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.

1) In terms of Kx, the gravitational potential energy of the planet-satellite X system is
(A) -2Kx
(B) -Kx
(C) -Kx/2
(D) Kx/2
(E) 2Kx

2) 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
(A) -Kx/2
(B) -Kx/4
(C) 0
(D) +Kx/4
(E) +Kx/2

Show the work for each question.

The gravitational potential energy of the planet-satellite X system is [tex]2 Kx[/tex] and the work done on satellite X by the force is zero. The option E and C is correct.

The kinetic energy of satellite [tex]X[/tex]:

[tex]KE= \dfrac {GMm}{2(3r)}[/tex].........................2

Gravitational energy of the planet,

[tex]GE= \dfrac {GMm}{3r}[/tex]..........................1

From equations 1 and 2,

[tex]Kx = \dfrac 12 GE \\\\GE = 2 Kx[/tex]

Thus, the gravitational potential energy of the planet-satellite X system is [tex]2 Kx[/tex].

2) The direction of the gravitational force and the displacement of the satellite in circular orbit are perpendicular to each other.

Therefore, the work done on satellite X by the force is zero because displacement is not caused by the force.

To know more about gravitational potential energy:

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