Question

A satellite circles the moon at a distance above its surface equal to 2 times the radius of the moon. The acceleration due to gravity of the satellite, as compared to the acceleration due to gravity on the surface of the moon is

Answer 1

The orbit is a circle whose radius is 3 times the radius of the surface
(both measured from the center of the moon). So the acceleration due
to gravity at the orbital altitude is

1/3² = 1/9 = 11.1% of its value on the surface.

Answer 2

From Newton's Law of Universal Gravitation:

F = GMm / r^2. ( When at the surface).

G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Moon
m = Mass of Satellite
r = distance apart, between centers = in this case it is the distance from Moon to the Satellite.

Recall: F = mg.

mg = GMm / r^2
g = GM / r^2.........................(i). When at surface.

Note when the satellite is at a distance 2 times the radius of the moon.
Therefore, the distance between centers = 2r + r = 3r.
Note, when need to add radius of the moon, because we are measuring from center of the satellite to center of the moon.

From (i)
g = GM / (3r)^2. The distance r is replaced with 3r

g = GM / 9r^2 = (1/9) * GM / r^2

Therefore gravity on the satellite is (1/9) times that on the Moon.