orbital velocity of satellite unit 5 new.pptx

orbital velocity unit 5 grade 9

1. ARTIFICIAL SATELLITES . ARTIFICIAL SATELLITES An object that revolves around a planet is called a satellite . Types of Satellite Natural Satellite: The moon revolves around the Earth so moon is a natural satellite of the Earth. Artificial Satellite: Scientists have sent many objects into space. Some of these objects revolve around the Earth. These are called artificial satellites. Most of the artificial satellites, orbiting around the Earth are used for communication purposes. Artificial satellites carry instruments or passengers to perform experiments in space

2. Communication satellites They take different time to complete their one revolution around the Earth depending upon their distance h from the Earth. Communication satellites take 24 hours to complete their one revolution around the Earth. As Earth also completes its one rotation about its axis in 24 hours, hence, these communication satellites appear to be stationary with respect to Earth. It is due to this reason that the orbit of such a satellite is called geostationary orbit.

3. Motion of Artificial Satellite A satellite requires centripetal force that keeps it to move around the Earth. The gravitational force of attraction between the satellite and the Earth provides the necessary centripetal force Consider a satellite of mass m revolving round the Earth at an altitude h in an orbit of radius r with orbital o velocity vo . The necessary centripetal force required is given by equation R+h r = R+h R> > >h r = R

4. Numerical 5.1:Find the gravitational force of attraction between two spheres each of mass 1000 kg. The distance between the centres of the spheres is 0.5 m. m 1 = 1000kg m2 = 1000kg d=0.5m F= 6.673 Nm 2 kg -2 x 10 -11 x1000kgx1000kg 0.5mx0.5m F= 26.67x 10 -5 N F =2.67x 10 1 x10 -5 N F = 2.67x10 -4 N

5. Numerical 5.2:The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N.Find their masses. F=0.0 06673 N F=0 . 0 0 6 6 7 3 N F= 6.673x 10 -3 N 6.673x 10 -3 N = 6.673x 10 -11 Nm 2 kg -2 x mass xmass 1mx1m

6. The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N.Find their masses. F=0.0 06673 N 6.673x 10 -3 N = 6.673x 10 -11 Nm 2 kg -2 x mass x mass 1m x1m 6.673x 10 -3 N x 1m 2 = mass 2 6.673x 10 -11 Nm 2 kg -2 10 8 kg 2 = mass 2 Taking Under root 10 4 kg = mass

7. ASSIGNMENT 5.2:The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N.Find their masses. F=0.0 06673 N 5.1:Find the gravitational force of attraction between two spheres each of mass 1000 kg. The distance between the centres of the spheres is 0.5 m.

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9. 14 th Sep,21

10. VARIATION OF G WITH ALTITUDE Equation shows that the value of acceleration due to gravity g depends on the radius of the Earth at its surface. The value of g is inversely proportional to the square of the radius of the Earth. But it does not remain constant. It decreases with altitude . Altitude is the height of an object or place above sea level. The value of g is greater at sea level than at the hills.

11. VARIATION OF G WITH ALTITUDE Consider a body of mass m at an altitude h as shown in figure 5.5. The distance of the body from the centre of the Earth becomes R + h. Therefore, using equation (5.6), we get

12. According to the above equation, we come to know that at a height equal to one Earth radius above the surface of the Earth, g becomes one fourth of its value on the Earth. Similarly at a distance of two Earths radius above the Earth's surface, the value of g becomes one ninth of its value on the Earth .

13. Mass of Mars = M m = 6.42 × 10 23 kg Radius of Mars = R m = 3370 km = 3370 × 1000 m = 3.37 × 10 8 m Acceleration due to gravity of the surface of Mars = g m =? g m = G M m /R 2                 or       g m = 6.673 × 10 -11 × 6.42*10 23 /(3.37*10 6 ) 2                                               = 6.673*10 -11 * 6.42*10 23 / 11.357  2 3370 km. = 42.84/11.357 = 3.77 m -2 Find the acceleration due to gravity on the surface of the Mars. The mass of Mars is 23 6.42x10 kg and its radius is 3370km

14. Rough work

15. A satellite requires centripetal force that keeps it to move around the Earth. The gravitational force of attraction between the satellite and the Earth provides the necessary centripetal force Consider a satellite of mass m revolving round the Earth at an altitude h in an orbit of radius r with orbital o velocity vo . The necessary centripetal force required is given by equation MOTION OF ARTIFICIALSATELLITES g h r = v 2

16. g h = GMe (R+2R) 2 g h =GMe (3R) 2 g h = GMe 9 R 2 g h = 1 g :g=GMe 9 R 2 g=GMe R 2

17. g h = GMe (R+R) 2 g h = GMe (2R) 2 g h = GMe 4 R 2 g h = 1 g :g= GMe 4 R 2

18. √ g h r = √ v 2 √ g h r = v r = R+h R> > >h r = R √ g h R= v g = g h √ g R = v A satellite revolving around very close to the Earth, has speed v nearly 8 km/s or 29000 km/h