what is the orbital period of an imaginary satellite orbiting just above the moons surface?

Answer 1

v^2 = a * r
(orbital velocity ^2 = accelleration * radius from center)

a = .1654 earth's, so * 9.8 (m/s^2) * .1654 = 1.62 m/s^2
r = 1,738.14 km, which = 1,738,140 m (keep units the same)
therefore v = 1,678 m/s, or ~1.7 kps

Answer 2

For a body (satellite) with negligible mass compared with the massive body it is orbiting:
T = 2*pi * squareroot[a^3 / (G*M )]

a is length of orbit's semi-major axis
G is the gravitational constant;
M the mass of the central body

in this case a is (just above) the radius of the moon: 1,737.5 km
and G * M for the moon is 4902 .8 km^3/s^2

The period is close to 6500 seconds, which is 1.8 hours

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