At what velocity would a satellite orbit a planet of mass...?

Question

At what velocity would a satellite orbit a planet of mass 20 * 10^24 kg at a distance of 39700 km from the center of the planet?
•What would be the orbital KE of the satellite if its mass was 790 kg?
•How much would PE and KE of this satellite change as the satellite moved from a circular orbit of radius 39700 km to a circular orbit of radius 43670 km?

Answer 1

If an airborne object is closer to the Planet, it will experience a strong grav. force of Planet and would be forced fall onto the Planet. The satellite (m, v) uses the rocket to cross the strong grav. field of Planet, and is placed in rarefied grav force field.

At very great distances from Planet's center, force of gravity becomes appreciably small so that, any object placed there would be forced to move in a circular orbit.

Now the sufficient and adequate condition for satellite to orbit the planet is that the centripetal force required to remain in the circle should be fully provided by the grav. Force between satellite and planet (M) at that distance ( r ); so net force on satellite is zero. F(grav) – F(centripel) =0

G M m / r^2 = m v^2 / r v^2 = G M / r ----- (1)
Part (a): v = sqrt (G M / r) = sqrt [6.67*10-11* 20*10^24 / 39700*1000]
v = 5796.72 m/sec Orbital velocity
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(b) orbital KE = (1/2) m * v^2 = 0.5*790*5796.72*5796.72
= 1.327 * 10^10 joule = G M m /2r

PE = - G M m / r (attractive) = - 2.654 * 10^10 joule

Total Energy =PE + KE = - G M m / r + G M m /2r = - G M m /2 r
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(c) satellite goes from r radius orbit to r1 radius (bigger orbit)
(KE)r = G M m /2r and (KE)r1 = G M m /2r1
(KE)r1 = (KE)r [39700 / 43670 ] = 0.909 times earlier KE
= Reduces to = 1.206 * 10^10 joule
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(PE)r1 = - G M m /r1 = - [G M m /r] (r/r1)
= - 2.654 * 10^10 * 0.909 = joule
= - 2.627 * 10^10 * 0.909 = joule
(PE)r1 = increased ( or become lesser negative or twars origin (0.0). PE is zero (maximum at r = infinity) - when satellite goes bigger orbits ( >> infinity) PR increases