Question

A satellite orbiting the moon very near the surface has a period of110 min. What is free-fall acceleration on the surface of the moon?

Answer 1

1.54 m/s²

Step-by-step explanation:

The free-fall acceleration is calculated as

g = w²r

Where w is the angular velocity of the satellite and r is the radius of the moon.

The angular velocity can be calculated as

`w=(2\pi)/(T)`

Where T is the period, so

T = 110 min = 110 x 60 s = 6600 s

Then,

`w=\frac{2\pi}{6600\text{ s}}=9.52*10^(-4)\text{ rad/s}`

Finally, the radius of the moon is r = 1.7 x 10⁶ m, so the free-fall acceleration is

`\begin{gathered} g=w^2r \\ g=(9.52*10^(-4))^2(1.7*10^6) \\ g=1.54\text{ m/s}^2 \end{gathered}`

Therefore, the answer is 1.54 m/s²