A satellite orbiting the moon very near the surface has a period of 110 minutes. What is the...

Question:

A satellite orbiting the moon very near the surface has a period of {eq}110 \ min {/eq}.

What is the moon's acceleration due to gravity?

Period in a circular orbit

The period {eq}T {/eq} of body orbiting another more massive body of mass {eq}M {/eq} in a circular orbit with radius {eq}R {/eq} is

{eq}T = \dfrac{2\pi R^{3/2}}{\sqrt{GM}}, {/eq}

where {eq}G {/eq} is the gravitation constant.

Answer and Explanation:

Let the surface of the moon be a distance {eq}R {/eq} from its center. Hence, we assume that the satellite orbiting the moon very near the surface is with circular orbit of radius {eq}R {/eq}. Recall that the acceleration due to gravity near the surface is

{eq}g = \dfrac{GM}{R^2}, {/eq}

where {eq}M {/eq} is the mass of the moon. Given {eq}T = \dfrac{2\pi R^{3/2}}{\sqrt{GM}} = 110 \text{ mins} = 6600 \ \text{s} {/eq}, we can write

{eq}\begin{align} T &= \dfrac{2\pi R^{3/2}}{\sqrt{GM}} \\ \Rightarrow \quad T^2 &= \dfrac{(2\pi)^2 R^3}{GM} \\ &= \dfrac{(2\pi)^2 R}{GM/R^2} = \dfrac{(2\pi)^2 R}{g} \\ \Rightarrow \quad g &= \dfrac{(2\pi)^2 R}{T^2}. \end{align} {/eq}

Now, for the moon {eq}R = 1.74 \times 10^6 \ \text{m} {/eq}, hence

{eq}g = \dfrac{(2\pi)^2 (1.74 \times 10^6 \ \text{m})}{(6600 \ \text{s})^2} = 1.58 \ \text{m}/\text{s}^2. {/eq}

Therefore, the moon's acceleration due to gravity near the surface is {eq}1.58 \ \text{m}/\text{s}^2 {/eq}.


Learn more about this topic:

Loading...
Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
42K

Related to this Question

Explore our homework questions and answers library