2. 1 SATELLITESSATELLITES

3. 2 Newton’s Law of Gravitation M1M1 M2M2 r F F

4. 3 CIRCULAR MOTION r F CENTRIPETAL FORCE

5. 4 SATELLITES r m v Equation of Motion M

6. 5 SATELLITE VELOCITY v r m M

7. 6 v R m For an orbit CLOSE to the surface F = mg v = √ r g v = √ 6.4x10 6 x 10 = 8000 ms -1 v = 8 km/s

8. 7 Example 3: A 110kg satellite is placed in orbit about the Earth at a height of 35900km. (The Earth has a mass of 6.0 x 10 24 kg and a radius of 6370km. 110kg Solution: The distance from the satellite to the centre of the Earth is: 35900 + 6370 = 42270 km = 4.2270 x 10 7 m Calculate: The force of gravity on the satellite. The orbital speed. The period of the orbit in hours

9. 8 a.F = GMm  r 2 = 6.67 x 10 -11 x 6.0 x 10 24 x 110  (4.2270 x 10 7 ) 2 = 25N b. The force of gravity on the satellite is the centripetal force: F c = F g mv 2 /r = GMm/r 2 v =  (GM/r) v =  (6.67 x 10 -11 x 6.0 x 10 24 / 4.2270 x 10 7 ) v = 3100ms -1

10. 9 c. The distance for one orbit d = 2  r d = 2 x  x 4.2270 x 10 7 = 2.6559 x 10 8 m The time taken for one orbit is the period, T = d/v. T = 2.6559 x 10 8 / 3077 = 86316 s = 24hours This satellite will orbit the Earth once every 24hours, the same time as the rotation of the Earth. This means that the satellite could remain in position over the same place on the Earth’s surface if it is above the equator. This is called a geosyncronous orbit and is used for communication satellites. Because the speed of a satellite depends only on the radius of its orbit, all communications satellites have the same orbital radius.

11. 10 GE0SYNCHRONOUS COMMUNICATIONS SATELLITE TO REMAIN OVER ONE PLACE ON THE EARTH’S SURFACE, THE PERIOD HAS TO BE THE SAME AS THE EARTH’S DAY.

12. COMPLETE EXERCISES PAGE 53 RUTTER 11