Lesson 3 - Kepler's Laws of the Planetary Motion.pdf

This material contains the three laws of planetary motion which were proposed by Johaness Kepler.

1. KEPLER’S LAWS OF PLANETARY MOTION
Physical Science
KarenMarielR.Sable

2. Let’s review!
Guessing Game

3. Guess the model!
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Hint:
LTAF
Answer:
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4. Guess the model!
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Hint:
CONIPERCUS
Answer:
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5. Guess the model!
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Hint:
GOPYTHARAS
Answer:
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6. Guess the model!
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Hint:
HEBRA
Answer:
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7. OBJECTIVES
❑solve the ratio of the cube of the mean distance of
the planet from the Sun and the square of its
revolution;
❑determine the time needed for the planet to
revolve around the Sun using the Kepler’s laws of
planetary motion; and
❑explain the importance of Kepler’s Laws of
Planetary Motion in the scientific field in the
present days.
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8. Exercise 1: Draw it out!
On a given paper, draw where do you think
is the correct position of the sun in the
planetary orbit. You have 30 seconds to
draw the sun! Aja!
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9. Kepler’s Laws of
Planetary Motion
LESSON 3
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10. Johaness Kepler
Brahe hired Kepler as a sort of "research
assistant" primarily to prove that Brahe's model the
geoheliocentric model is the right model. Kepler's task
is to fit in the data collected by Brahe into the model
he proposed by doing the mathematical calculation.
Unfortunately, Brahe died before his model is proven.
Kepler inherited a vast set of data that will prove
crucial for developing his Three Laws of Planetary
Motion later.
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11. 11

12. The planets move in elliptical orbits
with the Sun at a focus (F1). The
other focus (F2) is empty.
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First Law of Planetary Motion:
Law of Elliptical
Orbit or Law of
Ellipses (1609)

13. As the planets orbit around the sun, the planets
cover equal areas in equal times. For this to
happen, as shown in the figure below the point
A to B when the planets are nearest to the Sun
it moves and lowest at point C to D when the
planets are farthest from the Sun.
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Second Law of Planetary
Motion:
Law of Equal
Areas (1609)

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Second Law of Planetary
Motion:
Law of Equal
Areas (1609)
When the planet is nearest to the Sun, it is
called perihelion. When it is farthest from
the Sun, it is called aphelion.

15. The ratio of the squares of the
periods (the time needed for one
revolution about the Sun) of any of
the two planets revolving around the
Sun is equal to the ratio of the cubes
of their mean distances from the Sun.
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Third Law of Planetary
Motion:
Law of Periods
(1619)

16. That is if T
1 and T
2 represent the
periods for any two planets, and r
1
and r
2 represent the mean distances
from the Sun, then
It can be rewritten as
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Third Law of Planetary
Motion:
Law of Periods
(1619)

17. Meaning: is the same for
each planet
To determine the value of
proportionality constant k, the value
of Earth’s known orbit could be used:
T
Earth = 365.24 days
r
Earth = 149 million kilometres or 149
600 000 km or
149.6 x
10
6

km
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Third Law of Planetary
Motion:
Law of Periods
(1619)

18. T
Earth = 365.24 days
r
Earth = 149 million kilometres or 149
600 000 km or
149.6 x
10
6

km
Thus, the ratio of the cube of the
mean distance of Earth from the Sun
and the square of its revolution is
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Third Law of Planetary
Motion:
Law of Periods
(1619)

19. Practice makes it easy!
Practice problem 1. Find of planet DW.BL-218x
where its:
T
DW.BL-218x = 432.21 days
r
DW.BL-218x = 136.1 x
10
6

kilometers
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20. Practice makes it easy!
Practice problem 1. Find of planet DW.BL-218x
where its:
T
DW.BL-218x = 432.21 days
r
DW.BL-218x = 136.1 x
10
6

kilometers
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21. Practice makes it easy!
Practice problem 2. Using the third law of planetary
motion, calculate the number of days needed for
planet BW-512 to complete its revolution around the
sun. The mean distance of the said planet from the sun
is
144.7 x
10
6

kilometers. Use the given
proportionality constant assuming that the planet is
part of the solar system.
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22. Practice makes it easy!
Practice problem 2.
Formula:
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Where
r
1 = mean distance of planet Earth from the Sun
r
2 = mean distance of planet BW-512 from the Sun
T
1 = days of revolution of planet Earth around the Sun
T
2 = days of revolution of planet BW-512 around the Sun

23. Practice makes it easy!
Practice problem 2.
Solution:
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We can use the calculated proportionality
constant, k, to substitute with

24. Practice makes it easy!
Practice problem 2.
Solution:
=
From the problem,
144.7 x
10
6 km is the mean distance of
planet BW-512 from the Sun. So,
=
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25. Practice makes it easy!
Practice problem 2.
Solution:
= km3

= km3
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26. Practice makes it easy!
Practice problem 2.
Solution:

Thus, it takes 347 days for planet BW-512 to complete its revolution
around the sun.

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27. Practice makes it easy!
Practice problem 3. Find of a certain planet
where:
T
Planet = 4,332.5 days
r
Planet = 483.6 x
10
6

kilometers
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28. Practice makes it easy!
Practice problem 4.
Mars takes 686.98 Earth days
to complete one orbit around
the sun. What is the mean
distance of Mars from the sun?
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29. Summary
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•Johaness Kepler formulated the laws of the planetary
motion with the help of the collected data of Tycho Brahe.
•First Law of Planetary Motion: Law of Elliptical Orbit or Law
of Ellipses (1609)
•Second Law of Planetary Motion: Law of Equal Areas(1609)
•Third Law of Planetary Motion: Law of Periods(1609)

30. Exercise 2: Fish and Board!
Instruction: Find the answer of each given problem using
the given formula. Then, write it on your board. The first
one to raise their board with the correct answer will have
a chance to fish a mystery score in the fish bowl. The
number of the points you will get will be your
corresponding points.
Formula:
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31. Exercise 2: Fish and Board!
First problem. Find of Neptune where:
T
Neptune = 60,195 days
r
Neptune = 4,497.1 x
10
6

kilometers
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32. Exercise 2: Fish and Board!
First problem. Find of a Neptune where:
T
Neptune = 60,195 days
r
Neptune = 4,497.1 x
10
6

kilometers
Answer: 0.00001078
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33. Exercise 2: Fish and Board!
Second problem. Find of planet XYW
where:
T
XYW = 455.5 days
r
XYW = 155.2 x
10
6

kilometers
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34. Exercise 2: Fish and Board!
Second problem. Find of planet XYW where:
T
XYW = 455.5 days
r
XYW = 155.2 x
10
6

kilometers
Answer: 1.80
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35. Exercise 2: Fish and Board!
Third problem. Find of Mars where:
T
Mars = 686.651 days
r
Mars = 227.9 x
10
6

kilometers
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36. Exercise 2: Fish and Board!
Third problem. Find of Mars where:
T
Mars = 686.651 days
r
Mars = 227.9 x
10
6

kilometers
Answer:
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37. Exercise 2: Fish and Board!
Fourth problem. Find of planet X where its:
T
planet X = 575.32 days
r
planet X = 165.4 x
10
6

kilometers
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38. Exercise 2: Fish and Board!
Fourth problem. Find of planet X where its:
T
planet X = 575.32 days
r
planet X = 165.4 x
10
6

kilometers
Answer:
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39. Exercise 2: Fish and Board!
Fifth problem.
The revolution of Saturn around
the sun takes 10,758.5 Earth
days. What is the mean distance
of Saturn from the sun?
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40. Exercise 2: Fish and Board!
Fifth problem.
The revolution of Saturn around the sun
takes 10,758.5 Earth days. What is the
mean distance of Saturn from the sun?
Answer: 1.43
x
10
9

kilometers
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41. Assessment: ½ crosswise paper
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According to the Law of periods that the ratio of is the
same for each planet, so what do you think is the period
of revolution of an imaginary planet if its mean distance
from the Sun is
337.9 x
10
6

km? Comparing its period of
revolution to Earth, is the imaginary planet near or far
from the Sun? Show your solution.

42. Assignment: The Masterpiece!
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Share what you have learned in the lesson by making a
masterpiece out of your learning about Kepler’s Laws of
Planetary Motion. You can freely express yourself. It can
either be a song, dance, poem, story, artwork or an
essay representing the knowledge you have learned in
this lesson. You can use any materials for your work.

43. Thank you
for listening!
Prepared by :Ms. Karen Mariel R. Sable