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

Answer:

1.54 m/s²

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

$w=\frac{2\pi}{T}$

Where T is the period, so

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

Then,

$w=\frac{2\pi}{6600\text{ s}}=9.52\times10^{-4}\text{ rad/s}$

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

$\begin{gathered} g=w^2r \\ g=(9.52\times10^{-4})^2(1.7\times10^6) \\ g=1.54\text{ m/s}^2 \end{gathered}$

Therefore, the answer is 1.54 m/s²