Question

Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is

Answer 1

Answer:

The work down on satellite X by the force in terms of Kx is $\dfrac{-K_x}{4}$.

Explanation:

The work done is given as in terms of

$W=\Delta TE$

Where ΔTE is the change in total energy.

This is given as

$W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}$

Rearranging it in  terms of K_x gives

$W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}$