Question

A satellite orbiting the moon very near the surface has a period of 110 minutes. What is the free fall acceleration of the moon (Gmoon) ? Answer has to be with one decimal and 2 sig figs with m/s2 unit

Hint: being close to the surface means that the height H above the surface is negligible.

Answer 1

write the expression for centripetal acceleration.

a = v^2/r

Where v is the velocity of the satellite and r is the radius of the orbit.

Since we are near the surface we can take the altitude of the satellite to be 0 so the radius of it's orbit will be the same as the radius of the moon.

Circumference = 2*pi*r = 2*pi*1.74x10^6 = 1.093x10^7 m
Period = 110 minutes = 6600 seconds
v = C/Period = 1.093x10^7/6600 = 1.656x10^3 m/s

a = (1.656x10^3)^2/1.74x10^6 = 1.57 m/s^2

So the acceleration at the moon's surface would be approximately 1.57 m/s^2