of diameterd2 and pupil of diameterd1 If
meter. 1 diameter of telescope
a and mm 5 diameter pupil a witheye. naked at power gathering Light Compare
diameter power gathering Light or
diameter
4
π
power gathering Light
diameter
4
π
circle a of Area
Hence, circular. are objectives the of Most collect. can it light muchhow determines telescope a of objective an
of area The power. gathering its called is light more collect to telescope a of Ability
4
2
3
2
2
2
2




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
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



:1 Example
:Power gatheringLight The*
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

95. 95
Brightness of an image
*
f/15. ratio focal than image brighter give
willf/3 ratio Focal example For
ratio. focal large than image brighter
a in results ratio focal small small Thus
ratio. focal the to alproportioninversely
is image an of brightness The
ratio. focal called is f/d

1

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
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:Telescope Of Length Focal The*
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

96. 96
Julian day number
•ForallmodernAstronomicalcalculations,12noonof1
st
January4713BCisselectedasanepoch.The
numberofdayselapsedfromtheepochuptothedesiredGregoriandatearecalculated,resultingiscalled
JulianDayNumberorJDinshort.
•TheJDcorrespondingto1
st
January2000isthensubtractedfromtheJDofthedesireddateandresultis
dividedbythenumberofthedaysinJuliancenturyof36525days.ThisratioisthenrepresentedasTand
usedforallAstronomicalcalculations.JeanMeeushasgivenasimplemethodtofindJDofarequired
GregorianDate.Thestepsofthemethodareasfollows.
•1.Y=Year,M=Month,D=Day
•2.IfM>2,YandMremainunchanged
•3.IfM=1or2(JanuaryorFebruary)replaceYY–1andMM+12
•4.Calculate
• A=INT(Y/100)andB=2–A+INT(A/4)
•5.JulianDayNumberisgivenby
• JD=INT{365.25(Y+4716)}+INT{30.6001(M+1)}+D+B–1524.5
•Example:FindJDfor19
th
Nov.2001
•1.Y=2001,M=11,D=19
•2.SinceM>2YandMremainunchanged
•3.Skip
•4.A=INT(2001/100)=20,B=2–20+INT(20/4)=-13
•5.JD=INT{365.25(2001+4713)}+INT{30.6001(12)}+19–13–1524.5
• =2453384.0+367+19–1537.5=2452232.5
• T=(JD–2451545.0)/36525=(2452232.5–2451545.0)/3625=0.018822724
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

97. 97
Epact: Phase of Moon on 1
st
January
•EpactmeansthephaseoftheMoonatzerohourson1
st
Januaryofanyyear.Knowing
EpactprobabledatesoffullandnewMoonfortheentireyearcanbeestimated.Every19
yearsphasesoftheMoonrepeat.TheMetoniccycleprovesthisfact,fromwhichthe
phaseoftheMoonon1
st
Januarycanbejudged.Thecalculationisquiteapproximate;
hencetherecanbeadifferenceofonedayinestimatedphaseoftheMoon.Stepwise
methodoffindingepactisasfollows.
•Step1:ThephasetheMoonon1
st
JanuaryAD.Willgetrepeatedintheyears20,39,58,
etc.Hencedividetherequiredyearby19andnotedownonlytheremainderofthe
division.
•Step2:SinceSolaryear(365days)andLunaryear(354days)differby11days.Every
yearthes11dayswillgetaccumulated.Hencemultiplytheremainderofsteponeby11.
•Step3:Lunarphasecycleisof30days,hencedividedthemultiplicationofstep2by30
andagainfindtheremainderonly.Theresultantremaindergivesroughlythephaseofthe
Moonon1
st
January.
•Step4:TheformulatodecidetheEpactisasfollows

 
81523
13
stJanuary1MoononPhaseofthe








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

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
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MOD
30
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MOD
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MOD Epact
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2008 Jan 1st on Moon the of phase the find us let example an As
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MODJanuary 1st on Moon the of PhaseEpact
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

98. 98
Epact: Phase of Moon on 1
st
January
•Thuson1
st
January2008therewas8
th
tithiofdarkfortnight.InIndianAlmanacs9
th
tithi
isgiven,because8
th
endedat4:52morningand9
th
started.Wecanexpectadifference
ofonetithibecauseofextensionandreductionofthetithes.Alsotithiatsunriseis
assumedtothetithioftheentireday.
•KnowingthephasesoftheMoonon1
st
January,severaleventscanbeestimated.
1.If1
st
January2008is8
th
tithiofdarkfortnightthentherewasfullmoon8daysbefore.
Thus24
th
December2007wasfullmoonday.
2.On15
th
DecembertheSuncompletes8signsandentersinNiryanSagirrariushenceon
1
st
January2008Sun’sIndianlongitudewas8030+15=255degrees.
3.InthedarkfortnightangulardifferenceintheSunandtheMoon=180-tithi12=180–96
=84deg.(nearly)
4.LongitudeoftheMoon=LongitudeoftheSun–Angulardifference=255–84=171
deg.(nearly).AAsterism=13.3deg.171/13.3=12.8ThustheMoonhascrossed
UttaraFalguniandenteredinHastaNakshtra.AccordingtoAlmanactheMoonwas
Spica.
5.Ifon1
st
January2008Sunrisewasassumedtobeat6AM.TheMoonrisemustbe
84/15=5.6hoursbeforesunrise.WhichmenstheMoonrisewasafter12.30inthenight.
Onthatdaymoonrisewasabout1AM.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

99. 99
Phase of the Moon
•TheoftheMoon’ssurfacethatisilluminatedbythesunlightiscalledphaseoftheMoon.
FromnewtofullMoonthePhaseoftheMoonincreaseswhileitdecreasesfromtonew
Moon.ThephaseoftheMoondependsonitselongation.Theelongationisanangle
betweentheSun,theEarthandtheMoon.TithiofanIndiancalendarisindirectly
elongationoftheMoon.ThusknowingthetithiofthedayphaseoftheMooncanbe
determined.


25.0



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2
120cos1
fortnight bright of tithi 5th Moon the of Phase
12060-180d 60deg.512 Tithi 5th deg. 12 Tithi One
fortnight. bright of tithi 5th on Moon the of phase the find s let example an As
e-180d
eelongation-180 d e Wher
2
dcos1
Moon the of Phase
follows, as is Moon the of phase the for formula The
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

100. 100
Phase of the Moon
*
lighted. be willportion easter
ghtdarkfortni the in whilelighted be willdisk the of portion westernfortnight bright the
In reversed. be willdisk sMoon' the of part lighted theonly same be willfortnight dark and
bright of tithi 5th on phase the Thus fortnight. dark the in reapeated be willphase The
2
36cos1
fortnight bright of tithi 12th Moon the of Phase
deg. 36144-180 d and
deg. 144 1212 tithi 12th have wefortnight bright of tithi 12th on gain A
east'.' elongation the case this In .elongation represents represntsdirectly tithi fortnight bright For
9.0
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

101. 101
To find the phase of the Moon Conway’s method
•JohnHartenConwayiswell-knownPrincetonUniversitymathematician.Heisfamousforhis
mathematicalprocedurecalled‘GameofLife’.Hismospopularworkis‘WiningWays’.Abookon
mathematicalgamesintwovolumes.Inthesecondvolumeofthatbook,Conwayhasgiven
somemathematicaltricksforcalendarcalculations.“TofindphaseoftheMoon”isoneofthem.
PhaseoftheMoonisanothernameof“Tithi”ofHinduCalendar.Usingthismethodphaseofthe
Moononanydayoftheyearcanbefound.Theanswermaydifferbyonedaybecauseofsome
assumption’s.
•Conway’smethodtofindthephaseoftheMoonisasfollows.
•Step1:Dividedthelasttwodigitsoftheyearby19andfindtheremainder.Iftheisgreaterthan
9subtract19fromtheresultsothatremainderisbetween-9to+9
•Step2:Multiplytheresultobtainedinstep1by11anddividethemultiplicationby30.Ifthe
remainderislessthan30subtractitfrom30andassumetheresulttobe–ve.Thustheanswer
obtainedwillbebetween-29to+29.
•Step3:Adddaynumberandmonthnumberintheanswerobtainedinstep2.butforJanuaryand
Februaryselectthenumbers3and4respectively.Forothermonthsusualnumbersaretogiven,
forexampleMarch=3,April=4……..Etc.
•Step4:Subtractthenumbergiveninfollowingtablefromtheanswerobtainedinstep3.
Century Numbertobesubtracted
1700to1799 6+2/3=6.66
1800to1899 1+1/3=1.33
1900to1999 4
2000to2099 8+1/3=8.33
•Step5:Dividedtheanswerobtainedinstep4by30andgetaremainderbetween0to29.the
finalanswerindicatesthephaseoftheMoonorthetithiofHinduCalendar.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

102. 102
To find the phase of the Moon Conway’s method
• Example 1: 12
th
December 2002
1. 02/19 R = +2
2. +2 11 22, 22/30 R = -8
3. -8 + 12 +12 = 16 Month No. = 12, Day No. = 12
4. 16 –8.3 = 7.8 8 21
st
Century -8.3
5. 8 8
th
tithi of white fortnight –( Shudha Astami) Phase of the Moon about ½ .
• Example 2: 29
th
January 2006
1. 06/19 R = +6
2. +6 11 66, 66/30 R = +6
3. 6 + 3 +29 = 38 January = 3, Day No. = 29
4. 38 –8.3 = 29.7 30 (21
st
Century -8.3)
5. 30 New Moon Amavasya
• Example 3: 4
th
July 2008
1. 08/19 R = 8
2. 8 11 88, 88/30 R = 28
3. 28 + 4 + 7 = 39 (July = 7, Day No = 4)
4. 39 –8.3 = 30.67 31 21
st
Century -8.3
5. 31/30 R =1
Pratipada (Almanac shows Dvitya because Partipada was omitted)
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

103. 103
Useful information to estimate Moon rise
•WiththehelpofTithiconceptofIndianAlmanacandeverydayMoonrisesabout50minutes
late,wecanveryroughlyestimatetimeoftheMoonrise.InonedaytheMoonmovesthrough
360/27.3=13.2degreestotheeast.Butduetotheellipticalorbit,thisangulardistancechanges
from11.5to15.5degrees.AverageangularspeedsoftheMoonandtheSunare13.2degrees
and1degreerespectively.ThuswithrespecttotheSuntheaverageangularspeedoftheMoon
is13.2-1=12.2degreesornearly12degrees.Thisangulardistanceiscalledatithi.Whenthe
angulardistanceoftheMoonincreasesby12degreeswithrespecttotheSunonetithiissaidto
becompleted.AgainbecauseofellipticalorbitoftheMoontimeintervalofatithicanchange
between20to28hours.Inadditiontothissometimesaparticulartithiisomittedoradvanced.
DuetothesecomplicationsitisverydifficulttocorrectlyestimatethetimeoftheMoonrise.With
someassumptionwecanstillveryroughlyestimatethetimeofMoonrise.Atleastitcangive
someideaapproximatelywhentheMoonwillrise.
ToestimatethetimeoftheMoonrisefollowingfactsareveryuseful.
•EverydaytheMoonrisesabout50minuteslate,becausewithrespecttothesun,themoon
movesthroughanangulardistanceofabout12.2degrees.SincetheEarthrotatesthrough1
degree4minutes.Tocoveradistanceof12.2degreesthemoonrequires12.24=48.8
minutesorroughly50minutes.Ifatcertainnomoonday(Amavasya)SunandMoonrisesat
6.30aM.NextdaytheMoonwillriseat6.30+50=7.20AM.
•Indarkfortnight(Krishnapaksha)MoonrisesinthenightbetweenSunsetandSunrise.
•Inwhitefortnight(Shuklapaksha)MoonrisesinthedaybetweenSunrisetandSunset.
•OnnomoondaylongitudesoftheSunandtheMoonarenearlyequal.
•Onfullmoondaythereisadifferenceof180degreesbetweenLongitudesoftheSunandthe
Moon.
•InwhitefortnightelongationbetweentheSunandtheMoon=Tithi12degrees.
IndarkfortnightelongationbetweentheSunandtheMoon=(180-Tithi)12degrees.
ElongationisananglebetweentheSun,theEarthandtheMoon.
•InwhitefortnightwesternsideoftheMoonisbright,whileindarkfortnighteasternsideofthe
Moonisbright,
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

104. 104
Useful information to estimate Moon rise
•AccordingtoIndianStandardTimeroughlythetimingsofsunriseandsunsetareas
follows,
Month Sunrise Sunset
November,December,January 7 6
February
March,April,September,October 6.30 6.30
May,June,July,August 6 7
#Theremaybeadifferenceof15minutesinthesetimings.
•On8
th
tithi(Ashtami)ofblackandwhitefortnight,phaseofthemoonisabouthalf.
•On1
st
dayofwhitefortnight(ShuklaPratipada)ifthedistancebetweentheSunandthe
Moonismorethan12degreesthenonlytheMoonisvisible.
•On4
th
JanuarydistancebetweentheSunandtheEarthisminimum.TheEarthisat
Perihelionofitsorbitandthelengthofthedayissmallabout11hours.
•On4
th
JulydistancebetweentheSunandtheEarthismaximum.TheEarthisat
Aphelionofitsorbitandthelengthofthedayismorelabout13hours.
•OrbitoftheMoonmakesanangleofabout5degreeswithelliptichencewithreferenceto
celestialequatortheMoonisat23.5+5=28.5degreestowardsNorthor23.5–5=18.5
degreestowardssouth.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

105. 105
Very Approximate estimate of Moon rise
•Wehavetwowaystoapproximatemoonrise,oneisIndianTithiandanotherisabout50minutesdelay
ofmoonriseeveryday.ItisnecessarytoknowwithwhichEnglishmonthsrequiredIndianmonthis
associated.ThenthedayofIndianmonthforwhichestimateofmoonriseistobemade,notedownthe
approximatetimingofsunrise.
•EstimateofmoonrisewiththehelpofIndianTithi.
•Whitefortnight(ShuklaPaksha):Moonrisebetweensunriseandsunset
Example1:4
th
Tithi(ShuklaChaturthi)
Sun-Moonelongation=412=48deg.
Moonriseaftersunrise48/153hours.
Ifsunriseisatabout6:30AM,moonrisewillbeinthemorningatabout6:30+39:30AM.
Example2:10
th
Tithi(ShuklaDashami)
Sun-Moonelongation=1012=120deg.
Moonriseaftersunrise120/158hours.
Ifsunriseisatabout6:30AM,moonrisewillbeinthemorningatabout6:30+82:30PM.
•Darkfortnight(KrishnaaPaksha):Moonrisebetweensunsetandsunrise
Example1:3
rd
Dark(KrishnaTritiya)
Sun-Moonelongation=(180-312)=144deg.
Moonriseaftersunset36/152.5hours.(180–144=36)
Ifsunsetisatabout6:30PM,moonrisewillbeinthenightatabout6:30+2:309PM.
Example2:12
th
Dark(ShuklaDwadashi)
Sun-Moonelongation=(180-1212)=36deg.
Moonriseaftersunset144/159:5hours.(180–36=144)
Ifsunsetisatabout6:30PM,moonrisewillbeinthenightatabout6:30+9:3016hours,thatis
4’Oclock(12+4)earlymonringor2.5hoursbeforesunrise.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

106. 106
Very Approximate estimate of Moon rise
•Moonrise:Using50minutesdelayofmoonriseeveryday.
ForthismethoditisnecessarytonotedownthenoMoondays(Amavasya)oftheyear.Adding15
daysinthosedaysfullmoondayscanbefound.FromnoMoondaytofullMoondayofIndianmonthis
calledwhitefortnight(ShuklaPaksha)andfromfullMoontonoMooniscalldarkfortnight(Krishna
Paksha).
•Whitefortnight(ShuklaPaksha)
Example1:Estimateofmoonriseon7
th
September2008
nomoondaywas30
th
August2008.30
th
Augustto7
th
Septemberno.ofdays=8.
dueto50minutesdelayofmoonriseaccumulatedminites=850=400minutes
400minutes=400/606hrs40min.
Ifsunriseisatabout6:30AM,moonrisewillbeaftersunrise6:30+6:40=1hr10minPM.
Almanacgivesmoonriseat12:48PM.
•Darkfortnight(KrishnaPaksha)
Example2:Estimateofmoonriseon24
th
November2008
nomoondaywas28
th
October2008,hence13
th
November(28+15=43)wasfullNoonday.From
From13
th
Novemberto24
th
NovemberNo.ofdays=11.dueto50minutesdelayofmoonrise
accumulatedminutes=1150=550minutes=9hrs10min.
Ifonthatdaysunsetwasat6intheeveningthenmoonrisewillbeaftersunset6:00+9:103in
themorning.Almanacgivesmoonriseat2:55inthemorning.
•AlternativeMethod
OrsimplycountnumberofdaysafterNoMoonDay,inthiscase,26days.
No,ofaccumulatedminutes=2650=1300minutes=21hrs40min.
Morning6toNextmornig6=24hrs.
24hrs-21hrs40min=2:20earlymorning.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

107. 107
Local sidereal time (LST)
•GreenwichSiderealTimeisthelocalsiderealtimefortheGreenwichMeridian.As
wemovetoeastorwestfromthezerolongitudeofGreenwich,thehourangleofthe
VernalsiderealtimeasadifferencebetweentheGreenwichsiderealtimeandthe
longitudeofthelocationdividedby15,becausetheEarthroutesthrough15degrees
inonehour.ThewestlongitudesgivesLSTearlierthanGSTandeastlongitudes
later.
•IngeneralLocalSiderealTimeisgivenby,
•LST=GST–GST/15
•LetuscalculateLSTatMumbaion19
th
November2001atzerohours.
•ForMumbai,Longitude=-72.833deg.=-72.8333/15hrs=-4.8555hrs.
•On19
th
November2001,GST=3.872875hrs.
•MumbaiLST=3.872875–(-4.8555)hrs=8.728375hrs=8hr43min42.15sec.
•
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

108. 108
Greenwich sidereal time (GST)
•TwosuccessivepassagesoftheSunacrossthemeridianiscalledtheSolarday,whichis
assumedtobe24hours.InsteadoftheSunifastarisselectedasthereference,thetime
intervalbetweentwosuccessivepassagesofthestaracrossthemeridianisnot24hoursbutit
isabout23hours56minutes4second.SincethestarsarequitefarawayfromtheEarththan
theSun,theStararrivesonthemeridianbeforetheSun.InthistimeintervaltheEarthmoves
throughabout1degree.ThustheSunappearsprogressivelydisplacedagainstthebackground
stars.
•Whentimeismeasuredwithreferencetotwosuccessivepassagesofthestaracrossthe
meridian,itiscalledSiderealTime.ThistimemeasuredatGreenwichmeridianitiscalled
GreenwichSiderealTime(GST)
•InsteadofastarVernalEquinoxisselectedasthereference.Ingeneralthehourangleofthe
VernalEquinoxisdefinedasthesiderealTime.
•Around22
nd
SeptembertheSunarrivesattheAutumnalEquinox,henceonthatday12o’clock
nightVernalEquinoxcrossesthemeridian,thisistakenaszerohourGST.Afterthatsidereal
timewilladvanceby3minand56sec.overtheUniversalTime(UT).On21
st
MarchGSTwillbe
12hours.
•RightAscensions(RA)ofstarsaremeasuredfromVernalEquinox,hencewecanfindoutwhich
starswillcrossthelocalmeridianat12o’clocknight.TheLocalSiderealTime(LST)caneasily
calculatedifthelongitudeoftheplaceisknown.
•WiththehelpoffollowingequationswecanfindtheGST.
•1)Find T=(JD–2451545.0)/36525
•JDJulianDayNumberoftherequireddate.
•2)GST=6hr41min50.54841sec+8640184.812866secT+0.093104secT
2
-0.0000062secT
3
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

109. 109
Greenwich sidereal time (GST)
•Example:FindGSTon19
th
November2001
•1)On19
th
November2001 JD=2452232.5
T=(2452232.5–2451545.0)/36525=0.018822724
•2)GST=6hr41min50.54841sec+8640184.812866sec(0.018822724)
+0.093104sec(0.018822724)
2
•SinceTisverysmallfourthtermoftheequationisneglected.
•GST=6hr41min50.54841sec+21hr10min31.81sec+0(thirdtermofthe
equationisalsoneglected)
•GST=27hr52min22.35sec.
•GSTshouldbebetween0and24hours,hence24issubtractedfromtheresult.
•GST=3hr52min22.35sec.GST=3.872875hrs..
• Greenwich meridian
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

110. 110
Approximate calculation of GST
•InsteadofrigorouscalculationwecanestimatetheGSTwithinfewsimplesteps.
•1.Weknowthat22
nd
SeptemberGST=0hrs
•2.Countthenumberofdayselapsedfrom22
nd
Septemberuptorequireddate.
•3.WethatthedifferencebetweenSolardayandSiderealdayisapproximately3
minutes56secondsor3.93minutes.
•4.Multiplythenumberofdayselapseduptotherequireddateandthedifference
betweensolarandsiderealday.Converttheresultintohour,minutes,andseconds.
ThisthenistheGSToftherequireddate.
Example:EstimatetheGSTon19
th
November2001
•No.ofdayselapsedfrom22
nd
Septemberto19
th
November2001=
Sept=8,Oct=31,Nov=19Total58days
•Noofdayselapsed3.93
=583.93minutes=227.94minutes=3.799hours=3hrs47min56.4sec.
ThisisapproximatevalueofGSTon19
th
November.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

111. 111
Rising and setting of the Sun
•KnowinglatitudeoftheplaceandthedeclinationoftheSunontherequireddate,wecan
easilycalculatethesunriseandsunsettimingsatthatplace.InIndianAlmanacsthe
declinationoftheSunisgivenforeverydayoftheyear.
•TheHourAngleatriseindicatesthedurationofhalfday.Followingformulaisusedtofind
theHourAngle.
•cos(HourAngle)=-tan(latitude)tan(declinationofSun)
•Letuscalculatethetimingsofsunriseandsunseton19
th
November2001atDhaka
•LatitudeofMumbai=19degNorth
•DeclinationofSunon19
th
November2001=-19.4222deg.
•cos(HourAngle)=-tan(19)tan(-19.4222)=0.124
•HourAngle=83.02669579deg.=83.02669579/15hours=5.535hours
•Thisishalfdaydurationoftheday
•Durationoftheday=25.535hours=11.07hrs=11hrs4.2min.
•TimeofSunrise=12–HourAngleinhours=(12–5.535)hrs=6hrs27min.54sec.
•Thisislocaltimeofsunrise.InordertogettimeofsunriseinIndianStandardTime(IST)we
havetoadd38min.40sec.
•ISTiswithreferenceto82.5degEastlongitudes.
•Thedifference=9.667deg.
•TheEarthturnsthrough15deg.In1hror60min.
•Toturnthrough9.667deg.Itwilltake38.668minor38min40sec(nearly).
• SinceMumbaiistothewestof82.5deg.eastlongitudewehavetoadd38min.40sec
inthelocaltimetogetIST.
•TimeofsunriseatMumbai=6hrs27min54sec+0hr38min40sec=7hrs06min34
secIST
•TimeofSunset=12+HourAngleinhours=12+5.535=17.535=17hr32min
•TimeofSunsetinIST=17hr32min+0hr38min40sec=18hr10min40sec
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

112. 112
Rising and setting of the Sun
•Asananotherexampleletusfindoutsunriseandsunsettimingson19
th
November2001
London.AtLondonlocaltimeisGMTandhencenocorrectionisrequired.Thelatitudeof
Londonis51.5deg.North.
•Cos(HourAngle)=-tan(51.5)tan(-19.4222)=0.4432678
•HourAngle=63.6874deg=63.6874/15hours=4.2458hours
•Thisishalfthedurationoftheday
•Durationoftheday=24.2458=8.4916hrs=8hrs29min29.76sec.
•TimeofSunrise=12–HourAngleinhours
• =12–4.22458=7.7542hrs=7hrs45min15.12sce
•ThetimeofSunset=12+HourAngleinhours
• =12+4.2458=16.2458hrs=16hrs14min44.88sce
•Sunsetwasatabout4o’clockintheevening.Thedurationofthedaywasonlyabout8.5
hours.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

113. 113
Metonic cycle
•ThetimeintervalbetweentwosuccessivecrossingsoftheVernalEquinoxbytheSunis
called“TropicalYear”.OneTropicalYearconsistsof365.2422days.
•TheTimeintervalbetweentwosuccessiveconjunctionsoftheSunandtheMooniscalled
‘SyndonicMonth”.Itis29.53059dayslong.
•Intheyear433BC,theancientGreekastronomerMetonfoundarelationshipMonth.He
observedthatthenumberofdaysin19Tropicalyearsarealmostequaltothenumberof
daysin235Synodicmonths.Therelationshipcanbeprovedasfollows.
•19Tropicalyears=19365.2422=6939.602days
•235SynodicMonths=23529.53059=6939.689days.Thedifferenceisonly0.087days,
thatis2hrs5min16sec.Therelationshipsuggeststhatafterevery19yearstheSunand
theMooncomeintosameconfiguration.Inotherwords,afterevery19yearsthephasesof
theMoonarerepeated.TheremaybeadifferenceofonedaydependingontheNumberof
LeapyearsinthePeriodof19years.TherepetitionofthephaseoftheMoonafterevery19
yearsiscalled‘MetonicCycle’.FollowingtableprovestheMetonicCycle.
Date Phase Special Event
28
th
May 1900 New Moon Solar Eclipse
29
th
May 1919 New Moon Solar Eclipse
29
th
May 1938 New Moon Solar Eclipse
27
th
Oct 1901 Full Moon Lunar Eclipse
27
th
Oct 1920 Full Moon Lunar Eclipse
28
th
Oct 1939 Full Moon Lunar Eclipse
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

114. 114
The Equation of time
•TimekeepingismostimportantaspectofdailylifeandinAstronomyaswell.TheSunis
mostusefulcelestialobjectfortimekeeping.EventhoughtheEarthmovesalongthe
ellipticalorbitaroundtheSun;foranobserverontheEarthitappearsthatthesunmoves
inellipticalorbitaroundtheEarth.ThespeedoftheSunkeepsonchangingthroughoutthe
year.ThusfortimekeepingthetruetheSunisnotuseful.
•ThesolutiontothisproblemistoassumeameantheSunmovingalongtheCelestial
equatorwithmeanspeed.ThemeanspeedofthemeanSunisgivenby,
n=360/365.25=0.9856deg/day
•ThetimeiskeptaccordingtothisfictitiousSun.Theequationoftimeisdefinedas
differencebetween,RightAscensionofthemeanSunandRightAscensionofthetrue
Sun.
•Equationoftime=RAofMeanSun–RAofTrueSun
•RAofmeanandtrueSuncanbefoundasfollows.
•RAOfMeanSun
•RAofMeanSun=n(t–t
0)–(360–w)
•Where,
•n=0.9856deg/dayspeedofmeansun
•t
0=ThedaywhentheSunisatperigee.Around3
rd
or4
th
JanuaryeveryyeartheSunisat
perigee.
•t=Thedayonwhichequationoftimeisrequired.
•W=Longitudeofperigee,whichcanbefoundfromAlmanac.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

115. 115
The Equation of time
•RAOfTrueSun
•Where,
•Longitude=LongitudeoftheSunongivenday,obtainedfromAlmanac.
•Inclination=Inclinationoftheeclipticwithrefencetocelestialequator=23.45deg.
•LetuscalculateEquationofTimeon19
th
Nov.2001
•RAofMeanSun=n(t–t
0)–(360–w)
•t
0=ThedaywhentheSunisatperigee.Around3
rd
or4
th
JanuaryeveryyeartheSunis
atperigee.
•t=19
th
Nov.2001andt
0=3
rd
Jan.2001
•t–t
0=320days(Jan=25,Feb=28,Mar=31,Apr=30,May=31,Jul=31,Aug=
31,Sept=30,Oct=31,Nov=19)
•LongitudeoftheSunon3
rd
JanJan2001=w
•IndianNiryanAlmanacgivesw=8s18deg48min17sec.=258.8047deg.
•ToconvertitintoSayanand23.8783deg.Ayanansha
•258.8047+23.8783=282.6830Sayan  
 










longitudecos
nonclinatiocoslongitudesin
tanSun True ofRA
1
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

116. 116
The Equation of time
•RAofMeanSun=0.9856(320)–(360-282.6830)=238.075deg.=238.075/15hrs
=15.8716hrs
•On19
th
Nov2001LongitudeoftheSun=7sign2deg48min17sec=212.8388889
Niryan
•FromIndianAlmanac
•AddtheAyanansh23.8783
•212.8388889+23.8783=236.7171889Sayan
•ItisnecessarytoputtheSuninproperquadrantasfollows.
•WehavebothNumeratoranddenominatorwith-vesigns.HencetheSunmustbeinthe
3
rd
quadrant.Thus180deg.Shouldbeaddedtotheresult.
•RAofTrueSun=54.4168+180=234.4168deg.
• =234.4168/15hrs=15.6277hrs.
•Theequationoftimeon19
th
Nov.2001
•RAofMeanSun–RAofTrueSun=15.8716–15.6277=0.2439hrs
• =0.243960min=14.634min
•+vesignmeansMeanSunisfasterthanTrueSun.  
 










longitudecos
nonclinatiocoslongitudesin
tanSun True ofRA
1   
 

















0.54877-
0.76699-
tan
89l236.71718cos
23.45cos236.717889sin
tanSun True ofRA
11
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

117. 117
The Equation of time
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

118. 118
Horizontal sundial hour angle
•HorizontalSundialismostcommonandveryeasyto
constructforthelocationshavinglatitudeslessthan15
degreesnorthorsouththeHorizontalSundialisnot
veryattractive;becausetheangleofGnomonisvery
smallfortheselatitudes.
•Therareverysimpleformulatolayoutthehourlines
ontheHorizontalSundial.Theformulaisasfollows.
•Tan(D)=tan(t)sin(L)
•Where,DTheangleofhourlinewithmeridianline
i.e.12o’clockline.
•tHourangleoftheSun,LLatitudeofthelocation
•HourangleoftheSunis90deg.WhentheSunisnear
horizon.
•Weshallcalulatethehourlinesforthelocationhaving
latitudeof50deg.North.
•Thetablegivesthevaluesofhour’slineswiththe12
o’clockline.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

119. 119
Horizontal sundial hour angle
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

120. 120
Approximate estimate Of the position of the Sun
•NauticalAlmanacgivethevaluesofRightAscensionandDeclinationoftheSun.Itis
comparativelyveryeasytoestimatethesevalues.Therearesomestandardformulae.In
additiontothatwehavetocalculatetheJuliandayoftherequireddate,andthenfind
thevalueofTgivenbyfollowingrelation.
•T=(JD–2451545.0)/36525
•ProceduretofindtheJulianDayNumberisgivenelsewhere.Calculationswillbe
simplifiedbyneglectinghigherordertermsofT.
•FollowingsequenceofformulaeleadtotheRightAscensionandDeclinationoftheSun.
•1.GeometricmeanLongitude
L
0=280.46645+36000.76983T+0.0003032T
2
•2.MeanAnomaly
M=357.5291+35999.05030T-0.0001559T
2
•3.EquationOfCenter
C=+1.9146sinM
•4.TrueLongitude
Longitude=L
0+C
•5.RightAscensiontan(RA)=tan(longitude)cos(i)
•6.Declinationsin(declination)=sin(longitude)sin(i)
iAnglebetweenEclipticandCelestialEquator=23.439deg
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

121. 121
Approximate estimate Of the position of the Sun
•Example:FindRAandDeclinationoftheon19
th
November2001.
•JDon19
th
November2001=2452232.5
•T=(2452232.0–2451545.0)/36525=0.018822724
•1.GeometricmeanLongitude
L
0=280.46645+36000.769830.018822724+0.00030320.018822724
2
=238.099deg.
•2.MeanAnomaly
M=357.5291+35999.050300.018822724-0.00015590.018822724
2
=315.129288deg.
•3.EquationOfCenter
C=+1.9146sin(315.129288)=-1.350768
•4.TrueLongitude
Longitude=238.099+(-1.350768)=236.748deg.
•5.RightAscensiontan(RA)=tan(236.748)cos(23.439)=1.3993
RA=54.442993.6299hrs=3h37m47s
•6.Declinationsin(declination)=sin(236.748)sin(23.439)=-0.3326
Declination=-19.4293deg.=19deg25min45sec.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

122. 122
To find day of the week
•UsingtheJulianDayNumbers,thedayoftheweekofanydayofanyearbetween1700to
2200ADcanbeeasilyfound.JulianDayNumberisthenumberofdayselapsedsince
Monday1
st
January4713BC.JulianDayNumberisacontinuouscountingofthedaysupto
therequireddayofanyear.Themethodisasfollows.
•Step1:A=JulianDayNumbersoftheselectedcenturyyear
CenturyYears(AD) JulianDays
1700 2341972
1800 2378496
1900 2415020
2000 2451545
2100 2488069
2200 2524593
•Step2:B=Selectedyear365
•Step3:C=INT(SelectedYear/4) {Takescareofaccumulatedleapdays}
•Step4:D=No.ofdaysuptoselectedday
•Step5:Day=MOD{(A+B+C+D)/7} {MODmeansfractioninti7orremainder}
•If,Day=0:MondayDay=1:TuesdayDay=2:WednesdayD=3:Thursday
Day=4:FridayDay=5:SaturdayDay=6:Sunday
•Example1:Authorofthisbookwasbornom5
th
December1938,whatwastheday?
•A=2415020,B=38365=13870,C=INT(38/4)=9,
D=No.ofdaysupto5
th
December=339
•Day=MOD{(2415020+13870+9+339)/7}=MOD(2429238/7}=0(Remainder=0)
• 5
th
December1938wasMonday
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

123. 123
Day of the week from Gregorian calendar
•Stepwiseprocedureisasfollows.
•1.A=(14–month)/12 month=JanuarytoDecember1To12
•2.y=year–A year=4digitvalue
•3.M=month+12A–2
•4.D=[day+y+y/4–y/100+y/400+(31M)/12]MOD7
•Consideronlyintegervaluesoftheratios.
•MOD7meansonlyconsidertheremainderafterdividingby7.
•Accordingtoremainderdayistofixedasfollows
•D=0:Sunday D=1:Monday D=2:Tuesday D=3:Wednesday
•D=4:Thursday D=5:Friday D=6:Saturday
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

124. 124
Day of the week from Gregorian calendar
•Example1:Authorofthisbookwasbornom5
th
December1938,whatwastheday?
•1.A=(14–month)/12=(14–12)/12=2/12Integer=0
•2.y=year–A=1938-0=1938 year=4digitvalue
•3.M=month+12A–2=12+120–2=10
•4.Intheformula:y/4=1938/4=484 Integer
y/100=1938/100=19 Integer
y/400=1938/400=4 Integer
(31M)/12=(3110)/12=25Integer
•5.D=[5+1938+484–19+4+25]MOD7=(2437)MOD7=1(remainder)
• 5
th
December1938wasMonday
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

125. 125
Very approximate estimate of the position of Sun
•AncientIndianshaveacceptedapointoppositetothestarSpica(Chitra)asthestarting
pointoftheNiryanZodiac.EventodayIndianCalendarsacceptthisstartingpointofsignof
Zodiac.Modernsystemallots30degreestoeachsignstartingfromtheVernalEquinox.
•VernalEquinoxhasaretrogrademotion.Itmovesbackthrough1degreein72years.In
theyear280AD,VernalEquinoxwasconsideringwiththepointoppositethestarspica.
Sincethenitisshiftedbackbyabout24degrees.HenceinordertoconvertNiryan
longitudetoSayanlongitude,wehavetoaddabout24degrees.Thisdifferencewill
steadilyincrease.
•Itisobservedthatroundabout13,14,or15
th
ofeverymonththeSunentersnewNiryan
sign.ForconvinceweshalltakeentryoftheSuninnewNiryansignon14
th
dayoftheSun
entersinTaurusetc.14
th
Aprilisaheadof21
st
Marchby24days.Atpresenteveryyear
theSunenterstheVernalEquinoxon20
th
or21
st
March.
•Withtheaboveinformationwecanprepareatableofveryapproximatepositionofthesun
forniryanandsayancalendar.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

126. 126
Very approximate estimate of the position of Sun
*
Day Niryan NiryanSign Sayan
and Longitude Longitude
Month (deg.) Entry (deg.)
14thApril 0 Aries 24
14thMay 30 Taurus 54
14thJune 60 Gemini 94
14thJuly 90 Cancer 114
14thAugust 120 Leo 144
14thSeptember 150 Virgo 174
14thOctober 180 Libra 204
14thNovember 210 Scorpio 234
14thDecember 240 Sagittarius 264
14thJanuary 270 Capricornus 294
14thFebruary 300 Aquarius 324
14thMarch 330 Pises 354
21thMarch 360/0
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

127. 127
Very approximate estimate of the position of Sun
•WecanalsoestimateRightAscensionandDeclinationofthesun
•Example1:ApproximatelyfindthedeclinationoftheSunon19
th
November.
ApproximateNiryanLongitudeon14
th
November=210degrees.
ApproximateNiryanLongitudeon19
th
November=210+5=215degrees.
ApproximateSayanLongitudeon19
th
November=215+24=239degrees.
InclinationofEcliptic=23.439degrees
sin(declination)=sin(Logitude)sin(inclination)=sin(239)sin(23.439)=-0.34095
Declination=-19.935degrees ActualDeclination=-19.5degrees.
ActualNiryanLongitude=213.75degrees.
Day and Month R.A. R.A. Day and Month R.A. R.A
(hrs) (deg) (hrs) (deg)
21th March 0 0/360 21th September 12 180
21th April 2 30 21th October 14 210
21th May 4 60 21th November 16 240
21th June 6 90 21th December 18 270
21th July 8 120 21th January 20 300
21th August 10 150 21th February 22 330
1 hr = 15 degrees
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

128. 128
Very approximate estimate of the position of Sun
•Example2:ApproximatelyfindthedeclinationoftheSunon10
th
June.
ApproximateNiryanLongitudeon14
th
June=60degrees.
ApproximateNiryanLongitudeon10
th
June=60-4=56degrees.
ApproximateSayanLongitudeon10
th
June=56+24=80degrees.
InclinationofEcliptic=23.439degrees
sin(declination)=sin(Logitude)sin(inclination)=sin(80)sin(23.439)=0.391729
Declination=23.062degrees ActualDeclination=23deg.3min43sec.
ActualNiryanLongitude=55.25degrees.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

129. 129
Approximate calculation Of Indian first point Of Aries and
vernal equinox
•IndianalmanacsgivelotofAstronomicalinformation.Thisisonlyhandyandcheap
databook,easilyavailableeveryyeartoIndianArmatureAstronomy.Indian
AlmanacsonlygiveLongitudes(Bhog)ofsun,Moon,andplanets.Latitudes(Shar)are
givenatsomeintervaloftime.Problemis,theLongitudesaregivenwithreferenceto
IndianstartingpointofZodiacorfirstpointofaris.Thepointexactlyoppositetothe
starSpicaisselectedasthestartingpointofZodiacbyancientindianastronomers.
Thealmanaccalculatedwithreferencetothisfixpointiscalled‘Niryan’.ThusIndian
signsofZodiacarefix.
•Intheyear280AD,TheVernalEquinox(VasantSampata)wascoincidingwiththe
indianfirstpointofAris.ButVernalEquinoxhasretrogademotion.Itshiftsthrough
about50.2arcsecondseveryyearotthrough1degreeinabout72years.Nowthe
differencebetweenVernalEquinoxandIndianfirstpointofarishasaccumulatedto
about24degrees.AnAlmanaccalculatedwithreferencetoVernalEquinoxiscalled
‘Sayan’.WesterncountriesfollowSayanAlmanac.TheirsignsofZodiacaremovable.
•FromIndianalmanactogetlongitudeswithreferencetoVernalEquinox,itis
necessarytoaddtheangulardifferencebetweenVernalEquinoxandindianfirstpoint
ofaris.Thisdifferenceiscalled‘Ayanansh’.ThevalueofAyananshissteadily
increasing.ApproximatemethodofcalculationofAyananshisasfollows,
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

130. 130
Approximate calculation Of Indian first point Of Aries and
vernal equinox
Step1:Findouthowmanyyearselapsedsince1900ADupto1
st
Januaryofthe
selectedyearcalledthisnumber‘t’
Step 2: Use followingrelationto findthe Ayanansh
Ayanansh=22deg27min43.1sec+50.2564tsec.
Example1:FindAyananshof1
st
January2009
Step1:Numberofelapsedyearssince1900AD
t=2009–1900=109
Step2:Ayanansh=22deg27min43.1sec+50.2564109sec.=23
0
:59’:1.51”
IndianAlmancgivestheAyananshof4
th
January2009as23
0
:59’:12”Example
Example2:IndianAlmancgivesvaluesofLongitudeoftheSunandMoonat5:30AM
as,
Sun:8:16
0
:42’:46”
Moon:10:05
0
:46’:26”
FindcorrespondingSayanLongitude
SayanLongitude=NiryanLongitude+Ayanansh
SayanLongitudeoftheSun=8:16
0
:42’:46”+23
0
:59’:1.51”=9:10
0
:41’:47.51”
SayanLongitudeoftheMoon=10:05
0
:46’:26”+23
0
:59’:.51”=10:29
0
:45’:27.51”
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

131. 131
Elementary Astronomical Calculations:
Telescope and Calendar ( Lecture –3)
Thank You
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

132. 132
Elementary Astronomical Calculations:
Eclipses and Planets ( Lecture –4 )
•By Sukalyan Bachhar
•Senior Curator
•National Museum of Science & Technology
•Ministry of Science & Technology
•Agargaon, Sher-E-Bangla Nagar, Dhaka-1207, Bangladesh
•Tel:[phone](Off), Contact: 01923522660;
•Websitw: www.nmst.gov.bd; Facebook: Buet Tutor
&
•Member of Bangladesh Astronomical Association
•Short Bio-Data:
First Class BUET Graduate In Mechanical Engineering [1993].
Master Of Science In Mechanical Engineering From BUET [1998].
Field Of Specialization Fluid Mechanics.
Field Of Personal Interest Astrophysics.
Field Of Real Life Activity Popularization Of Science & Technology From1995.
An Experienced TeacherOf Mathematics, Physics & Chemistry for O-,A-, IB-&
Undergraduate Level.
Habituated as Science Speaker for Science Popularization.
Experienced In Supervising For Multiple Scientific Or Research Projects.
17 th BCS Qualified In Cooperative Cadre (Stood First On That Cadre).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

133. 133
Angular diameters of the solar and lunar disks
•EventhoughthediametersoftheSunandthemoonareconsiderabletheyappearintheformof
disksfromtheearth.Theang;ethatthesunortheMoonsubtendsatoureyeiscalledtheangular
diametersoftheirdisks.Theaveragevalueofthisangleisabouthalfadegreeorabout30arc
minutes.Howevertheangulardiameterdependsontheirdistancesfromtheearth.
•Anangleisdefinedastheratioofanarcandtheradius.Mathematicallywehave,
angle=(arc/radius)radians
•Butthisvalueisintermsofradians.Knowingthethatradiansareequalto180degrees,wehave
tomultiplytheaboveequationby180/togetangleindegrees.
angle=(arc/radius)180/degrees.
•ForcalculatingtheangulardiametersofthedisksweshalltaketheactualdiameteroftheSunor
themoonasanarcandthedistancefromtheEarthasradius.
•AngularDiameteroftheMoon’sdisk.
Wehave,DiameteroftheMoon=3500km(asanarc)
DistanceoftheMoon=384000km(average);356000km(minimum);405000km(maximum)
AngularDiameteroftheMoon’sDisk=(3500/384000)180/=0.5222deg.=31.33arcmin(average)
AngularDiameteroftheMoon’sDisk=(3500/405000)180/=0.495deg.=29.7arcmin(minimum)
AngularDiameteroftheMoon’sDisk=(3500/356000)180/=0.5633deg.=33.8arcmin(average)
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

134. 134
Angular diameters of the solar and lunar disks
•Angular Diameter of the Sun’s disk.
We have, Diameter of the Sun = 1392000 km (as an arc)
Distance of the Sun = 149600000 km (average)
= 147200000 km (minimum)
= 152100000 km (maximum)
Angular Diameter of the Sun’s Disk = (1392000/149600000) 180/= 0.533 deg. = 32 arc min
(average)
Angular Diameter of the Sun’s Disk = (1392000/152100000) 180/= 0.524 deg. = 31.46 arc min
(minimum)
Angular Diameter of the Sun’s Disk = (1392000/147200000) 180/= 0.54 deg. = 32.5 arc min
(maximum)
For total solar eclipse angular diameter of the Moon’s disk should be greater than the Sun’s disk.
For annular solar eclipse angular diameter of the Moon’s disk should be smaller than the Sun’s disk.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

135. 135
Horizontal equatorial parallax
•ForthecalculationofsolarEclipsetheexactdistancesoftheSunandtheMoonmustbe
known.ThesedistancecanbeestimatedfromtheHorizontalEquatorialParallaxoftheSun
andtheMoon.Thesevaluesarealwaysgivenwiththeothereclipsedata.
•ImagineanobserveronthesurfaceoftheEarthandletthedisksoftheMoonandtheSun
areinhislineofsightonhishorizon.Forhimtherewillbeasolareclipsebutforan
observeratthecenteroftheEarththerewillbenosolarEclipse,becausetheSunandthe
Moonwillnotbeinhislineofsight.
•TheanglethatEarth’sradius(r)subtendsatthedistanceoftheSun(S)ortheMoon(M),
whentheyareobservedatthehorizonfromaparticularplaceonthesurfaceoftheEarthis
calledHorizontalEquatorialParallax(p).
•FortheSunandtheMoonthisparallaxismaximum.Asthesecelestialobjectsmovefrom
easttowestthevalueofparallaxdecreasesanditbecomeszerowheneitheroftheobjects
reachesthezenith.
•Fromtrigonometry,
•Sin(p)=r/RR=r/(sin(p))RDistanceofSunorMoon
•ForAugust1999totalSolarEclipsefollowingvaluesaregiven.
•HorizontalEquatorialParallaxoftheMoon=0.9789555deg.
•HorizontalEquatorialParallaxoftheSun=0.002411deg.
•DistanceoftheMoon=6378/(sin(0.9789555))=373306.31km.
•DistanceoftheSun=6378/(sin(0.002411))=1.5156810
8
km.
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136. 136
The eclipse year
•ThepointsofintersectionsoftheapparentpathoftheMoonandtheSunarecalled
AscendingandDescendingnodes.ForEclipsethesetwopointsareveryimportant.
Thenodeshavegotretrogrademotion.TheAscendingorDescendingNode
Completesoneroundinabout18.6yearsor6798.3days.Aftertheconjunctionofthe
SunandtheAscendingnode,theystartmovingawayfromeachother.SincetheSun
isfastcomparedtotheAscendingnodetheymeetagainafterabout346.62days.This
timeintervaliscalled‘EclipseYear’.
•PeriodictimeoftheSunisnearly365.25days,whiletheperiodictimeofAscendingis
6798.3daysornearly18.6years.Usingthesetwovaluesthetimeintervals,the
Eclipseyearcanbefound.
•1/EclipseYear=1/PeriodictimeoftheSun–1/PeriodictimeofAscendingNode
=1/365.25–1/(-6798.3)=0.00288494
EclipseYear=1/0.00288494=346.62689Days
•EclipsesarepossiblewhentheSunisneartheAscendingorDescendingNode.From
aboveitisclearthat,iftheSunisattheAscendingNode,afterabout13daysitwill
arriveattheDescendingnode.EhavetofindthepositionoftheMoonwhenthesunis
atAscendingorDescendingNodetoestimateprobabilityofanEclipse.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

137. 137
The saros cycle
•Thetimeintervalbetweentwosuccessivecrossingsoftheascendingornodebythe
Sunis346.62dys.Thistimespaniscalled‘EclipseYear’.
•ThetimeintervalbetweentwosuccessiveconjunctionsoftheSunandtheMoonis
29.53059days.Thistimeintervaliscalled‘SynodicMonth’.
•About750yearsBCKhaldianscametoknowtherelationshipbetweentheEclipse
YearandtheSynodicMonths.Theyfoundthatnumberofdaysin19Eclipseyearsare
nearlyequaltothenumberofdaysin223Synodicmonths.Therelationshipcanbe
provedasfollows,
19EclipseYears=19346.62=6585.78days
223SynodicMonths=22329.53059=6585.32days
Thediffernceisonly0.46days.
•Thetimeintervalof223SynodicMonthsor6585.32daysiscalledSaros.Itcomesout
tobeequal18yearsand11.32days.SinceEclipsesarepossibleonlywhentheSun
andtheMoonareneartheAscendingorDescendingNode.Sarosisdirectlyrelatedto
theeclipses.AfteraperiodofoneSarostheeclipsesarerepeated.
•Ifthereare4LeapyearsintheSarosperiod,thetimeintervalis18yearsand11.32
days,butiftherefiveLeapyearsintheSarosperiod,thetimeintervalbecomes18
yearsand10.32days.Eachseriescontains70ormoreeclipses.TotalSolarEclipseof
22
nd
July2009wasrepetitionoftheEclipseof11
th
July1991.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

138. 138
The saros cycle
•Sarosseriesarenumberedaccordingtothetypeeclipseandwhethertheyoccuratthe
Moon.sascendingordescendingnode.OddnumbersareusedforSolareclipses
occurringneartheascendingnode,whereasevennumbersaregiventodescending
nodeSolareclipses.
•AsanexampleletusconsiderSarosseriesnumber136.
•SincethenumberiseventheSolareclipsesinthisseriesoccurnearthedescending
node.
Firsteclipsewason14
th
June1360
Lasteclipsewillbeon30July2622
DurationofSarosnumber136=1362.11years.
Therearetotal71eclipsesinthisseries.
FollowingaresomeoftheeclipsesintheSarosseries136,
18
th
May 1901 22
nd
July 2009
29
th
May 1919 02
nd
August 2027
08
th
June 1937 12
th
August 2045
20
th
June 1955 24
th
August 2063
30
th
June 1973 03
rd
September 2081
11
th
July 1991 14
th
September 2099
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

139. 139
Length of the Moon’s shadow
•FortheSolarEclipse,thelengthoftheShadowofMoonisimportant.Whentheshadowfallson
thesurfaceoftheEarth,TotalSolarEclipseResults.IftheMoonisveryclosetotheEarththe
diameteroftheshadowfallingonthesurfaceoftheearthismaximumandtimeintervalof
totalitymaximum,butnotmorethan7.5minutes.Iftheapexoftheshadowisatsomedistance
abovethesurface,AnnularSolarEclipseresults.
•WeshallfirstfindoutageneralequationofthelengthoftheMoon’sshadowandthenapplythe
resultstosomeSolarEclipses.
•ThelineABVisdrawntangentialtothesunandthemoon changes. also Moon the and Sun the between Distance
constant. not isshadow its of length Earth, the around orbot elliptical in moves Moon the Since

Moon the of Radius - Sun the of Radius
distance Moon-Sun Moon the of Radius
L
r-R
Dr
LShadow sMoon' of Length
L
r
LD
R
D Moon the and Sun the between Distance SM L,Shadow sMoon' of LengthMV
r, Moon the of Radius MB R, Sun the of Radius SA Here,

MV
MB
SV
SA
write,can weHence
similar; are ΔΔMV&ΔSAV common isV ΔΔMVB& ΔΔSAVfor Also90MBVSAV
0










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

140. 140
Length of the Moon’s shadow
•FortheSolarEclipse,thelengthoftheShadowofMoonisimportant.Whentheshadowfallson
thesurfaceoftheEarth,TotalSolarEclipseResults.IftheMoonisveryclosetotheEarththe
diameteroftheshadowfallingonthesurfaceoftheearthismaximumandtimeintervaloftotality
maximum,butnotmorethan7.5minutes.Iftheapexoftheshadowisatsomedistanceabove
thesurface,AnnularSolarEclipseresults.
•WeshallfirstfindoutageneralequationofthelengthoftheMoon’sshadowandthenapplythe
resultstosomeSolarEclipses.
•ThelineABVisdrawntangentialtothesunandthemoon changes. also Moon the and Sun the between Distance
constant. not isshadow its of length Earth, the around orbot elliptical in moves Moon the Since

Moon the of Radius - Sun the of Radius
distance Moon-Sun Moon the of Radius
L
r-R
Dr
LShadow sMoon' of Length
L
r
LD
R
D Moon the and Sun the between Distance SM L,Shadow sMoon' of LengthMV
r, Moon the of Radius MB R, Sun the of Radius SA Here,

MV
MB
SV
SA
write,can weHence
similar; are ΔΔMV&ΔSAV common isV ΔΔMVB& ΔΔSAVfor Also90MBVSAV
0










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

141. 141
Length of the Moon’s shadow
•TotalSolarEclipse:22
nd
July2009
•WehavefollowingdataEquatorialHorizontalParallaxoftheSun=Ps=0.00241666
•eg.EquatorialHorizontalParallaxoftheSun=Ps=0.00241666deg.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

142. 142
Length of the Moon’s shadow
•FortheSolarEclipse,thelengthoftheShadowofMoonisimportant.Whentheshadowfallson
thesurfaceoftheEarth,TotalSolarEclipseResults.IftheMoonisveryclosetotheEarththe
diameteroftheshadowfallingonthesurfaceoftheearthismaximumandtimeintervalof
totalitymaximum,butnotmorethan7.5minutes.Iftheapexoftheshadowisatsomedistance
abovethesurface,AnnularSolarEclipseresults.
•WeshallfirstfindoutageneralequationofthelengthoftheMoon’sshadowandthenapplythe
resultstosomeSolarEclipses.
•ThelineABVisdrawntangentialtothesunandthemoon changes. also Moon the and Sun the between Distance
constant. not isshadow its of length Earth, the around orbot elliptical in moves Moon the Since

Moon the of Radius - Sun the of Radius
distance Moon-Sun Moon the of Radius
L
r-R
Dr
LShadow sMoon' of Length
L
r
LD
R
D Moon the and Sun the between Distance SM L,Shadow sMoon' of LengthMV
r, Moon the of Radius MB R, Sun the of Radius SA Here,

MV
MB
SV
SA
write,can weHence
similar; are ΔΔMV&ΔSAV common isV ΔΔMVB& ΔΔSAVfor Also90MBVSAV
0










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

143. 143
Length of the Moon’s shadow
TotalsolarEclipse:22
nd
July2009
•Wehavefollowingdata.
•EquatorialHorizontalParallaxoftheSun=Ps=0.00241666deg.
•EquatorialHorizontalParallaxoftheMoon=Pm=1.02221666deg.
•RadiusoftheMoon=r=1738km.,RadiusoftheSun=696000km.
•RadiusoftheEarth=6378km.
•InordertofindthedistanceoftheSunandtheMoonfromtheCenteroftheEarthfollowing
formulacanbeused.
•Sincethelengthofthemoon’sshadowisgreaterthanthedistancebetweentheEarthandthe
moon,theshadowfallsonthesurfaceoftheEarth.
•DurationoftheEclipseislongbecausetheMoonisveryclosetotheearth.Nearestdistanceof
theMoonfromtheEarthisabout356000km.forthiseclipse,distanceis357509km.The
maximumdurationofthiseclipseisabout6.5minutes.  
 
 
km. 377650
1738-696000
1508563491738
shadow sMoon' the of Length
2009July 22 On
km. 150856349357509-151213858Moon the and Sun the between Distance
km. 357509
1.02221666sin
6378
Moon the of Distance
km. 151213858
0.00241666sin
6378
Sun the of Distance
2009July 22 On
Parallax Horizontal Equatorialsin
6378
Earth the from Distance







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

144. 144
Length of the Moon’s shadow
AnnularSolarEclipse:15
th
January2010
•Wehavefollowingdata.
•EquatorialHorizontalParallaxoftheSun=0.00247222deg.
•EquatorialHorizontalParallaxoftheMoon=0.9014722deg.
•EclipseisAnnularbecausetheapexoftheMoon’sshadowisnotreachingtheEarth’s
surface.Itisatadistanceof405389–369023km.fromthesurface.
•ThedistanceoftheMoonfromtheEarthis405389km.Itsmaximumdistanceisnearly
406000km.  
 
km. 369023
1738-696000
147410301738
shadow sMoon' the of Length
2010January 15 On
km. 1474101303405389-147815519Moon the and Sun the between Distance
km. 405389
0.9014722sin
6378
Moon the of Distance
km. 147815519
0.00247222sin
6378
Sun the of Distance






Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

145. 145
To estimate the instant of maximum solar eclipse from
Indian almanac
•IndianAlmanacgivedailyNiryanlongitudesofthesunandtheMoon,alsotheirdailyangular
speeds.Thelongitudesaregivenat5:30amISTor00:00hrsUT.Itisnotnecessarytoconvert
NiryanlongitudesintoSayan,becausewhiletakingsubtractionthedifferenceswillgetcancelled.
Withthisinformationwecanestimatetheinstantofmaximumsolareclipse.Followingstepsare
required.
•1.FindthedifferenceinlongitudesoftheSunandtheMoonindegrees.
•2.FindtheangularspeedoftheMoonwithrespecttothesunindegreeperhour.
•3.Theinstantofmaximumeclipsecanbefoundfromthefollowingformula.
InstantofMaximumEclipse=(Differenceinlongitudes)/Relativespeedhours
•4.TheinstantofmaximumeclipsewillbeinUT.
•5.Add5:30hourstoconvertitintoIST.
•Example1:TotalSolarEclipse:22
nd
July2009
•Sun3:05:20:3995.344166deg.
•Speed=57min.17sec57.28330.954722deg.V=0.954722/24=0.03978deg/hr.
•Moon03:03:48:2293.80611deg.
•Speed=445min.03secin12hrs.7.584166deg.In12hrsV=0.6320deg/hr.
•Differenceinlongitudes=95.344166-93.80611=1.5380deg.
•Relativespeed=0.6320–0.03978=0.5922deg/hr.
•InstantofMaximumEclipse=(1.5380/0.5922)=2.597hrs.=2hrs35min.49.2sec.
•NASAbulletingives:2:35:12.4UT
•InstantofmaximumeclipseinIST=2hrs:35min.:49.2sec.+5hrs:30min=8hrs:5min:49.2sec
•Themaximumeclipseat144deg8.4minEastlongitude144.14/15=9.6hrs.
•Themaximumeclipseinlocaltime9hrs36min+2hrs35min.49.2sec.12hrs11min.49.2
sec.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

146. 146
To estimate the instant of maximum solar eclipse from
Indian almanac
•Example2:AnnularSolarEclipse:15
th
January2010
•Sun09:00:42:55270.7152deg.
•Speed=61min.08sec61.1333min1.0188in24hrs.V=1.0188/24=0.04245deg/hr.
•Moon08:27:27:34267.4594deg.
•Speed=355min.37sec355.61665.9269deg.In12hrsV=0.4939deg/hr.
•Differenceinlongitudes=270.7152–267.4594=3.2558deg.
•Relativespeed=0.4939–0.04245=0.45145deg/hr.
•InstantofMaximumEclipse=(3.2558/0.45145)=7.2118hrs.=7hrs12.7min.UT
•NASAbulletingives:7hrs:6min:22.9secUT
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

147. 147
Width of the shadow when total solar eclipse is maximum
* 2r- Shadow of Diameter get, weside both to 2by gMultiplyin
negative is answer r- r'
shadow. the of radius get weequations above the Solving
point selected andshadow the of Apex the between distance X
desired idshadow of radius whereMoon the behind distance S
Moon the and Sun the between distance D
shadow the of radius r' Moon the of radius r Sun the of radius R Where,

XS
r
X
r'
&
XS
r
XSD
R
equations. following writecan weThen .90QCV PBV OAV
and common is V angle since similar areQCV andPBV OAV, triangles figure the In
0






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D
S
1
D
S
R2
D
S
1
D
S
R
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

148. 148
Width of the shadow when total solar eclipse is maximum
*  
 
nearly 6750 Wgives BulletinNASA
km 672434843240
D
S
12r
D
S
2RWPenumbra the of widthThe
km 258 gives BulletinNASA
244km34843240
150856349
351131
13476
150856349
351131
1392000 shadow the of Diameter
km. 3511316378357509S surface sEarth' the and Moon the between Distance
km. 150856349D Moon the and Sun the between Distance
km. 357509
1.02221666sin
6738
Moon the of Distance
deg. 1.02221666 Moon the ofParallax Horizontal Equatorial
km. 151213858
0.00241666sin
6738
Sun the of Distance
deg. 0.00241666 Sun the ofParallax Horizontal Equatorial
km 3476 Moon the of Diameter 2r km 1392000 Sun the of Diameter 2R





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

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
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

2009 July 22nd :Eclipse Solar Total :Example Earth the of radius - Moon the of distance S
Earth, the of surface the onshadow the of diameter the get To

Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

149. 149
Speed of the shadow in the total solar eclipse
•EverypointontheEarthhassameangularvelocitybutdifferent
linearvelocity.Ingenerallinearvelocityatanylatitudeisgivenby,
•v=2Rcos(latitude)/24km/hr.
•SincetheMoonorbitsaroundtheEarth,ithasalsolinearvelocity,
givenbyV
m=2(distance)/(27.324km/hr.
•SincetheEarthrotatesfromwesttoeastandtheMoonalso
orbitsaroundtheEarthfromwesttoeast.
•Thenwehavethespeedoftheshadow,V=V
m-v
•Example:TotalsolarEclipse:22
nd
July2009
•Forthiseclipsetheshadowismovingonanaveragealong24degreesnorthlatitudes.
•LinearvelocityoftheEarthat24deglatitude=[26400cos(24)]/24=1530km/hr.
•DistanceoftheMoonfromtheEarth=357509km
•LinearspeedoftheMoon=[2357509]/(27.324)=13428km/hr.
•Speedoftheshadow=3428–1530=1898km/hr.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

150. 150
Duration of totality other than central line
•Supposethat,
•D=Widthofthepathatcentralline
•d=Perpendiculardistanceoftheobserverfromcentralline
•C=Lengthpfthechordatd
•Thenwehave,
•Lengthofthechord=C=
•If,T=Durationoftotalityatthecentrallineand t=Durationoftotalityatthe
distanced
•Wehave.
•Example:Foracertaintotalsolareclipse22
d4D 22
d4D
D
T
D
C
Tt  minutes 3.2754250
250
min 4
t
line? central from kilometers 75 distance a attotality of duration the is What
km. 250D and minutes 4T
22


Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

151. 151
Length of the Earth’s shadow
•ThelunarEclipsedependsonthewaytheMoonenterstheEarth’sshadow.Manytimes
thepathoftheMoonmakescertainanglewiththeeclipticandtheMoontravelsonlya
partofUmbraoritonlypassesthroughPenumbra.Inthefirstcasewecanhaveeither
totalorpartialLunarEclipse.InthesecondcasewehavePenumbraLunarEclipse.
ProceduretofindthepathoftheMoonduringtheeclipseisbitcomplicated,leavingit
asidewecanestimatesomeimportantfactsofLunarEclipsefromEarth’sshadow.
•FormulaforthelengthoftheEarth’sshadowissameaslengthoftheMoon’sshadow.We
havetoreplacetheMoonbyEarthintheformula. 
)nearly.(k1387282







6378-696000
101.56378
L Then,
shadow sEarth' the of Length L distance, Earth - Sun D
km. 6378Earth the of Radius r km, 696000 Sun the of Radius R have, We
Earth Of RadiusSun of Radius
distance EarthSunEarth of Radius
Shadow sEarth' the of Length
8
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

152. 152
Length of the Earth’s shadow
•Total Lunar Eclipse: 21
st
December 2010 
night. the in 46:2 about at i.e.
08:46:230:5-8:16:8 is eclipse the of middle India For
hours 30:5by UT of ahead is Time Standard Indian
UT 8:16:8
as. given is eclipse middle the of Time
Moon the of distance the at found are rsThediamete
km. 16655
180
π3765041.26732
Penumbra of Diameter
Penumbra of Diameter
9390km.
180
π3765041.4290
180
πradiusAngle
UmbraDiameterof
376504km.themoondistanceofradius
umbradiameterofarc
1.4290deg.0.71452meterAngularDia
deg.
π
180
radius
arc
Angle
deg. 0.7145radius Angular
Umbra of Diameter
376504km.
0.97064sin
6378
moonDistanceof
0.7145deg.Umbra of radius Angular
deg. 1.2673 Penumbra of radius Angular
deg. 0.97064 Moon the ofParallax Horizontal Equatorial
given. are parameters Following










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









Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

153. 153
Diameter of the Earth’s shadow at the distance of Moon
–Bhaskaracharya’s method
•Bhaskarachaya,inhisSiddhantashiromonihasgivenacorrectmethodtofindthediameter
oftheEarth’sshadowatthedistanceofMoon.Thisvalueisessentialtodeterminethetime
intervalofthetotallunareclipse.TheBhaskarachaya’sformulaisasfollows,
 
 
astronomy. modern in equation the to similarexactly is equation This
P1S1Pr have, weThus
S1(say)Sun of radius angular(S/2)/s
P(say)Moon ofparallax horizontal(E/2)/m
rconeshadow of radius angulard/2m
E/2sS/2mE/2md/2m
2mby Dividing
E)/s-m(SEd
ES2EB2SA2SG But,
2SG)/SE(ED2EBSG)/SE(2EB2EBd
EF)2(EB (nearly) 2BF2CDd
orbit. lunar at coneshadow of diameter2CD Hence,
orbit lunar the indicates CD
SG/SEEDEFhence, EF/EDSG/SE
similar are EDF triangle and SEG Triangle
Moon of distancem Sun of distances
Earth of diameterE diameterS Where,
m/sE-S-Eddistance sMoon' at coneshadow the of Diameter






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





Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

154. 154
Diameter of the Earth’s shadow at the distance of Moon
–Bhaskaracharya’s method
•Total Lunar Eclipse: 21
st
December 2010 
 
km. 4695 (ππ/1800.7145376520
(PI/180)angleradius arc coneshadow the of Radius
arc/radiusangle using, distance into it Convert
deg. 0.7145 radius Angular
data, From
km. 4635 coneshadow of Radius
9270km 3530- 12800
km 10000012800)/147-2000376520(139 - 12800coneshadow the of Diameter
have, weformula sBhaskara' Using
)m.(assumed147100000kSunDistanceof
December, late in Earth the to nearvery is Sun the Since
376520km.0.97066400/sinMoon the of Distance
Moon of nce6400/dista0.9706sin
deg. 0.9706 Parallax Horizontal
0.7145deg. conesshadow the of Radius Angular












Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

155. 155
Total solar eclipse from Moon
•ThetotalsolareclipsemustbeafantasticviewfromtheMoon.Dependingonyourlocationon
theMoontheEarthwillappeartobealmostatthesamepositioninthelunarsky.Thesizeof
theEarthwillbealmostfourtomesthesizeoftheMoonasviewedfromtheEarth.TheEarth
isgoingtoeclipsethesun.Hencethedurationoftheeclipsewillbequitelong.OntheEarth
maximumdurationoftotalityisnevergreaterthan7.5minutes.Wecanestimatethelongest
durationoftotalityonmoon.
•ForlongesteclipsetheEarth-Moondistanceshouldbeminimumandtheeclipseshouldbe
central,i.e.theSunshouldpassalongtheequatorialdiameteroftheEarth.
•WeshallfirstestimatethemaximumangulardiameteroftheEarthasseenfromtheMoon.
•Wehave,
•Angle=(arc/radius)(180/)
•Here,arc=DiameteroftheEarth=12800km.
•Radius=MinimumdistancebetweenEarthandMoon=356000km.(nearly)
•AngulardiameterofEarth=(12800/356000)(180/)=2deg.(nearly)
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

156. 156
Total solar eclipse from Moon
•Inordertoestimatemaximumdurationoftotality,we
havefollowingdata,
AngulardiameterofSun=0.5deg.(nearly)
Sun’sangularspeedacrossthesky=360/29.53=12.2
deg/day
Sunmoves1degreein1/12.2=0.082day=2hours
(nearly)
•Weassumethateclipseiscentral.At1
st
and4
th
contactstheSin,sdisktouchesandleavestheEarth’s
diskrespectively.From1
st
and4
th
contactsSun
travelsanangulardistanceof0.25deg.+2deg.+
0.25deg.=2.5deg.
•Hencepartofeclipsewilllast22.5=5hours
•From2
nd
and3
rd
contactSun’sdiskiscompletely
behindtheEarth’sdisk.Inthispartoftheeclipsesun
travelstheangulardistanceof1.5degreesand
durationoftotalitywillbe,22.5=3hours.Thisis
longestdurationoftotality.
•IftheSunpassesaboveorbellowtheequatorial
diameteroftheEarth,thedurationoftotalitywillbe
lessthan3hours.
•TheSunrayswillpassthroughtheatmosphereandill
bedeviatedanddispersed.Henceaglowwillappear
aroundtheEarth’sdiskduringtotality.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

157. 157
Eclipses and Planets
Elementary Astronomical Calculations: Lecture-04
Thank You
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

158. 158
Elementary Astronomical Calculations:
Magnitudes Of The Stars ( Lecture –5 )
•By Sukalyan Bachhar
•Senior Curator
•National Museum of Science & Technology
•Ministry of Science & Technology
•Agargaon, Sher-E-Bangla Nagar, Dhaka-1207, Bangladesh
•Tel:[phone](Off), Contact: 01923522660;
•Websitw: www.nmst.gov.bd; Facebook: Buet Tutor
&
•Member of Bangladesh Astronomical Association
•Short Bio-Data:
First Class BUET Graduate In Mechanical Engineering [1993].
Master Of Science In Mechanical Engineering From BUET [1998].
Field Of Specialization Fluid Mechanics.
Field Of Personal Interest Astrophysics.
Field Of Real Life Activity Popularization Of Science & Technology From1995.
An Experienced TeacherOf Mathematics, Physics & Chemistry for O-,A-, IB-&
Undergraduate Level.
Habituated as Science Speaker for Science Popularization.
Experienced In Supervising For Multiple Scientific Or Research Projects.
17 th BCS Qualified In Cooperative Cadre (Stood First On That Cadre).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

159. 159
Albedo
•AmountofsunlightreceivedbyaplanetdependsonitsdistancefromtheSun.Whatever
sunlightaplanetreceives,apartofitisabsorbedandremainingpartisreflectedbythe
planet.Thepercentageofthesunlightreflectedbyaplanetdependsontheicecover,
clouds,natureofthesurface,presenceofoceansetc.Duetolargedarkareasverylittle
amountofsunlightisreflectedfromtheMoon.AsagainstthattheVenusiscompletely
coveredbythickclouds,hencealargepercentageofsunlightisreflectedfromVenus.
•ThefractionofincomingsunlightthataplanetreflectsiscalleditsAlbedo.Following
tablegivesthealbedooftheplanets.
Planet Albedo
Mercury 0.10
Venus 0.76
Earth 0.39
Moon 0.07
Mars 0.16
Jupiter 0.51
Saturn 0.61
Uranus 0.31
Neptune 0.35
Pluto 0.40
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

160. 160
Oblateness of the Planets
•Everyplanetrotatesaboutitsaxis.Therotationcreatescentrifugalforce,whichis
proportionaltothedistanceofaparticlefromtheaxisofrotationandangularvelocityofthe
planet.Hencemagnitudeofthecentrifugalforceismaximumontheequatoranddecreases
towardsthepoles.Onthepolescentrifugalforceiszero.Duetounevenmagnitudeuneven
magnitudeofthecentrifugalforcetheplanetgetsflattened.Itsequatorialdiameterisusually
greaterthanitspolardiameter.Theflatteningofaplanetiscalledoblateness.Fromthe
followingequationoblatenesscanbeestimated.
•Foraperfectsphereofoblateneesiszero.FastrotatingplanetslikeJupiterandSaturn
oblatenessisconsiderable.Followingtablegivestheoblatenessoftheplanets.
Planet Equatorial Polar Oblateness
Diameter Diameter
(km.) (km.)
Mercury 4879 4879 0.0
Venus 12102 12102 0.0
Earth 12756 12712 0.00345
Mars 6792 6752 0.00589
Jupiter 142984 133708 0.0648
Saturn 120536 120536 0.098
Uranus 51118 49946 0.023
Neptune 49528 48682 0.017
Pluto 2390 2390 0.0
Moon 3476 3472 0.00115Diameter Equatorial
Diameter Polar Diameter Equatorial
Oblateness


Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

161. 161
Orbital velocities of the planets

(nearly)km/sec 30 km/sec 29.86 V
sec. 360024 365.25
km. 101.52π
Earth of Velocity Orbital
seconds 360024 365.25 days 365.25 timePeriodic T
km. 101.5 Sun the from distance AverageR
Earth, For
Earth. the ofvelocity orbital the find first us Let
2
T
T
R
R
VV
T
T
R
R
V
V


T
R2π
V Earth of Velocity Orbital

T
R2π
VPlanet of Velocity Orbital
derieved. be can formula Following found. be can planets other of velocities
Orbital ,comparisonby then and Earth, the ofvelocity orbital the calculate First
1
T
2ππ
Velocity Orbital
TimePeriodic
orbit the of nceCircumfere
Velocity Orbital
follows, as is planet a ofvelocity orbital average find to formula simple
very A velocity. orbital least has Pluto planet farthest the and highest hasMercury planet nearest
Thus Sun. the from planet the of distance the to alproportioninversely isvelocity Orbital
8
8























Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

162. 162
Orbital velocities of the planetsplanets. all of velocities angular average gives table following The
(nearly) km/sec. 24 24.32
1.88
1
1.52430 Mars ofvelocity Orbital
1.88
1
T
T
and 1.524
R
R
Mars Gor
Mars. ofvelocity
orbital the find us Let formula. second from found be can planets the of velocities orbital
the Earth, the withplanets the of ratios timeperiodic and distance the Knowing






Planet Distance Ratio Periodic Tine Ratio Orbital Velocity
(R’/R) (T/T’) ( km/sec)
Mercury 0.387 4.15 48
Venus 0.723 1.625 35
Earth 1.000 1.000 30
Mars 1.524 1/1.88 24
Jupiter 5.204 1/11.9 13
Saturn 9.582 1/29.5 9.7
Uranus 19.201 1/84 6.8
Neptune 30.047 1/164.8 5.47
Pluto 39.236 1/247.77 4.75
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

163. 163
Elongation of the Earth for outer planets
•InthecaseofEarthelongationisananglebetweentheSun,EarthandPlanet.Asseen
fromtheearth,mercuryandVenusareinnerplanets,theirmaximumelongationsareabout
28and47degreesrespectively.Hencetheseplanetscanbeobservedthroughoutthenight.
ForallplanetsfromMarsonwards,theearthisaninnerplanet.Hencefromtheseplanets
theEarthcanonlybeobservedbeforesunriseoraftersunset.Butiftheangleofelongation
oftheEarthislessthanabout10degrees,theEarthcanneverbeobserved,becauseit
cannotbeseenintheglareofSun.
•Wedefinetheangleofelongationasfollows,Elongationwithreferencetoaplanet
•Marscompletesonerotationin24.6hours,whichmeans360degreesarecoveredin
•24.6hours.Hencetocover37.59degrees2.57hoursarerequired.Inotherwordsinthe
skyofMars,theearthcanbeseenabout2.5hpursbeforesunriseoraftersunset..deg59.37
.deg
180
105.1524.1
105.1
ofEarthElongation
.km105.1524.1cetanMarsDisSun
.km105.1cetanEarthDisSun
fromMars
thasviewedonoftheEarheelongatiLetusfindt
180
8
8
8
8













ancePlanetDistSun
nceEarthDistaSun
Angle
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

164. 164
Elongation of the Earth for outer planets
•KnowingtheaveragedistancesoftheplanetsfromtheSun,theelongationsoftheEarth
fortheouterplanetscanbecalculated.Theyarecompiledinthefollowingtable.
•Itisveryclearfromthetable;theEarthcanneverbeseenfromSaturn,Uranus,Neptune
andPluto,sinceitwillbealwaysveryclosetotheSun.EvenfromJupiter,theEarthcan
bejustseenforaverylittlespanoftime.
Planet Distance of the Planet Elongation of Earth
(AU.) (Deg.)
Mars 1.524 37.59
Jupiter 5.2 11.00
Saturn 9.54 06.00
Uranus 19.18 03.00
Neptune 30.06 01.90
Pluto 39.44 01.45
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

165. 165
Titus –Bode Law
•In1772GermanAstronomerYohanTitusfoundaverysimpleruletofindthedistancesof
theplanets,fromtheSun,intermsofaveragedistancebetweentheSunandtheEarth.
•Afterfewyears,anotherGermanAstronomerYohanBode,popularizedtherule.Since
thentheruleisknownas‘Titus–BodeLaw’.Thelawcanberepresentedbythefollowing
formula.
•Thevaluesof‘n’fortheplanetsareasfollows,
•Mercury:n=-infinityVenus:n=0Earth:n=1Mars:n=2Asteroids:n=
3Jupiter:n=4 Saturn:n=5Uranus:n=6Neptune:n=7 Pluto:
n=8
•Substitutingthevaluesintheabovelaw,wecanfindtheaveragedistancesoftheplanets
fromtheSun.Comparisonoftheactualdistancesoftheplanetsandthoseobtained.
•
•fromtheTitus–BodeLawareasfollows,
•ItisclearfromthetablethatNeptuneandPlutodonotobeytheTitus–BodeLaw,butto
rememberthedistancesoftheplanetsfromtheSunthelawisuseful.
•UsingthislawWilliamHershelfoundtheplanetUranus.
•AsteroidbeltisbetweenMarsandJupiter.Thelawgivestheaveragedistanceofthebelt
fromtheSun.
•Thereisnomathematicalproofforthelawn
20.3 0.4 Sun the from Planet a of Distance 
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

166. 166
Titus –Bode Law
•
Planet Actual Distance Distance by Titus-Bode law
(AU) (AU)
Mercury 0.39 0.4
Venus 0.72 0.7
Earth 1.00 1.00
Mars 1.52 1.6
Asteroids 2.8 2.8
Jupiter 5.2 5.2
Saturn 9.55 10.00
Uranus 19.2 19.6
Neptune 30.1 38.8
Pluto 39.5 77.2
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

167. 167
The Earth day of Venus
•ForVenustheEarthisanouterplanet.HenceEarthwillriseandsetintheskyofVenus.ButEarthwill
riseinwestandwillsetintheeast.DuetoverythickcloudyatmosphereofVenus,theEarthcannotbe
seenfromthesurface.EarthwillbeseenasquiteabrightobjectfromoutsidethecloudcoverofVenus.
Earth’srise,setandrise,onVenuscanbecalledtheEarthdayofVenus.Wewanttofindthelengthof
thisday.
•PeriodictimeofrotationoftheVenusandperiodictimeofrevolutionofEartharecomparable.Venus
completesonerotationfromeasttowestin243.1days,whiletheEarthcompletesonerevolutionaround
theSunin365.25days.InordertofindtheEarthdayofVenuswehavetofindrelativeangularspeedsof
EarthandVenus.Thisgivenbyfollowingrelation.
•Relativeangularspeed=angularspeedofVenus–AngularspeedoftheEarth 
days. Earth four be
willthere timesynodic one of interval the during that suggests This 586).146(4
day earth the times 4 is whichdays, 584 is Venusof timesynodic the thatknow We*
Venus.ofsky the in seen be not willEarth the days 73 next For days. 73146/2
aftereast, the in sets and westthe in rises Earth the that indicates sign negative The
(nearly) days 146days 145.95
0.006851
1
Venusofday Earth
0.0068510.00273780.004113
365.25
1
-
243.1
1
Day Earth
1
(Nearly) days. 365.25 Earth of yearSidereal
west)to east from is (Rotation 243.1days. Venusofday Sidereal
Earth of yearSidereal
2π
-
Venusofday Sidereal
2π
Day Earth
2π







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

168. 168
Length of a day on Mercury and Venus
•Iftherotationperiodofaplanetiscomparabletoitsperiodofrevolution,thedayand
nightcycleonsuchplanetbecomesunusual.ThistypeofsituationexistsonMercury
andVenus.ThelengththedayonMercuryis176Earthdays,whileonVenusitis
about117Earthdays.InthecaseofEarthandallotherplanetstheperiodof
revolutionsareverylargecomparedtotheirperiodofrotations,hencethedaylengths
ontheseplanetsonlydependontheirperiodofrotations.Forexample,theEarth
completesonerotationin24hours,whileitsperiodofrevolutionis365.25daysthatis
whywehave12hoursofdayand12hoursofnight,approximately.
•WecanfindthelengthofadayonMercuryandVenus,fromtheirrelativeangular
velocitiesofrotationandrevolution.Ingeneralwehave, Venus.andMercury onday a of length the find can weformula this Using
T2
1
-
T1
1
T
1

revolution of Period T2 rotation of Period T1day the of Length T Where,
T2
2π
-
T1
2π
T
2π

revolution ofvelocity Angular-rotation ofvelocity Angularvelocity angular Relative




Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

169. 169
Length of a day on Mercury and Venus
* 
starts. night of days 58.5 Then days. Earth 58.5nearly
after east the in sets and westthe in rises Sun the that indicates sign negative The
(nearly) days 117 days 116.74
0.008565
1
Venusonday of Length
-0.0085650.0044500.004115
224.698
1
-
243.1
1
day of Length
1

have, wevalues, these ngSubstituti
days 224.698 T2 and days 243.1- T1 Venus,For
days. Earth 88 forMercury ofsky
in be willSun days. Earth 88 of night and days Earth 88 ofday is thereMercury on Thus
(nearly) days 176 days 175.97
0.005682
1
day of Length
0.0056820.011360.01705
87.97
1
-
58.65
1
day of Length
1

have, wevalues, these ngSubstituti
days 87.97 T2 and days 58.65 T1 Mercury, For






: Venuson day the of Length
:Mercury on day the of Length
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

170. 170
Rising and setting of the satellites of Mars
•TwosatellitesrevolvearoundMars,thenearestiscalledPhobosandthefarthestiscalledDeimos.
ThesesatellitesareverysmallcomparedtoMars.DiameterogPhobosisabout24kilometersand
thatofDeimosiaonlyabout12kilometers,whilediameterofMarsisabout6800kilometers.
•PhoboscompletesonerevolutionaroundMarsisonly7.65hours,whileDeimosrequiresabout
30.3hourstocompletesonerevolution.Butitdoesnotmeanthattherisingandsettingofthese
satellitesrequirehalfthetimeoftheirrevolutions.Sincetherevolutiontimesofthesesatellitesare
comparabletotheperiodictimeofrotationofMars,itisinstructivetofindthetimeintervals
betweenrisingandsettingofthesesatellitesintheMartiansky.
•WehavetouserelativeangularspeedsofrotationofMarsandrevolutiontimesofthesatellites.satellite. the of rotation of timePeriodic T2
T2
1
-
T1
1

T
1

Mars of rotation of timePeriodic T1
Mars to respect day with Satellite T Where,
/T2 Wspeed angular But

T2
2
-
T1
2

T
2

W2W1- W
satellite the of revolution of speed Angular W2
Mars of rotation of speed Angular W1speed angular relative WLet,
satellite the of rotation of speed Angular- Mars of rotation of speed Angular
speed angular Related
have, We













Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

171. 171
Rising and setting of the satellites of Mars 
 
 
 
hours. 66nearly after westthe in sets and east the in rises Deimos
whilehours, 5.5 after east the in sets and westthe in rises Phobossky Martian the In
hours. 66nearly after westthe in sets and east the in rises Deimos Thus

0.007548
1
Deimos
0.0075480.0330.04055
30.3
1
-
24.66
1
day Deimos
1

Deimos ofday one of interval time T
hours 30.3 T2
Mars of revolution of Period hours 24.66 T1 :Deimos For
hours. 5.5 about after east the
in sets andsky Martian the the of westthe in rises Phobos that indicates sign Negative
Phobos. ofday one of interval time T
nearly hours -11.1
0.090
1
Phobos ofday One
0.0900.130720.04055
7.65
1
-
24.66
1
day Phobos
1

hours 7.65 T2
Mars of revolution of Period hours 24.66 T1 :Phobos For
nearlyhrs132hours48.132Onedayof 









:Deimos on day One
:Phobos on day One
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

172. 172
The angular size of the Sun and Charon from Pluto
•EventhroughPluroisnowremovedfromtheplanetarylistoftheSolarSystem,itisinterestingtofind
outangularsizesofsolarandCharondisksfroPluto.CharonisasatelliteofPluto.Sincethe
eccentricityofthePluto’sorbitiscamparativelylargerthanotherplanetsofthesolarsystem,its
maximumandminimumdistancesfromtheSunwillappeartochangeappreiably.Wehavethe
followingdataaboutorbitofPluto.
•Averagedistancefromthesun=6,000,000,000km.
•Maximumdistancefromthesun=7,400,000,000km.
•Minimumdistancefromthesun=4,400,000,000km.
•DiameterofPluto=2400km.
•Diameterofcharon=1200km.
•Pluto-Charondistance=19,600km.
•DiameteroftheSun=1,392,000km.
•TodeterminetheangularsizesofthedisksofSunandZharonfromPluto,weusetherelationofan
angle.
•
•
•min.arc 0.6
π
180
7400000000
1392000
Pluto from Sun of diameter Minimum
min.arc 1
π
180
4400000000
1392000
Pluto from Sun of diameter Maximum
min.arc 0.78
π
180
6000000000
1392000
Pluto from Sun of diameter Average
answers. following get willWe
Pluto from Charon or Sun of Distance Radius
Charon or Sun of Diameter Arc Here,
deg.
π
180
radius
Arc
Angle






.deg
.deg
.deg
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

173. 173
The angular size of the Sun and Charon from Pluto
•IfweassumethattheaveragediameterofSun’sdiskfromtheEarthtobe32arc
minute,asviewedfromPlutotheaverage,maximumandminimumsizeofthesun’sdisk
willbe,1/38
th
,32
nd
and1/50
th
respectively,oftheEarth’sdisk.
•LetusnowfindthesizeofCharonfromPluto,wehave,
•Earth. the from viewed disk Lunar than larger times 14 be to appear willdisk sPluto, Charon From
deg. 7
π
180
19600
2400
Charono from Pluto of size Angular
have, weCharon, from viewed as Pluto of size find to einstructiv also is It
Earth. the from observed disk Lunar than larger times 7 be willdisk sCharon' means Which
deg. 3.5
π
180
19600
1200
Pluto from Charon of size Angular


.deg
.deg
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

174. 174
Resonance revolutions of the planets
•TheplanetsrevolvearoundtheSunindifferentintervalsoftime.Theperiodictimeofa
planetdependsonitsdistancefromthesun.Ifweexaminetheperiodictimesofthe
planets,itappearsthattheyhavesomerelationamongsteachother.Thisiscalled
‘ResonanceRevolution’oftheplanets.Followingtablegivestheperiodictimesofthe
planets.
4revolutionsofMercury=40.2144=0.9649years
1revolutionofEarth=1.000years
Difference=0.0356years
5revolutionsofMercury=50.2144=1.2055years
2revolutionsofVenus=20.6156=1.2312years
Difference=0.0257years
Planet Periodic time Planet Periodic time
(years) (years)
Mercury 0.2411 Saturn 29.46
Venus 0.6155 Uranus 84
Earth 1.000 Neptune 164.78
Mars 1.88 Pluto 248
Jupiter 11.86
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

175. 175
Resonance revolutions of the planets
3 revolutions of Venus = 3 0.6156 =1.8468 years
1 revolution of Mars = 1.88 years
Difference = 0.0332 years
3 revolutions of Venus = 3 0.6156 =1.8468 years
1 revolution of Mars = 1.88 years
Difference = 0.0332 years
6 revolutions of Mars = 6 1.88 =11.28 years
1 revolution of Jupiter = 11.86 years
Difference = 0.42 years
5 revolutions of Jupiter = 5 11.86 = 59.3 years
2 revolutions of Saturn = 229.46 = 58.92 years
Difference = 0.38 years
2 revolutions of Uranus = 2 84 = 168 years
1 revolution of Neptune = 164.78 years
Difference = 3.22 years
3 revolutions of Neptune = 3 164.78 = 494.36 years
2 revolutions of Pluto = 2248 = 496 years
Difference = 1.66 years
Compared to total numbers of years involved, the difference are very small, hence the planets can be
assumed to be revolving in resonance around the Sun.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

176. 176
The Roche limit
•ItisnowclearthatJupiter,Saturn,Uranus,andNeptuneallhaverings.Thequestion
offormationofringswassolvedbyEduardRoche.Hesuggestedthat,ifasatellite
comeswithinacertaindistancefromaplanet,duetotidalforcesitbreaksintopieces.
Thepiecesthusformaringaroundtheplanet,Aspectacularexhibitionofthis
phenomenonwasobservedin1994.Twentyonepiecesofthebroken‘Shoemake-
Levy-9’cometcolidedwithJupiterfrom16
th
to22
nd
July1994.
•Thedistancefromaplanetatwhichasatelliteorcometbreak,iscalled‘RocheLimit’.
WithfewsimpleassumptionsandmathematicswecanderiveaformulafortheRoche
Limit.
•Imagineaplanetofmass‘M’andradius‘R’.Assumethattherearetwosmallspheres
AandB,mass‘m’andradius‘r’.Alsoweassumethatthespheresareincontact,so
thatthedistancebetweentheircentersis‘2r’.Letthesphere‘A’beatadistance‘D’
fromtheplanet,whichisverylargecomparedto‘r’.Thedistanceofsphere‘B’fromthe
planetwillbe‘D+2r’.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

177. 177
The Roche limit
 
 

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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

178. 178
The Roche limit


R5.2D
R



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

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


d dp Let
separated. be willsphere the
d
dp
2 D if, or
D
d
dp
R2
r
dπr
3
4
D
dpπR
3
4
2
have, wevalues, these ngSubstituti
d planet ofdensity πr
3
4
m
dp planet ofdensity πR
3
4
M But,

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m
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separated. be willspheres the spheres
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

179. 179
The Roche limit
•Thusifasatelliteorcometcomeswithin2.5timestheradiusoftheplanetitbreaksdueto
tidalforces.
•TheRocheLimitofthesolarsystemaregiveninthefollowingtable.
Planet Radius (km) Roche limit (km)
Mercury 2450 6125
Venus 6050 15125
Earth 6400 16000
Mars 3410 8525
Jupiter 71680 179200
Saturn 60480 151200
Uranus 25665 64162
Neptune 24830 62075
Pluto 1200 3000
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

180. 180
Temperature of a planet
•Everyplanetreceivesenergyfromthesun.Theamountofenergyreceivedisinversely
proportionaltothesquareofthedistanceoftheplanetfromtheSun.Aplanetabsorbs
energyanditssurfacegetsheated.Theplanetthenradiatestheenergybacktospace.
Iftheenergyreceivedbytheplanetisequaltotheenergyemitted,itstemperature
remainssteady.
•Inordertofindthesurfacetemperatureofaplanetwehavetousethesolarluminosity.
Thisisgivenby,
•Solarluminosity=Ls=410
26
Jules/sec.
•Theamountofenergyreceivedperunitareaperunittimeatadistance‘r’fromtheSun
iscalledsolarflux,itsexpressionis,
•4r
2
istheareaofasphereofradius‘r’
•SolarFlux=Ls/(4r
2
)
•Thesolarfluxencountersthediskofaplanetofradius‘R’.
•Energyreceivedbyaplanet=R
2
SolarFlux=R
2
Ls/(4r
2
)
•Energyreceivedbyaplanet={R
2
/(4r
2
)}Ls
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

181. 181
Temperature of a planet
•Treatingplanetasablackbody,theenergyemittedbytheplanetisgivenby,
•Energyemittedbytheplanet=SurfaceareaStefan’sconstant4
th
Powerofabsolute
temperature
•=4R
2
(5.6710
-8
)T
4
•Ifthetemperatureoftheplanetissteady
•Energyemitted=energyreceived
•4R
2
(5.6710
-8
)T
4
=R
2
/(4r
2
)}Ls
•desired. is effect this extent certain Upto Effect'. House Green' the
of because is increase The celsius. deg 15 is Earth the of etemperatur average Actual
C8K281
105.67π
104
101.4962
1
Earth the of etemperatur Surface
Joule/sec.104Ls & m101.496rdistance Earth, For
1/4
8
105.67π
Ls
r2
1
Tplanet the of eTemperatur
00
1/4
8
26
11
2611
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

182. 182
Temperature of a planet
*
•desired. is effect this extent certain Upto Effect'. House Green' the
of because is increase The celsius. deg 15 is Earth the of etemperatur average Actual
(nearly) CK
105.67π
104
101.4969.5832
1
Earth the of etemperatur Surface
Joule/sec.104Ls & m101.496rdistance Saturn, For
00
1/4
8
26
11
2611
18390
583.9

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Saturn. the of etemperatur surface the Find :2 Example
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

183. 183
Oppositions of Mars
•OppositionofMarsisquiteaninterestingphenomenon.AtthetimeofoppositionMarsis
beyondtheEarth,onSun-EarthlineanditisveryclosetotheEarth.DuringthistimeMars
canbeseenasaverybrightreddishobjectinthenightsky.SincetheorbitofMarsis
elliptical,thedistancebetweentheearthandMarsattimeofoppositionvaries,butvery
rarelythisdistanceisshortest.Forexample,on27
th
August2003,therewasanopposition
ofMars.OnthatdayMarscameveryclosetotheEarth,afterabout56,000years.Stillits
distancewasabout55.758millionkilometersfromtheEarthanditsapparentsizewas
25.13arcseconds.OnthatdayMarswasatitsperihelionandtheEarthwasnotveryfar
awayfromitsaphelion.
•Ifwestartwithcertainoriginalopposition,wecanexaminehowsun-Earth-Marslineshifts
andafterhowmanydaysbeforeoraftertheoppositiontakesplacewithreferenceto
acceptedoriginalopposition.Tosimplifythecalculationsweshallacceptthelengthofthe
yeartobe365.25daysandMartianyeartobe687days.Foraccuratecalculationssidereal
yearisrequired.Ayearalsorepresentsthethetimeofrevolutionofaplanetaroundthe
Sun.
•Exactsynodictimeofmarsis779.84days.Fromcertainopposition,nextoppositionwill
takeplaceafter779.84/365.25=2.13536years.DuringthistimetheEarthcompletes2
revolutionsandadvancesthroughanangleof0.13536360=48.73degrees,fromthe
originalopposition.SinceaveragespeedoftheEarthis360/365.25=0.9856deg/day,48.73
degreesareequivalentto,48.73/0.9856=49.6days.Thusfromoriginaloppositionnext
oppositionwilltakeplace48.73degreeaheadandafter49.4days.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

184. 184
Oppositions of Mars
•Itcanbeshownthat,7,15,22,37,133,170,and303oppositionsofMarsarealmost
intergralmultiplesoftheEarthandMarsrevolutionsoryears.Consider7oppositions,
•7779.94=5459.58days
•5449.58days=(5459.58/365.25)=14.9515years(revolutions)(nearly)
•5459.58days=(5459.58/687)=7.958Martianyearsorrevolutions(nearly)
•After7oppositionsSun-Earth-Marslinewilladvancethrough,
•748.73=34.11341degrees(nearly).
•Thismeansitwillbeabout,360-341=19degreesbehindtheoriginaloppositionand7
th
oppositionwilltakeplaceabout19daysearlier.
•Followingtablegivestheresultsofallabovementionedoppositions.Fromthetableitwillbe
clearthatafterfirstoriginalacceptedopposition,about647Earthyearstheoppositionof
marswilltakeplacealmostatthesameposition.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

185. 185
Oppositions of Mars
•After79,284,and363orafter37,133and170oppositionsrespectivelytheoppositions
arealmostneartheoriginalopposition.
•+vesignmeansaheadand–vesignmeansbehind.
No. of Earth Martian Sun-Earth-Mars line Days after or
Oppositions (years) (years) after or before original before original
(days) (days)
001 002 001 +48.70 +49.40
007 015 008 -19.00 -19.30
015 032 017 +10.70 +10.80
022 047 025 -08.30 -08.50
037 079 042 +02.35 +02.38
096 205 109 -03.60 -03.70
133 284 151 -01.28 -01.30
170 363 193 +01.00 +01.08
303 647 344 -00.23 -00.23
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

186. 186
Oppositions of Mars
•Minimumdistancesofoppositions(Minimumdistanceslessthan56millionkilometersare
onlygiveninthetable.
Date Earth-mars distance Apparent Mars diameter
(Million km. (arc sec)
25 August 1719 55.951 25.05
13 August 1845 55.839 25.10
18 August 1845 55.803 25.11
22 August 1924 55.777 25.12
27 August 2003 55.758 25.13
15 August 2050 55.957 25.04
30 August 2082 55.884 25.08
19 August 2129 55.842 25.10
24 August 2208 55.769 25.13
28 August 2287 55.688 25.16
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

187. 187
Visibility of rings of Saturn from Saturn
•RingsystemofSaturnisitscharacteristics.Othergasgiantshavealsoringsbutthey
arenotprominentasSaturn.RingsofSaturnarealsovisiblethroughanordinary
telescope.Actuallytherearethreeprominentrings,theyarenamedasA,B,andC.A
isoutermost,BisinmiddleandCistheinnermostring.TheringDisinsidetheringC,
butitisverydim.SimilarlytherearethreemoreringsbeyondA,buttheyarealsovery
dim.Wewanttoinvestigatevisibilityoftherings,ifsomeonestandsonSaturnitself.
•FromthenorthandsouthofSaturnnoringwillbevisible.Iftheobserverstandsonthe
higherlatitudesofSaturntheringswillnotalsobevisible.Wecanfindoutfromwhich
latitudevisibilityoftheringsstart.Roughlythedimensionsoftheringsaregiveninthe
followingtable.Thevaluesarenotexact.AlltheringsareinplaneofSaturn’sequator.
•
Ring Outer radius (km) Inner radius (km) Width (km)
A 136000 120000 16000
B 115000 90000 25000
C 88000 72000 16000
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

188. 188
Visibility of rings of Saturn from Saturn
•RadiusofSaturn=60000km
•LLatitude
•ForOutermosttipof‘A’ring.
•Iftheobserverstandson,64
•Cos(L)=(60000/136000)=0.441176
•L=arccos(0.441176)=63.82degs.64degs.
•DegreesnorthorsouthlatitudeofSaturn,thetipofthering‘A’willbevisible.Toobserveentire‘A’
ring,
•Fromlatitudeof60degreesnorthorsouthlatitudeofSaturn,entire‘A’ringwillbevisible.
•Cos(L)=(60000/120000)=½L=60
0
.
•Similarlywecanfindthevisibilityofrings‘B’and‘c’.
•Visibilityofring‘B’
• Outermosttipwillbevisiblefromlatitude58.55degrees.
• Innermosttipwillbevisiblefromlatitude48.2degrees.
•Visibilityofring‘B’
• Outermosttipwillbevisiblefromlatitude58.55degrees.
• Innermosttipwillbevisiblefromlatitude48.2degrees.
•Thismeans,anobserverhastostandat33.5degreesnorthorsouthlatitudetoobserverall‘A’,
‘B’and‘C’rings.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

189. 189
Resonance revolution of satellites of Jupiter
•Atpresent63satellitesofJupiterareknown.Io,Europa,Ganymede
andCallistohaveconsiderablediameters.Allothersareverysmall.
Allothersareverysmall.DiameterofGanymedeisevengreaterthan
planetMercury.Thestatisticsofthesefourplanetsisasfollows.
•CasuallookatorbitalperiodsofIo,EuropaandGanymedereveals
thattheyrevolvearoundJupiterinresonance.Ifweindicatetheorbital
periodsbyT1,T2andt3respectively,wehave,
•T2=2T1 (3.55=21.77nearly)
•T3=4T1=2T2(7.16=41.77and7.16=23.55nearly)
•Angularspeedsofthesesatellitesarealsorelated.
•Therelationshipisfoundtobeasfollows,
•n1–3n2+2n3=0(nearly)
•203.3898–3101.4084+250.2793=-0.2
•Ifexactvaluesoforbitalperiodsareconsideredtheanswercomesout
tobezero.
Satellite Orbital period Angular Speed
(days) (Angle/day)
Io 1.77 360/1.77 = 203.3898 (n1)
Europa 3.55 360/3.55 = 101.4084 (n2)
Ganymede 7.16 360/7.16 = 50.2793 (n3)
Satellite Diameter (km) distance (km) Orbital period (days)
Io 3630 422000 1.77(T1)
Europa 3138 671000 3.55(T2)
Ganymede 5262 1070000 7.16(T3)
Callisto 4800 1883000 16.69
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

190. 190
Rough estimate of rising and setting of the planets
•Knowingthelongitudeofplanetofaplanetandthesunonagivenday,wecanroughly
estimatethetimeofriseandsetoftheplanet.IntheIndianalmanacstheNiryanvaluesof
thelongitudesat5:30AMaregivenforeverydayoftheyear.Fromthelongitudesofthe
sunandtheplanet,wecanfindelongationoftheplanet.Fromtheelongationwecan
roughlyestimatetheriseandsettimingsoftheplanet.
•Theelongationofaplanetistheangulardistancebetweentheplanet,theEarthandthe
Sun.
•IfthelongitudeoftheplanetislessthanthelongitudeoftheSun,Theplanetisbehindthe
Sunandtheelongationistothe‘west’.Iftheelongationis‘west’,planetrisesbeforethe
sunriseandsetsbeforethesunset.
•IfthelongitudeoftheplanetisgreaterthanthelongitudeoftheSun,Theplanetisaheadof
theSunandtheelongationistothe‘east’.Iftheelongationis‘east’,planetrisesafterthe
sunriseandsetsafterthesunset.
• Westelongation=LongitudeoftheSun–Longitudeoftheplanet
• Eastelongation=Longitudeoftheplanet–LongitudeoftheSun
•Iftheelongationislessthanabout12degrees,theplanetwillbeintheglareoftheSun
andwillnotbeseenatall.
•WeknowthattheEarthrotatesthrough15degreesinonehour.Comparedtothe
rotationalmotionoftheEarthwecanneglectthemotionoftheplanets.
•WestElongation
•Timeofriseoftheplanet=elongation/15hoursbeforesunrise.
•EastElongation
•Timeofriseoftheplanet=elongation/15hoursaftersunrise.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

191. 191
Rough estimate of rising and setting of the planets
•Example:Findtherisingandsettingoftheplanetson5
th
December2008.
•Neglectingminutesandseconds,followingNiryanvaluesofthelongitudesoftheplanetsareobtained
fromtheIndianalmanac.
•1sign=30degrees
•Mercury:Elongation=234–229=05degrees‘east’.
•Sinceelongationislessthan12degreesMercurycannotbeseen.
•Venus:Elongation=272–229=43degrees‘east’.
• 43/15=2.866hrs=2hours52minutes
•Venuscannotbeseenthroughouttheday,butwillbeseenaftersunsetforabout2hours52minutes.
•Mars:Elongation=229–229=0degrees
•MarsisinconjunctionwiththeSunandwillnotbeseenatall.
•Jupiter:Elongation=269–229=40degrees‘east’.
• 40/15=2.66hrs=2hours40minutes
•Jupitercannotbeseenthroughouttheday,butwillbeseenaftersunsetforabout2hours40minutes.
•Saturn:Elongation=229–177=52degrees‘west’.
• 52/15=3.46hrs=3hours28minutes
• Saturnwillbeseenfor3hrs28minutesbeforesunrise.
Planet Sign Degrees Total Degrees
Sun 7 19 7 30 + 19 = 229
Mercury 7 24 7 30 + 24 = 234
Venus 8 02 9 30 + 02 = 272
Mars 7 19 7 30 + 19 = 229
Jupiter 8 29 8 30 + 29 = 269
Saturn 7 27 5 30 + 27 = 177
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

192. 192
Solar constants of the planets
•Solarconstantisamountofheatenergyinwattsreceivedbyonemetersurfaceofaplanet.
•InordertofindtheSolarconstants,itisnecessarytodecidethetotalamountofenergyemittedbythe
Suninalldirections.Thencalculatetheamountofenergyreceivedpersquaremeteratsurface,whose
radiusisthedistanceoftheplanetfromtheSun.Wehavefollowingrelations.
•Thetotalamountofheatenergy(E)emittedbytheSun,E
•=Stefan’sconstantSurfaceareaoftheSunFourthpowerofthesurfacetemperatureoftheSun.
•Where,Stefan’sconstant=5.6710
-8
Joule/m
2
-deg
4
.
•SurfaceareaoftheSun=4squareoftheradiusofSun=4(6.9610
8
m)
2
=6.08710
18
m
2
.
•SurfacetemperatureoftheSun=T=5800degKelvin.T=5800
4
=1.131610
15
.
•E=(5.6710
-8
)(6.08710
18
)(1.131610
15
)=3.910
26
watts.
•IncaseoftheEarthradiusofthesphere=1.510
11
met
•Areaofthesphere=4(1.510
11
m)
2
=2.82710
23
m
2
.
•SolarconstantofEarth=(3.910
26
)/(2.82710
23
)=1.37910
3
watts/sq.met.
Planet Distance (AU) Solar constant ( watt/m
2
)
Mercury 0.3871 3566
Venus 0.7233 2660
Earth 1.0000 1380
Mars 1.5237 595
Jupiter 5.2028 51
Saturn 9.5380 15
Uranus 19.191 3.8
Neptune 30.061 1.5
Pluto 39.529 0.9
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

193. 193
Distance to the horizon on the planets
•Ifyouarestandingatanelevationof‘h’metersaboveaflatsurfaceonaplanet.Youcan
estimatethedistancetothehorizon,providedtheradiusoftheplanetisknown.
•Thelineofsighttothehorizonisatangenttotheplanet.Alinewhichtouchesthesphereof
theplanetatjustonepointismarked‘B’inthedrawing.If‘O’isthecenterofthesphereof
theplanet,suchatangentisperpendiculartotheradius‘OB’.
•Inthefigure,
•Apositionoftheobserver
•helevationoftheobserver
•RRadiusoftheplanet
•ApplyingPythagorastheorem,wehave,
•(OA)
2
=(AB)
2
+(OB)
2
•(R+h)
2
=D
2
+R
2
•R
2
+2Rh+h
2
=D
2
+R
2
•D
2
=h(2R+h)
•Ifh<<2R,then,
•D
2
=2Rh
•D=(2Rh)
1/2
•Example1:ForEarthR=6400km,andLeth=1km.
• D=(264001)
1/2
=113kms.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

194. 194
Distance to the horizon on the planets
•Example 2:For Moon R = 1750 km, and Let h = 1.5 meter.
• D = (2 1750 1.5 10
-3
)
1/2
= 2.3 km.
•Example 3:Estimate the distance to the horizon for an astronot of height 1.5 meters, on
Mercury and Mars.
•1) For Mercury R = 2439 km
• D = (2 2439 1.5 10
-3
)
1/2
= 2.7 km.
•2) For Mars R = 3397 km
• D = (2 3397 1.5 10
-3
)
1/2
= 3.2 km.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

195. 195
The Magnitude scale
•InthesecondcenturyBC,Hipparchuscompliedacatalogueofaboutathousandstars.
Heclassifiedthestarsintosixcategoriesofbrightness,whicharenowcalled
magnitudes.Thebrightestappearingstarswereplacedbyhiminfirstmagnitude;the
fainteststarsthatthathecouldsee,wereofsixthmagnitude.Hipparchus’scalewasnot
continuous.
•In1856NormanPogsonproposedthescalethescaleofscaleofstellarmagnitudethat
isnowadopted.Henotedthatwereceiveabout100timesasmuchlightfromastarof
thefirstmagnitude,asfromoneofthesixthmagnitude.Adifferenceoffivemagnitudes
thereforecorrespondstoaratioinlightenergyof100to1.
•Thebrightnessofastarisequivalenttoenergyflux(f)perunitarea,perunittime,
receivedfromastar.Pogsonsuggestedthattheratiooflightfluxcorrespondingtoa
stepofonemagnitudebefifthrootof100.whichisabout2.512.thusafifthmagnitude
stargivesus2.512timesasmuchlightasoneofsixthmagnitudestar,andafourth
magnitudestar,2.512timesasmuchlightasfifthmagnitudestaror2.5122.512times
asmuchasasixthmagnitudestar.Fromthestarsthird,second,andfirstmagnitude,
receive2.512
3
,2.512
4
,2.512
5
(=100)timesasmuchlight,respectively,asfromasixth
magnitudestar.
•Iff1andf2areobservedenergyfluxesandm1&m2aremagniyudes,wehave,
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

196. 196
The Magnitude scale
•If f1 and f2 are observed energy fluxes and m1 & m2 are magnitudes, we have,
•Some visual magnitudes are compiled in the following table. 

 

 
 


 
 

 
1.9101010
f2
f1

.brightness in change relative find 7.8, to 7.1 from changes star a of magnitude Apparent
on. so and star magnitude second than brighter times 2.512 is star magnitudes first So
2.512100
f2
f1
1010
f2
f1
ratio.flux to scorrespond 6 to 1 of range Magnitude
f2
f1
2.5logm1m2 10
f2
f1
10100
f2
f1
0.28/2.57.17.8/2.5m1m2
52/2.516
/2.5m1m2/5m1m2
2/5m1m2













:Example
Object Magnitude
Sun -26.5
Full Moon -12.5
Venus (at brightest) -04.0
Jupiter, Mars (at brightest) -02.0
Sirius -01.5
Aldebaran, Altair +01.0
Naked eye limit +06.5
Binocular limit +10.0
15 cm. Telescope limit +13.0
Hubble Space Telescope +29.0
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

197. 197
Absolute magnitude scale
•Evenifallstarswereidentical,theywouldnotappeartohavesamebrightness,because
theyareatdifferentdistancesfromus.Theamountoflightthatwereceivefromastaris
inverselyproportionaltothesquareofitsdistance.SincetheSunisveryclosetousit
givesbilliontimesmuchmorelightthanotherstars.
•Onewaytocomparetheintrinsicratherthanapparentbrightnessofstars,istodetermine
whattheirmagnitudewouldbeiftheywereallatthesamedistancefromtheEarth.This
distanceisselectedas10parsec.Weknowthat1parsec=3.26lightyear.Hence10
parsec=32.6lightyears.Magnitudeofastarmeasuredat10parseciscalledits‘Absolute
Magnitude’.






















d1
d2
5logm1 - m2
d1
d2
2.5logm1 - m2
d
1
f
distance of Square
1
Flux Energy But
f2
f1
2.5logm1 - m2
relationbasic the from Starting
2
2
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

198. 198
Absolute magnitude scalestars. some of magnitudes absolute and apparent gives table Following
5. be willSun the of magnitude thepc 10 From
54.7731.572-26.86.63144
1
2062650
5log-26.8M
AU2062650 pc 10D AU,1d 26.8,mmagnitude apparentSun, For
Sun. the of magnitude absolute calculate us let example an As
d
D
5logmM
d
D
5logmM
D
d
5logM-m
parsec. 10Dd1 and dd2
magnitude absolute theMm1
star a of magnitude apparent themm2
Let,






























58.26
Star Apparent Magnitude Absolute Magnitude
Procyon 0.48 -3.0
Achernar 0.60 -0.9
Pollax 1.21 1.2
Alpha Centauri -0.01 4.4
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

199. 199
Absolute magnitude in terms of parallax    1.2673.733-52.153-5-1.580.430626-55-1.580.3715log5-1.58M
1.58- is magnitude visual
the and sec.arc 0.371 is whichofparallax Sirius, of magnitude absolute the Find
5logpmM
logp
p
1
loglogd
p
1
d But
5logd-5mM
5-5logd 5log10-5logd
10
d
5logM-m
have wevalues, these ngSubstituti
parsec 10 D star the of distance d Where,
D
d
5logM-m
have, wemagnitude, absolute the involving equationbasic the withstart us Let
star. a of magnitude absolute the find to fact this use can We
p
1
d
sec.arc inParallax
1
star a of distance
thatknow webecause distance, its gives indirectly star a ofParallax








:Example
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

200. 200
Absolute magnitude in terms of parallax 
 
 
 
 
years.light 8.733d
years.light 6 3.22.679parsec 2.679
3.732
10
10
10
10
10
10
10
d
1.4.M 1.46,m whichfor Sirius, of distance the Find
years.light 4.3d
years.light 6 3.21.3parsec 1.3
7.58
10
10
10
10
10
10
10
d
4.4.M 0.01,m whichfor Centauri, Alphaof distance the Find
10
10
d star a of distance
parsec. 10D But,
10
D
d star a of distance
10
D
d

5
mM
D
d
log mM
D
d
5log
D
d
5logM-m
equation.basic following the use can We
estimated.easily be can distance its kown is star a of magnitude absolute the If
0.5722.86/5/51.461.4
0.88054.41/5/50.014.4
/5mM
/5mM
/5mM

















:2 Example
:1 Example
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

201. 201
Bolometric magnitude
•Apparentmagnitudeindicatesthebrightnessofastarinvisiblerangeofthespectrum.But
allstarsalsoemitinvisiblelightlikeGammarays,Xrays,UltravioletandInfraredrays.
Evenwhiledeterminingtheabsolutemagnitude,theatmosphericeffectsoftheEarth
shouldbeconsidered.Manyluminousstarsappeardimbecausetheyemitmostofthe
lightininvisiblerange.Earth’satmosphereisopaquetoinvisiblelight.Henceitis
necessarytodefineanalternativedefinitionofthemagnitudeofthestars.Thenew
definitioniscalledbolometricmagnitude.
•BolometricMagnitude:IsanapparentmagnitudeofastarmeasuredaboveEarth’s
atmosphereandoverallwavelengthsoftheelectromagneticspectrum.
•Acorrectionisappliedtotheabsolutemagnitude,tofindtheBolometricmagnitude.Itis
calledBolometricCorrection(BC),Ingeneral,
•BolometricMagnitude=Mbol=M–BC
•BolometricCorrectionisverylargeforhottestandcooleststars.Thestars,having
temperature20000degreesKelvin,or3000degreesKelvin,valueofBC>3mag.
•GeneralformulaoftheBolometricmagnitudeisasfollows.
•Where,LLuminosityofthestarandLsLuminosityoftheSun






Ls
L
log5.272.4Mbol
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

202. 202
Bolometric magnitude
•Example: Find the Bolometric magnitude of Sirius and its luminosity.
•Given, m = -1.46, d = 2.7 pc, T = 10000 deg K, and BC = 0.6 mag.
•We have,
Sun. the rhan luminous times 40 about is Sirius Thus
Ls. 40 L 4010
Ls
L
1.6
2.5
0.84.72
Ls
L
log
Ls
L
2.5log4.72Mbol
0.80.61.4BCMMbol
1.40.275log1.46
10
d
5logmM
1.6
























Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

203. 203
Estimate of magnitudes of planets
•Visualmagnitudegivesanideaofapparentbrightnessofacelestialobject.Lightreceived
byaplanetfromtheSunispartiallyabsorbedandpartiallyreflected.Percentageoflight
energyreflecteddependsontheAlbedooftheplanet.Usingcomparisonoffluxreceived
fromthesunandaplanet,apparentmagnitudeofaplanetcanbefound,becausetheflux
ratioisrelatedtomagnitudedifference.Ourbasicrelationis,2
2
2
2
2
2
2
2
2
2
22
d4
1
Ar
4R1
Ls
fpEarth on received Planet fromFlux
Earth. and planet the between distance the is d'' and direction all in reflectedflux the If
Ar
4R1
Ls
Earth towards planetby reflectedFlux planet; the of Albedo AIf
r
4R1
Ls
πr
4ππR
Ls
planet theby dintercepteFlux
planet the of radius r Whereπr area of planet a of disk theonly seesflux Solar
Sun from planet of distance averageR1 Earth and Sun between distance averageR
4ππR
Ls
fp Planetby receivedFlux
4ππ
Ls
fe Earthby receivedFlux
have, weThen ; Sun of Luminosity Ls
Sun from planetby receivedflux fpf2 Sun from Earthby receivedflux fef1
Sun the of magnitude apparent msm2 planet a of magnitude apparent mpm1 Let,
f2
f1
2.5logm2m1-

















;
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

204. 204
Estimate of magnitudes of planets
*Thus we have to replace f2 by fp in the original formula  
 
 
  18.15538.1172.266152.4log5.272.26
104.635.0
10
122
16


















































 







2
22
8
122
22222
2
62
22
11
22
22
2
22
2
3.8444
2.5logmsme
0.35 Earth of Albedo A meters., 6400000 Earth of radius r and
meters., 384400000 Moon and Earth between distance d formula, above in R1R
same.nearly are sun from moon of distance and Earth the of distance have,
WeEarth. theby replaced be to is Planet and Moonby replaced be to is Earth case this In
Moon. from seen as Earth of magnitude apparent the is What:2 Example
4.6622.0626.728.82422.526.72106.672.5log26.72mp
106.050.84
101.50.6820.74
2.5log26.72
106.051AU0.84
0.682AU AU0.74
2.5logmsmp
Sun of magnitude -26.72ms
Venus)of (radius meters 6050000r 0.84 Venusof Albedo AU1R
elongation maximum at be to assumed is Venus AU0.682 d
meters. 101.5 AU AU0.7R1
venus. of magnitude apparent the estimate 1: Example

Earth. the from seen as planet a of magnitude apparent the estimate to formula general a is This
rAR
d4R1
2.5log
ALsr
4ππ4R1
4ππ
Ls
2.5log
f2
f1
2.5logm2m1-
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

205. 205
Radius of event horizon
•Astarisinequilibriumundertwoforces.Aforceduetoenergycreatedbynuclearfusionactsreadily
outward,tryingtoexpandthestar,whileforceduetogravityactsreadilyinward.Thisforcetriestocontract
thestar.Induecourseoftimethenuclearfusionreactioninsidethestarexhausted.Inthatcasegravity
takesover.Iftheoriginalmassofthestarismorethanthreetimethemassofthesun,thegravitational
forcecanshrinkthestartoapoint.Theendproductiscalled‘Blackhole’.Thegravitationalforceofblack
holeissoenormousthateverythingsurroundingitgetssucked.Notevenlightescapesfromtheblack
hole.Thelimitingradiusofaspherefromwhichlightcannotescapeiscalled‘Eventhorizon’.Henceatthe
surfaceoftheeventhorizontheescapevelocitymustbeequaltothevelocityoflight.Knowingmassofthe
starwecanfindtheradiusoftheeventhorizon.
•Forblackhole,(nearly) 30kmmeter1029.6R
Sun, of mass times 10 star a For
(nearly) 3kmmeter 102.96102101.485M101.485R
hole) black become never will(Sun Sun, For
M101.485R values, these ngSubstituti
Star the of MassM units. mks 106.67G
c
2GM
Rhorizon event of Radius g,Rearrangin
R
2GM
clight ofVelocity cityEscapeVelo
3
33027-27-
27-
11
2







Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

206. 206
Parallax
•Parallaxisausefulconceptwhichallowsustoestimatethedistanceofacelestialobjects.When
astarisviewedfromtwowidelydifferentpositions,thelocationofthestarappearstochangewith
referencetothebackgroundstars.Thisphenomenoniscalledparallax.Itisdefinedasfollows,
•Parallaxisarelativechangeinpositionofastarwithreferencetothetothebackgroundstars.
•Letusfindgeneralformulaforparallax.
•LetABbethebaseline.Itcanbediameteroftheearthortwicethedistancebetweenthesunand
earth.SincethestarsarequitefarawayfromtheEarthorSun,ABcanbetreatedasanarcofa
circle,theradiusofwhichisthedistanceofthestar.
•Let‘p’betheanglemadebythelineofsightsofthestarfromthepointsAandB.
•Fromthedefinitionoangle,wehave,radian
r
R
p
have, wevalues these ngSubstituti
rstar of distanceradius
distance Earth)(SunRarc
p,parallaxAngle
Here,
radius
arc
angle





Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

207. 207
Parallax
*
 
 
 
parsec
p
1
r
by given thus is parsec, in starany of distance
short inpc parsec 1 as defined is distance this AU206265 is star the of distance thesec arc 1p If
arcsec.1
AU
206265r
have, weformula, above the in values these ngsubstituti arcsec. 1 to equal be p'' Let
2AU)(AB km101.51AU
that,know Weshort). in (AU Unit alastronomic called is distance
average Earth.This the and Sun the between distance the be ABline base the let
parsec'.in' measured is star the of distance the secondsarc in measured isparallax of value the When
arcsec. p
R
20625
p/3600
360
2π
R
r
sec) 6060(1degsec
3600by divide degrees, into it convert to seconds,arc of terms in is P''parallax theGenerally
degp
360
2π
R
rstarDistanceof
degrees
2π
360
r
R
p
360/2πby gmultiplyin degrees into radian convert to order In
8









Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

208. 208
Parallax yearsLight 10.455pc 3.205parsec
0.312
1
Procyon of Distance
sec.arc 0.312Procyon ofparallax
yearsLight 4.3pc 1.3parsec
0.76
1
Centauri alpha of Distance
sec.arc 0.76Centauri alpha ofParallax
.parallaxes their from Procyon and Centauri alpha of distances the find us let
yearsLight 3.26parsec 1
109.46
101.5206255
parsec 1
by given isparsec 1 yearslight of term in hence
km. 109.46 yearLight 1 and
km 101.5206265parsec 1
But
12
8
12
8










Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

209. 209
Temperature and luminosity of the Sun
•LuminosityistheamountpfenergyemittedbytheentiresurfaceoftheSun.Luminositydependson
thesurfacetemperatureofthesun.Wien’sdisplacementlawgivesthesurfacetemperatureoftheSun
andStefan-Boltzmanlawgivesusanideaofamountofenergyemittedbyahotbodypersecond.
ActuallythelawisapplicabletotheblackBodies.TheSuncanbeapproximatedasablackbody.In
generalablackbodyidgoodabsorberaswellasgoodemitterofenergy. 
    Watts105.7Joules/sec105.8105.7m1074π LSun the of Luminosity
K5800 K
10500
102.898
TSun the of etemperatur Surface
units mks 105.7constant BoltzmanStefanConstant
m1074π4ππSun of areAWhere,
Tconst Asec per emittedEnergy of Amount Luminosity
that statesLaw Boltzman-Stefan
K5800 K5796 K
10500
102.898
TSun the of etemperatur Surface
T
102.898
10500
meters 10500meters nano 500ngthttedwaveleMaximumemieSun,Incaseofth
Kelvins in temperture the and meters in measured is length WaveWhere,
T
102.898
emitted length WaveMaximum
follows as writtenbe canlaw ntdisplaceme s Wien'The
2638
2
8
00
9
3
8
2
82
4
000
9
33
9
9
3


























Sun the of Luminosity
:Sun the of eTemperatur
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

210. 210
Radius of a star in terms of radius of Sun
•UsingStefan-Boltzmanlawwecanestimatethesizethesizeofastarintermsofradiusof
theSun. 
Sun. the of radius times 370 about is Betlegues of Radius
nearly 370
Rs
R
10
3000
5800
Rs
R
deg.Kelvin 3000Betlgues the of eTemperaturT
deg.Kelvin 5500Sun the of eTemperaturTs
magnitude Bolometric from Measured
10
Ls
L
Betlegues of case the In have, We
Ls
L
T
Ts
Rs
R

Ts
T
Rs
R
Ls
L
have, weequations, above the of ratio the taking
Sun the of etemperatur surfaceTs Star the of etemperatur surfaceT
Sun the of radiusRs Star the of radiusR Where,
Tsconst4ππRLSun the of Luminosity
Tconst4ππLStar the of Luminosity
4
2
4
242
42
42

































Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

211. 211
Lifetime of the Sun
•Whenastaronthemainsequence,itsbasicsourceof
energyistheconversionofhydrogenintohelium.Westart
withfourprotonsandendupwithoneheliumnucleus.Itis
observedthat0.007ofthemassofeachprotonisconverted
intoenergy.Withthispieceofinformationandknowingthe
Einstein’srelation‘E=mc
2
’wecanestimatethelifetimeof
theSun.
•In0.007ofthemassofeachprotoninthesunisconverted
intoenergyandifweassumethatmostofthemassofthe
Sunisinforofprotons,then0.007ofthesun’stotalmassis
availableforconversionintoenergy.Thetotalenergy
availableis. 
 
 
years.billion 5 is Sun
the of life remaining Thus years.billion 5 for livedalready has Sun the thatknow We
years.billion 10 years101Sun the of time Life
10. of factor aby estimate our lower to have weHence
place. takes reaction nuclear whereregion a in is Sun the of mass the of 10%only But
nearly years101sec 103.15sec
104
101.26
Sun the of time Life
Joules/sec 104Sun the of Luminosity
Luminosityby dividedenergy the is time life The
Joules 101.26Joules 1031020.007E
met/sec 103clight ofity Velockg, 102Sun the of Mass
ityLightvelocSun the of Mass0.007E
10
1118
26
45
26
45
2
830
830
2
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

212. 212
Meridian altitude of the Sun
•WhentheSunarrivesthelocalmeridian,itis12noonoflocaltime.Theangulardistanceof
theSunfromhorizontoitspositiononthemeridianiscalledmeridianaltitudeoftheSun.
Themeridianaltitudeofthesunchangesthroughouttheyear.Atallthelatitudesgreater
13.5north,theSunwillneverarriveatthezenith.Forallsuchplaceshavinglatitudeless
than23.5degrees,theSunwillarriveonthezenithtwotimesinayear.Forsomedays
meridianaltitudeoftheSunwillbegreaterthan90degrees.
•Themeridianaltitudeisgivenbythefollowingrelation,
•Meridianaltitude=(90–altitudeoftheplace)23.5degreesNorth
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

213. 213
Meridian altitude of the Sun
•AsanexampleletusconsiderMumbai.
•LatitudeofMumbai=19degreesNorth
•*22
nd
June
•DeclinationoftheSun=+23.5degreesNorth
•Meridianaltitudeofthesun=(90–19)+23.5=94.5degrees.
•*22
nd
December
•DeclinationoftheSun=-23.5degreesSouth
•Meridianaltitudeofthesun=(90–19)-23.5=47.5degrees.
•*21
st
Marchand22
nd
September
•DeclinationoftheSun=0degrees
•Meridianaltitudeofthesun=(90–19)+0=71degrees.
•*Duringnorthwardjourneyofthesun,therearetwodateswhenthedeclinationoftheSun
isexactlyequaltothelatitudeofMumbai,i.e.19degreesnorth.Thesetwodatescanbe
foundfromtheIndianalmanacs.About40daysbeforeandafter22
nd
June,wearrivenear
thefollowingdays.
•On16May,DeclinationoftheSun=19degreesnorth(nearly)
•On28August,DeclinationoftheSun=19degreesnorth(nearly)
•OnthesetwodaystheSunarrivesontheZenithoftheMumbai.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

214. 214
Tangential motion of the Stars
•Thestarsappeartobesteady,butthisisanillusion.Sincethestarsarequiteforawayfromuswecannot
sensetheirmotions.Thepropermotionofastarcanbedecomposedintotwocomponents.
•1)Theradialmotion:Thismotionofastarisourlineofsight.Itcanbemeasuredfromtheblueorthered
shiftofthespectrallinesinthestellarspectrum.
•2)Thetangentialmotion:Thismotionofastarisperpendiculartoourlineofsight.Itcanbemeasuredby
veryhighresolutiontelescope.
•Letusfindanexpressionforthetangentialmotionofastar.
•Theangleshouldbemeasuredinarcseconds.Wecanwrite, km/sec. 47.3 of speed tangential a has
yearpersec arc 0.1 is motion proper whoseand100pc is distance whosestar A :Example For
angle4.73r 10
10 6.52
angler
Vt

Then, km.). 10 3.084 parsec (One parsec. in measured is r'' distance the If
km./sec
10 6.52
angler
360036010 3.16
angler2
Vt
3600360
angler2
AD
Above,From
sec. 10 3.16 yearone Time
Time
AD
Vt
by, given isvelocity tangential The
star the of distance r
ntdisplaceme angular angle Where,
2ππ
AD
3600360
sec.arc in Angle
13
12
13
127
7
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

215. 215
Density of a pulsar
•In1967astronomerdetectedradiosignalsemittedbyacelestialobject.Thesignalswere
arrivingtowardstheEarthoneevery1.33730113seconds.Initiallythenatureofthatobject
couldnotbeconfirmed.Theywerethoughttobecomingfromsomeadvancedcivilization.
Butverysoonseveralsuchobjectswerelocated.Theseobjectswerethennamedas
‘PulsatingStars’,or‘Pulsars’.
•FromcarefulstudyofthePulsars,itwasconfirmedthattheywererapidlyrotatingNeutron
stars.WiththeknowledgeoftheperiodictimeoftheratiopulsesemittedbyPulsars,we
caneasilyfindtheirdensitiesandascertainthattheyarereallyNeutronstars.
• PeriodicTimeofaPulsar=2R/V
•Where,RRadiusofaPulsar,VVelocity(equatorial)
•ButaspherecanrotateonlyataspeedsuchthatthecentripetalaccelerationV
2
/Ratthe
equatorisjustequaltothegravitationalaccelerationGM/R
2
.3
dRG4
R3
dR4G
)d(densityR,where
R
GM
R
GM
R
V
23
3
2
2


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V
Pulsar the of RadiusR

3
4
Pulsar the of Mass M
V
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

216. 216
Density of a pulsar
•Periodic time T is given by, order. this of densities have stars Neutron
.kg/m 103.5d
102
103.74
d
is pulsar this ofDensity
sec. 102 sec. milli 2T period of Pulsar a Consider
T
103.74
d
d
103.74
d
1
106.67
3π
T
units. mks 106.67 G of value the ngSubstituti
d
1
G
3π
T
3
Gππ
2R
2ππ
3
d4GπG
2ππ
T
have, we V,of value the ngSubstituti
V
2ππ
T
316
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

217. 217
Energy generation in the interior of the Sun
•Theradiusofthecoreofthesunissupposedtobe1,50,000kilometers,wherethe
temperatureisnearly15milliondegreesCelsius.Thetemperatureissohighthat,Four
HydrogennucleicanfusetogethertoformoneHeliumnucleus.ThemassofoneHelium
nucleusisslightlylessthanthemassoffourhydrogennucleitakentogether.Theexcess
massisconvertedintoenergyaccordingtoEinstein’sequation.
•ProductionofHeliumtakesplaceinthefollowingsteps,
•1)FirsttwoHydrogennuclei,meansprotons,combinetoformonedeuteriumnucleus,
whichhasoneprotonandoneneutron.Thismeans,oneprotonisconvertedintoone
neutron.Thereactioncanbewrittenas,
•1Proton+1Proton1Deuterium+Positron+Neutrino
•ThePositronquicklycombineswithelectronandenergyisreleased.
•2)AProtonnowcombineswithDeuteriumtoproduceoneHelium3Nucleus.Helium3
consistsof,2protonsand1Neutron.InthisreactionGammarays,meansenergyis
liberated.
•Deuterium+ProtonHelium3+Gammarays(Energy)
•3)InthelaststagetwoHelium3nucleicombinetoformonHelium4Nucleus,itconsistsof
2Protonsand2Neutrons.Inthisreaction2Hydrogennucleiareliberated,andthe
reactionstartsagain.
•Helium3+Helium3Helium4+1Proton+1Proton
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

218. 218
Energy generation in the interior of the Sun
•Themassdeficitof4Hydrogenatomsand1Helium4atomcanbefoundasfollows,
•Massof1Hydrogennucleus(Proton)=1.007825units
•Massof4Hydrogennucleus(4Proton)=41.007825=4.03130units.
•Massof1Helium4nucleus=4.00268units
•Massdeficit =0.02862units
•Thismassdeficitis0.71percentofthemassofinitial4Hydrogennuclei.Thusif1kilogram
ofHydrogenisconvertedinto1Helium4atom,0.0071kilogramofmaterialisconvertedinto
energy.
•Energyreleasedbytheconversionof1kilogramofHydrogenintoHelium4,isgivenby,
• E=masssquareofvelocityoflight
• =0.0071(310
8
)
2
=6.410
14
=mass
• (410
26
)/(6.410
14
)=610
11
kg.
•LuminosityoftheSun=410
26
Joules.ToproducethisLuminosity,ofHydrogenmustbe
convertedtoHelium4eachsecond.But1kgofHydrogencorrespondsto0.007kgof
material.
•610
11
kgofHydrogen=610
11
0.0071=4.210
9
kgmatter
•ThusintheinterioroftheSun4.2billionkgofmattergetsconvertedintoenergypersec.
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

219. 219
Estimating surface temperature of a star
•Thereareseveralmethodsforestimatingsurfacetemperatureofstars.Inonemethod,
starsareassumedtobe‘BlackBodies’.ABlackBodyisanobjectwhichabsorbsradiation
ofallwavelengthsandemitsallwavelengtswhenitishot.
•Itisobservedthateverystaremitsmaximumenergyataparticularmaximumwavelength.
IfweassumeastartobeaBlackBody,wecanapplyWien’sDisplacementlawstatesthat,
•kelvin deg. 4140
10700
102.898
T
e?Temperatur its Whatnm. 700 atenergy maximum emits star A :Example
kelvin deg. 5796
10500
102.898
Sun the of eTemperatur
nanometer. 500 at
energy maximum emits Sun :example For
Kelvins. degree in eTemperatur and
meters in measured is length WaveWhere
T
102.898
length waveMaximum
9-
3-
9-
3-
-3
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

220. 220
Temperature and colour classification of stars
•Spectrumisoneofthetoolstoestimatethetemperatureofastar.Ingeneralapparent
magnitudeofastarisdecidedonthebasisofallwavelengthsappearinginitsspectrum.
Butastarisobservedtoemitmaximumenergyataparticularwavelength.Thisfactcan
beusedtoestimatethetemperatureofastar.Tofindatwhichwavelengthastaremits
maximumenergyfiltersareused.Abluefilterallowsonlythewaveofblueregionofthe
spectrum,whilevisualfilteronlywavesofgreen-yellowregion.Themagniyudesarecalled
BandVapparentmagnitudesofthestarrespectively.Thencolourindexofthestaris
found.
•ThecolourindexofastarisdifferencebetweenBandVmagnitudesofthestarColour
index=B-V
•Ifthecolourindexisnegativethestarishotandifpositivethestariscomparativelycool.
Colourindexvariesfrom-0.3forbluehotstarsto+2redcoolstars.
•Onthebasisoftheresponsetothewavelengthsinthespectrumthestarsareclassified
asfollows,
Spectral Class Temperature Colour
O 50000 –28000 Blue
B 28000 –10000 Blue-White
A 10000 –7500 White
F 7500 –6000 White-yellow
G 6000 –4900 Yellow
K 4900 –3500 Orange
M 3500 –2000 Red
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

221. 221
Temperature and colour classification of stars
•Example1:ThestarSpicahasapparentmagnitudes,
• B=0.7andV=0.9
• Whatiscolourindex?
• ColourIndexofSpica=B–V=0.7–0.9=-0.2
• Spicaisahotstar,sinceitscolourindexisnegative.
•Example2:Antareshasfollowingapparentmagnitudes,
• B=2.7andV=0.9
• WhatiscolourindexofAntares?
• ColourIndexofAntares=B–V=2.7–0.9=1.8
• Antaresisacoolstar,sinceitsColourIndexispositive.
•Colourindexofsomeofthestarsisgiveninthefollowingtable.
Star ColourIndex
Sigma Orion -0.24
Achernar -0.16
Vega +0.00
Procyon +0.42
Sun +0.65
Aldebran +1.54
Betelgeuse +1.85
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

222. 222
Estimating masses of the binary star system
•InBinarystarsystem,thestarsrevolveabouttheircommoncenterofmass.Theangular
velocityofboththestarsremainsthesame,butthelinearvelocitiesdependsontheir
distancesfromthecenterofmass,hencewecanwrite,

3
r2r
Gm2
r1r
Gm1
W
m2r2Wm1r1W
r
Gm1m2

.attraction nalgravitatio mutual theirby provided is whichforce,
lcentripeta aby upon acted is star each mass, of center the around srevolve star the Since
2
m2
m1
r2
r1
m2r2m1r1
have, wemass, of center
the at balanced be can stars the Since
mass of center the from B'' star of Distance r2
mass of center the from A'' star of Distance r1
m2 mass of B'' star ofvelocity Linear V2
m1 mass of A'' star ofvelocity Linear V1
r2 r1 stars two the between Distance r velocity, Angular W
Where,
1
V2
V1
r2
r1

r2
V2
r1
V1
W
22
2
22
2
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Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

223. 223
Estimating masses of the binary star system 


 
 
 
 
masses. solar 1.1342.0663.2m2
masses solar 2.066
3
6.2
3
3.22
m1
3.2
2
3m1
masses solar 3.2
2
m1
m1
2m2m1
m1
m2

2r1
r1

(2), equation Using A.Sirius as Barycenter the from far as twice is B Sirius that observed is It
masses solar 3.2
50
2.77.5
m2m1
2.77.5axis semimajor ual Act
years50T Period arcsec. 7.5 axis Semimajor
parsec. 2.67 Sun and Sirius of Distance b. and A Sirius of masses the Estimate :
m2m1
T
r
writecan wemass solar of terms In
Law Third sKepler' is Which
4π
m2m1G
T
r

r
m2m1G
T
4π

T
2π
WBut
r
m2m1G
W
have, we3 in 4 ngSubstituti
4
m2m1
m2r
r1
m2
m2m1r1
r1
m2
m1
r1r2r1r But
2
3
2
3
22
3
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1 Example
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

224. 224
Magnitudes of the stars and Stars
Elementary Astronomical Calculations: Lecture-05
Thank You
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

225. 225
Elementary Astronomical Calculations:
Universe and Space ( Lecture –6)
•By Sukalyan Bachhar
•Senior Curator
•National Museum of Science & Technology
•Ministry of Science & Technology
•Agargaon, Sher-E-Bangla Nagar, Dhaka-1207, Bangladesh
•Tel:[phone](Off), Contact: 01923522660;
•Websitw: www.nmst.gov.bd; Facebook: Buet Tutor
&
•Member of Bangladesh Astronomical Association
•Short Bio-Data:
First Class BUET Graduate In Mechanical Engineering [1993].
Master Of Science In Mechanical Engineering From BUET [1998].
Field Of Specialization Fluid Mechanics.
Field Of Personal Interest Astrophysics.
Field Of Real Life Activity Popularization Of Science & Technology From1995.
An Experienced TeacherOf Mathematics, Physics & Chemistry for O-,A-, IB-&
Undergraduate Level.
Habituated as Science Speaker for Science Popularization.
Experienced In Supervising For Multiple Scientific Or Research Projects.
17 th BCS Qualified In Cooperative Cadre (Stood First On That Cadre).
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

226. 226
Age of the universe
•BigBangtheorysuggeststhattheuniversewasbornoutofSingularity.Atthattime
everythingwascondensedintoaverysmallspace,wherethedensityandtemperature
wereextremelyhigh.Sincethattimeuniverseiscontinuouslyexpandingandaverage
densityandtemperaturearecontinuouslydecreasing.
•TherateofexpansionofuniverseisgivenbyHubble’slaw.In1929EdwinHubble
discoveredthatgalaxiesarerecedingfromeachother.Heobservedthatrateofseparation
ofanytwogalaxiesintheuniverseisdirectlyproportionaltothedistancebetweenthe
galaxies.
•Ifwehavetwogalaxiesatadistance‘R’apart,theirpresentrelativevelocityofseparationis
givenby,v
R
H
1
th
by, given is time Hubble
km. 1030.8396km. 109.46103.26parsec mega 1
km. 109.46 yearlight 1
yearslight 103.26parsec mega 1
yearslight 3.26parsec 1
parsec mega perkm/sec 71H
,constantis sHubble' of value acceptedCurrently
constant sHubble'H Where
HRv
18126
12
6
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
Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

227. 227
Age of the universe
•Hubble time gives us the estimate of the age of the universe. .elyapproximat yearsn13.7billio years1013.7
years101.37 years
31557600
100.4343
Universe the of Age
sec 31557600sec 360024365.25days 365.25 year1
sec 100.4343sec
102.3
1
Universe the of Age
sec 102.3
km 1030.8396
km/sec 71
parsec mega perkm/sec 71H But
H
1
Universe the of Age
9
10
18
18
18
118
18
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

228. 228
Observable universe
•Althoughtheuniverseisexpanding,wedonotknowactualsizeoftheuniverse.Thelightfrom
theboundaryoftheuniversehasnotyetreachedus.Sinceweknowtheapproximateageofthe
universe,wecanestimatethefarthestdistanceatwhichcelestialobjectscanbeobserved.This
distanceiscalled‘Horizon‘ofuniverseor‘Theobservableuniverse’.Thefarthestdistanceupto
whichtheuniversecanbeobserbedistermedas‘Hubble’sdistance.
•FromHubble’sLawwehave, (nearly). yearsLight billion 13 universe observable of Extent
yearsLight 1013d
yearsLight 101.3d
kms 10 yearlight 1 that assume us Let
kms 101.3mets 101.3
sec 104.3 met/sec 103d
sec 10 4.3th But,
thcduniverse observable the of Extent
d
th
1
c
have law we sHubble' on values these ngSubstituti
Universe. the of horizon the d R let and
time sHubble' th
H
1
thatknow We
c.by v replace shall Wemet/sec. 103cvelocity withtravels light the Since
HRv
9
10
13
2326
178
17
8
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
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229. 229
Critical density of the universe
•Wearenotsurewhetherouruniversewillkeeponexpandingoraftersometimeinfuture,itwillstart
contracting.Firsttypeofuniverseiscalled‘open’andsecondtypeiscalled‘closed’.
•Letusimaginethatentiremass(M)oftheuniverseisinsideasphereofradius‘R’.
•Letasmallmass‘m’isontheboundaryoftheuniverse.Dependingonwhethertheuniverseis‘open’or
‘closed’,thismasswillmoveawayortowardsthecenterwithvelocity‘v’.Wehave,

 
 nearly kg/m 102dcUniverse the ofdensity Critical
106.673.14168
102.323
8ππ
3H
dc
πGdc
3
4
H
2
1

dc'' bedensity critical let universe this For
universe. closed nor open neither have we0E
universe closed have we0 E
universe open have we0E If,
πGd
3
4
H
2
1
mRE
mdπGR
3
4
RmH
2
1
E
EpEkE m mass ofenergy Total
HRm
2
1
mv
2
1
Ek
by given isLaw sHubble' using ,m'' mass ofenergy Kinetic
mdπR
3
4
G
R
GMm
Epm mass ofenergy Potential nalGravitatio
d universe ofdensity πR
3
4
Msphere the in enclosed Mass
326
11-
2
18-2
2
22
222
22
2
3

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

230. 230
Estimate of the mass of Milky Way galaxy
•TheMilkyWaygalaxyconsistsofatleast200billionstarsofall
sizesandmasses.Henceitisverydifficulttoestimateitsmass,
knowingtheapproximatepositionoftheSunintheMilkyWay
anditsorbitalvelocity;wecanatleastjudgethemassofthe
MilkyWay,insidetheorbitoftheSun.furtherweshallassume
thatallstarsaresimilartoourSun.
•WeknowthattheSunisatdistanceofabout8.5kiloparsec
fromthecenteroftheMilkyWaygalaxy.Theorbitalspeedof
theSunisabout220km/sec.Thesetofactsareenoughto
estimatethemassofMilkyWaywithintheSun’sorbit.
•TokeeptheSuninitsorbitacentripetalforceshouldactonit.
ThisisprovidedbyhegravitationalforceoftheMilkyWay.Thus
wehave,  
 elyapproximat Sun the of mass times 100billionMilkyway of Mass
Sun the of Mass10kg101.9Milkyway of Mass
km 109.46 yearLight 1 and yearsLight 3.261parsec
106.67
109.463.26108.5102.2
G
Rv
MMilky Way of Mass
as, relation above writecan We
8.5kpcMilky Way of center from Sun the of DistanceR 220km/sec sun The of Speedv
Milky Way of massM Sun the of massm
Where,
Milky Way of force lGravitionaSun the on force lCentripeta
1141
12
11
153
2
52

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



231. 231
Doppler effect
•Imaginethatyouarestandingonaplatform,supposea
fasttrainentersandleavesthestation,withhigh
velocity.Insuchsituation,asaprecaution,thedriver
soundsatypicalrailhorn.Youfeelthatthepitchor
intensityofsoundincreasesasthetrainapproachesand
decreasesasthetraingoesawayfromyou.Thisisan
exampleofDopplerEffectineverydaylife.
•Changeintheintensityofsoundduetorelativemotionof
sourceandobserveriscalledDopplerEffect.
•TheDopplerEffectisobservedforsoundaswellaslight. Hz 4722
360
340
500
20340
0340
500f
away moving Source 2)
Hz 5312.5
320
340
500
20-340
0340
500f
towards moving Source 1)
Hz 500f met/sec 20v 0,vmet/sec 340v If

vv
v-v
ff
other each fromaway moving observer and source The 2)

v-v
vv
ff
other each towards moving observer and source The 1)
source ofvelocity v observer ofvelocity v
sound ofvelocity v frequency original f frequency Observed f If,
S0
S
0
S
0
S0














:Example

232. 232
Doppler effect

Red. as appear willcolour Blue
A6928 34000 L blue A4000 L If,
region. infrared in is which A12124 of
length wavea have to appears A7000 of light red the because shift Red called is This
A121243LL
3
1
L
c
1.5
0.5
L
c
0.5cc
0.5c-c
L
c
L
c

away moving Source 2)
colour. Blue the near is which A4041 of
length wavea have to appears A7000 of light red the because shift Blue called is This
A4041
3
A7000
3
L
L 3
L
c
0.5
1.5
L
c
0.5c-c
0.5cc
L
c
L
c

towards moving Source 1)

L length wave
c
frequency Angstrom7000λL
0.5cv met/sec, 10 3c If, :Example

vc
v-c
ff
away moving source The 2)

v-c
vc
ff
towards moving source The 1)
source ofvelocity v met/sec, 10 3 light ofvelocoty c
light. for effect Doppler The
00
0
0
0
0
0
0
0
8
8

















233. 233
Doppler effect
•Aspectrumofacelestialluminousobjectrevealsitsmanyproperties.Ifsuchsourceismoving
awayfromus,thespectrallinesappeartoshifttowardstheredendofthespectrum.Thisis
calledDopplerRedShift.In1842ChristianDopplerdiscoveredsimilareffectforsourceof
sound.Forsourceoflighttheshiftinspectrallinesdependonthevelocityofthesource.Similarly
iftheluminousobjectismovingtowardsus,thespectrallinesappeartoshifttowardstheblue
endofthespectrum.ThisiscalledBlueShift.Thusbymeasuringtheshiftinthewavelengthsof
thespectrallineswecanfindhowfasttheobjectismovingawayortowardsus.
•Manytimesthevelocityofthesourceiscomparabletothevelocityoflight.Inthatcasewehave
touseRelativisticDopplerEffect,forwhichtheformulaisasfollows,

234. 234
Doppler effect 
 
  
 
2139z
1124z


































1-1-
0.2
1.8
1-
0.8cc
0.8cc
z then0.8c v If, :3 Example
1-1-
0.4
1.6
1-
0.6cc
0.6cc
z then0.6c v If, :2 Example
0.1z then0.1c v If, 1: Example
neglected is term /cv
c
v
z
1
c
v
2
1
1
c
v
2
1
1z
have, weExpansion, Binomial the of term firstonly consider can wesmallvery isv/c Since
1v/c1-v/c11
v/c1-
v/c1
z
as, formula above themodify can we0.1c),
than (smaller light ofvelocity the to compared smallvery is source the ofvelocity the If
source ofvelocity c source ofvelocity v Where,
1
vc
vc
zRedshift
22
1/21/2
1/2
1/2

235. 235
Velocity of a galaxy from its redshift




 
 
 
 
km/sec 295000 3000000.984
300000
129
127
km/sec 300000
1128
1-128

km/sec 300000
1110.3
1110.3
v
velocity? its is what10.3, of redshift has Quaser A :2 Example
km/sec 188000 3000000.628
300000
5.37
3.37
km/sec 300000
14.37
1-4.37

km/sec 300000
12.09
12.09

km/sec 300000
111.09
111.09
v
1.09 is redshift whosegalaxy, a ofvelocity the Find 1: Example
c
11z
11z
v
c
v
11z
11z
have, wersdenominato and numerators the adding and gSubtractin
vc
vc
1
1z

vc
vc
1z
1
vc
vc
zRedshift
formula. followingby given is redshift the of valueic relativist The
2
2
2
2
2
2
2
2
2
2
2







































236. 236
Estimate of the temperature
•AtthetimeofBigBangtheUniversewasintheformofSingularity.Itwasveryhotandvery
dense.AfterBigBangtheUniversestartedexpandinganditstemperaturestartedfalling.We
canestimatethetemperatureoftheUniverseatanyinstantintimebyaformulasuggestedby
GeorgeGamow.TheformularelatestheageoftheUniverseanditstemperature.Gamow’s
formulaisasfollows,
•ItisassumedthatthePlankEraended10
-43
secondsafterBigBang,whenthetemperatureof
theUniverseattheendof10
-2
secondsafterBigBang.
•FollowingtablegivestheageandtemperatureoftheUniverse.
•Age(sec) Temperature(deg.K)
•10
-43
10
32
•10
-36
10
28
•10
-10
10
15
•10
-8
10
14
•10
-4
10
12
•10
-2
10
11
•10
-1
310
10
•1 10
10
•13.8min 3.47510
10
•35min 2.1810
8
•700000yrsn 2.1210
3Kelvin degrees
seconds in Age
10
TUniverse of eTemperatur
10
 Kelvin degrees 10
sec 10
10
T
11
2-
10

Sukalyan Bachhar, Senior curator, National Museum of Science & Technology, Bangladesh

237. 237
Hohmann transfer orbit
•Ifwewanttosendaspaceshipto,sayMars,itcannotbelaunchedwhenthedistancebetween
theEarthandMarsisshortest,i.e.whenMarsisinopposition.BothEarthandMarsarein
motion.Earthhasatangentialvelocityofabout30km.persecond,andtherockethastodo
workagainstthestronggravitationalforceofthesun.Thesearesomeofthereasonswhythe
shortroutemethoddoesnotwork.
• In1925WolfgangHohmannproposedaminimumenergytransferorbittosendaspaceshipto
otherplanets.Thismethodisnowcalled‘HohmannTransferOrbit’.Inthismethodthe
spaceshipissolaunchedthatitmovesinanellipticalorbitaroundtheSun.theperihelionpoint
ofsuchanorbitistheearthandAphelionpointistheplanetunderconsideration.
•WecanuseKepler’sThirdLawtoestimatethetimerequiredbythespaceshiptoreachthe
planet.orbit sspaceship' of axis major SemiR
spaceship of period OrbitalT
a1orbit sEarth' of axis major SemiRe
Earth periodof OrbitalTe
Re
R
Te
T
,thirdlawas
seKepler'anmodifythoorbitswecForthesetw
spaceship. the of orbit the is second and
orbit sEath' is one orbits; two consider shall We
RT
have, We
3
3
2
2
32






Sukalyan Bachhar, Senior
curator, National Museum of
Science & Technology,

238. 238
Hohmann transfer orbit 
 
 
 
Earth. the from degrees 44.445135.555-180 at be must Mars launch of time the at means Which
degrees. 135.555360
1.8822
0.70873
be will years0.70873 in marsby described Angle
degrees. 360 complete to required are years1.8822 Since
Launch. of time the at distance angular its estimate can weuniform isvelocity its and circular is
Mars the of orbit that assume weIf Sun. the around orbit one complete to years1.8822 takes Mars
Launch of time the at Mars of Location
nearly months 8.5 years0.70873
1.417454/2 orbit Transfer Hohmannby Mars reach to Time
required. is interval time this of Half Mars. reach to spaceship, the of period orbital is This
years1.4174541.261845 T
1.261845
2
2.523691
12
1.5236911
1T
?T 1.523691,a2 AU,1a1 year,1Te
have, weThen Mars. to sent be to s spaceshipi that assume us Let
get, weformula, above the in values these ngSubstituti
/2a2a1orbit sspaceship' the of axis major Semi And,
2a1
a2a1
TeT
a2a1orbit sspaceship' the of axis Major Then,
planet desired of axis major Semia2 If
3/2
3
33
2
3
22
































239. 239
Hohmann transfer orbit for inner planet
•WithreferencetoEarth,MercuryandVenusaretheonlyinnerplanets.ForouterplanetstheEarthshould
beattheperihelionatthelaunchofaspaceship.Inthecaseofinnerplanets,theEarthshouldbeatthe
aphelionatthetimeoflaunchandtheinnerplanetshouldbeattheperihelion,whenthespaceship
reachestheplanet.
•LetaspaceshipistobesenttoVenus.InthiscasetheSemimajoraxisoftheorbitwillbegivenby,0.16
0.7231
0.7231-
e
0.723Rv1AU,ReBut,
RvRe
Rv-Re
2a
Rv-Re
eorbit the ofty Eccentrici
Rv-Re2Rv-RvRe2Rv-2a be willfoci two
between distance the ,Perihelion at be to is VenusSince
axis major Semi
foci two between Distance
e
as, defined is orbit elliptical an ofty Eccentrici
AU0.723 Venusof axis major SemiRv
AU1Earth of axis major SemiRe
Where,
2
RvRe
a













240. 240
Hohmann transfer orbit for inner planet
 
months 4.8months 120.4
years0.4
2
0.8
2
T
Venusreach to Time
nearly ears0.8y T
0.86T
0.86a 1AU,ae 1year,Te
ae
a
Te
T
law. third sKepler' from flight of time total find can We
km/sec 15v
10 0.86
1
10 1.5
2
10 210 6.67v
0.86
2
0.7231
1
2
RvRe
a 1AURer
kgs. 10 2 Sun of MassMs units mks 10 6.67G Where,
a
1
r
2
GMsvlaunch Spaceship ofVelocity
32
3
3
2
2
1111
3011-2
3011-
2































241. 241
Distance Of geosynchronous satellite from the surface
of the Earth
•OrbitalperiodofageosynchronoussatelliteisexactlyequaltoperiodofrotationoftheEarth,i.e.24
hours.
•Wehave, 
Earth. the of surface the from km 36000 about at revolves Satellite nousGeosynchroA
km 36000km 35897h
6400-422975h
Earth of surface the above Height
km 42297.5meter 104.22975 r
1075.6736
4π
360024106106.67
r
T
4π
GM
r
r
GM
T
4π

r
GM
w
kgs 106Earth of massM
units mks 106.67constant nalGravitatioG
24hoursperiod OrbitalT
2ππ/satellite ofvelocity angularW
Earth of center the from satellite of distancer satellite of massm
Where,
r
GMm
mrw
Earth of attraction nalGravitatioforce lCentripeta
7
21
2
22411
3
2
2
3
32
2
3
2
24
11
2
2

















242. 242
Insertion velocity and period of a satellite
•Therearetworequirementsneededtoplaceasatelliteina
stableorbit.
•1)Bringthesatellitetotherequiredaltitude.
•2)givenecessaryorbitingvelocityofasatelliteequatethe
centripetalforcewiththegravitationalforce.Wehave,
 
min 28 hr 1sec
7800
65600003.14162
T km/sec 7.8
1600006400000
106106.67
v
kms. 160 height the at period its and satellite a ofvelocity insertion the Find
v
hR2π
Tby given is satellite the of Period
hR
GM
v VelocityInsertion
hR
GM
v
km. 6400R
surface, sEarth' the above heighth
Earth of center the from satellite of distancer Where,
hRr
units mks 106.67constant nalGravitatioG
velocity insertionv
kgs 106Earth the of massM
satellite the of massm Where,

r
GMm
r
mv
2411
2
11
24
2
2























:Example
Height (km) v (km/sec) T
320 7.715 1 hr31 min
1600 7.068 1 hr58 min
35880 3.070 24 hrs

243. 243
Characteristic velocity of a satellite
•Characteristicvelocity(vc)isthemaximuminitialvelocitythatasatellitemusthaveat
launch,aftertherockethasburnedallitsfuel,sothatithasthepropervelocityatthe
insertionpointC.
•TheenergyatthelaunchpointAshouldbeequaltothetotalenergyofthecircularorbitat
height‘h’.


hR
GM
v
hR
GMm
hR
mv
Earth the of force nalGravitatioforce lCentripeta
motion circular for but
2
hR
GMm
mv
2
1
E2
energy PotentialenergyKinetic C atenergy Total
velocitystic characteri vc Earth the of radiusR
constant onalMgravitatiG surface the from satellite the of heighth
Earth the of MassM satellite the of massm
Where,
1
R
GMm
mv
2
1
E1
energy PotentialenergyKinetic Aatenergy Total
C atenergy Total Aatenergy Total
2
2
2
2
2

















244. 244
Characteristic velocity of a satellite
*


 
 
   
 
 
km/hr. 32025km/sec 8.896met/sec 1.125107.9076
64000002
1600000
1
6400000
106106.67
vc
kms? 1600 oforbit an in placed be to isit if satellite a of velocity sticcharacteri the be should What :1 Example
R h for
2R
h
1
R
GM
vc
R
h
1
R
GM
R
h
1
R
2h
1
R
GM
vc
have, WeR;h sinceh/R of power higher the Neglecting
R
h
1
R
2h
1
R
GM

h/R1
1h/R12
R
GM

h/R1
1
2
R
GM
hR
GM
R
2GM
vc
hR
GM
R
2GM
vc
hR2
GMm
R
GMm
mvc
2
1
have, we3 and 1 Equating
3
hR
GMm
2
1
E2

hR
GMm
hR
GMm
2
1
E2
have, we,2 in value this ngSubstituti
3
2411
1/2
2
1
2
2
2















 

































































































245. 245
Orbital period and speed of the Chandrayan-1
•Chandrayan-1waslaunchedon22
nd
October2008anditstartedorbitingaroundtheMoon14
th
November
2008.Theorbitwaspolarandcircular.Thedistanceofthechandrayan-1fromthesurfaceoftheMoonwas
100kilometers.
•InordertokeepChandrayan-1incircularorbit,therequiredcentripetalforcewasprovidedbyMoon’s
gravitationalattraction.Thuswecanwrite, 

hr/km6000hr/km6.5889sec/km636.1sec/met1836
7105
18500002




















T
R2
v
timeperiodic
orbit the of ncecircumfere
v speed Orbital
min 118.4sec 7105
100.075106.6
1850000
2π
GM
R
2πTperiod orbital
R
GM
T
2π

R
GM
W
get, weequation, above grearrangin and m'' Cancelling
units MKS 106.6constant nalGravitatioG
2ππ/velocity AngularW
?period OrbitalT
km 18501001750orbit the of RadiusR
kg. 100.075Moon of massM
1Chandrayan of massm
Where,
R
GMm
mRW
Moon of attraction nalGravitatioforce lCentripeta
2411
33
33
11
24
2
2

246. 246
Thrust and specific impulse of a rocket
•RocketisanessentialcomponentofSpaceresearch.Since1957,
thousandsofsatelliteswerelaunchedbyvarioustypesofrockets.
VeryheavysatelliteslikeSpaceShuttle,HubbleSpaceTelescope,
SkyLab,arelaunchedbyverypowerfulrockets.Arocketworkson
theprincipleofNewton’sThirdLaw,whichstatesthat,actionand
reactionareequalandopposite.
•Inarocket,veryhotgasesarethrownoutthroughasmallnozzle.A
verystrongforcegeneratedashotgasesareejectedoutwithhigh
speeds.Thisforceofactionpushesthebodyoftherocketinopposite
direction.Theforcegeneratedinthiswayiscalled‘Thrust’.Asimple
relationofthenetthrustisasfollows,
 
ationnalaccelergravitatiog
ejectedpropellantgeofmassofrateofchan










dm/dt
impulseSpecific I ThrustF
Where,
1 g
dt
dm
IF
spthrust
spthrust  

time. unit per expelled is
propellant of weightunit whenobtained be can that thrust a is impulsespecific Thus,
2
gdm/dt
F
impulseSpecific
formula. above the from impulsespecific the
of equation writeancan Werocket. the of efficiency the of measure a is impulseSpecific
thrust



247. 247
Thrust and specific impulse of a rocket
* 
kg/sec 1212
g421
105
gI
F
dt
dm
second? per eaxpelled waspropellant of amount What
seconds. 421 wasImpulseSpecific
Its Newton. million 5 was
rocket Saturn the of stage second the of Thrust The
met/sec 4500 V 9.8459gI V
g
V
I
Velocity?Exhaust the is What
seconds. 459 is rocket Shuttle the of ImpulseSpecific
sec. 263I of ValueActual
nearlysec 267
9.813000
1034
I
kg/sec 13000time unit per expelled Propellant
Newton 1034 Newton million 34 Thrust
given rocket, 5-Saturn the of
the of stage the1 of impulsespecific the Find
VelocityExhaustV
g
V
I
relation. following from obtained be also can impulseSpecific
6
sp
thrust
espe
e
sp
sp
6
sp
6
st
e
e
sp















:3 Example
:2 Example
:Example1

248. 248
Elementary Astronomical Calculations:
Universe and Space ( Lecture –6)
Thank You