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

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.

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.