1. Mechanics S/L Option A.3 Orbital Motion

2. Orbital Motion Gravitation provides the centripetal force for circular orbital motion The behaviour of the solar system is summarised by Keplar´s laws Keplar´s law state 1. Each planet moves in an ellipse which has the sun at one focus 2. The line joining the sun to the moving planet sweeps out equal areas in equal times

3. Deriving the Third Law Suppose a planet of mass m moves with speed v in a circle of radius r round the sun of mass M The gravitational attraction of the sun for the planet is = G Mm r 2 From Newton’s Law of Universal Gravitation

4. If this is the centripetal force keeping the planet in orbit then G Mm = mv 2 ( from centripetal equation) r 2 r  GM = v 2 r

5. If T is the time for the planet to make one orbit v = 2  rv 2 = 2 2  2 r 2 T T 2  GM = 4  2 r 2 r T 2  GM = 4  2 r 3 T 2  r 3 = GM T 2 4  2  r 3 = a constant T 2

6. The square of the times of revolution of the planets (i.e. Their periodic time T ) about the sun are proportional to the cubes of their mean distance (r) from it. We have considered a circular orbit but more advanced mathematics gives the same result for an elliptical one. Keplar´s Third Law

7. Energy of Orbiting Satellites Potential Energy, E p A satellite of mass m orbiting the Earth at a distance r from its centre has Gravitational Potential, V = - G M e r Therefore the gravitational potential energy, E p = - G M e m r

8. Energy of Orbiting Satellite Kinetic Energy, E k By the law of Universal Gravitation and Newton’s Second Law G M e m = mv 2 r 2 r Therefore the kinetic energy, E k = ½mv 2 = G M e m 2r

9. Energy of Orbiting Satellite Total Energy, E p + E k Total Energy = - G M e m + G M e m r 2r Total Energy = - G M e m 2r Total energy is constant for a circular orbit.

10. Graphs of Energy Against Radius Potential Energy EpEp 1/r EpEp r i.e. E p  -1/ror E p = -k/r where k is the constant of proportionality = GM e m

11. Graphs of Energy Against Radius Kinetic Energy EkEk 1/r EkEk r i.e. E k  1/ror E k = k/r where k is the constant of proportionality = GM e m 2

12. Graphs of Energy Against Radius Total Energy E total 1/r E total r i.e. E total  -1/ror E total = -k/r where k is the constant of proportionality = GM e m 2

13. Weightlessness Consider an object of mass m hanging from a spring balance which is itself hanging from the roof of a lift mg T a

14. The body is subjected to a downward directed force mg due to the Earth, and an upwards force T, due to the tension in the spring. The net down ward force is (mg – T ), and by Newton´s second law mg – T = ma Where a is downward directed acceleration of the body

15. If the lift is stationary, or is moving with a constant speed, a = 0 and therefore T = mg i.e. The balance registers the weight of the body as mg However, if the lift is falling freely under gravity, both it and the body have a downward directed acceleration of g i.e. g = a

16. It follows from the equation that T = 0 i.e. The balance registers the weight of the body as zero It is usual to refer to a body in this situation as being weightless The term should be used with care, a gravitational pull of magnitude mg acts on the body whether it is in free fall or not, and therefore, in the strictest sense it has weight even when in free fall.

17. The reason it is said to be weightless is that, whilst falling freely, it exerts no force on its support. Similarly, a man standing on the floor of a lift would exert no force on the floor if the lift were in free fall. In accordance with Newton´s third law, the floor of the lift would exert no upwards push on the man and therefore he would not have the sensation of weight.

18. Orbital Weightlessness An astronaut in an orbiting spacecraft has a centripetal acceleration equal to g 1, where g 1 is the acceleration due to gravity at the height of the orbit The spacecraft also has the same centripetal acceleration The astronaut therefore has no acceleration relative to his spacecraft, i.e. he is weightless