# A satellite has a mass of 6135 kg and is in a circular orbit 3.80 105 m above the surface of a...

## Question:

A satellite has a mass of 6135 kg and is in a circular orbit 3.80 105 m above the surface of a planet. The period of the orbit is 2.08 hours. The radius of the planet is 3.85 106 m. What would be the true weight of the satellite if it were at rest on the planet's surface? W=

## Satellite Revolution Period:

If a satellite orbites a planet of radius {eq}R {/eq} at a height {eq}h {/eq} from its surface and time period of orbit be {eq}T {/eq}, then we can express acceleration due to gravity on the surface of the planet as:

{eq}\displaystyle{g^{'} = \frac{4\pi^{2}(R + h)^{3}}{T^{2}R^{2}}} {/eq}

True weight of the satellite of mass {eq}m {/eq} on the surface of the planet when it is at rest can be given by:

{eq}\displaystyle{w^{'} = m\, g^{'} = \frac{4\pi^{2}m(R + h)^{3}}{T^{2}R^{2}}} {/eq}

## Answer and Explanation:

Mass of the satellite: {eq}\color{blue}{m = 6135\, \rm kg} {/eq}.

Radius of the planet: {eq}\color{blue}{R = 3.85\times 10^{6}\, \rm m} {/eq}.

Height of the satellite from the surface of the planet: {eq}\color{blue}{h = 3.80\times 10^{5}\, \rm m} {/eq}.

Period of orbit: {eq}\color{blue}{T = 2.08\, \rm h = 7488\, \rm s} {/eq}.

Therefore, true weight of the the satellite on the surface of the planet will be given by:

{eq}\displaystyle{\begin{align*} \color{red}{W} &= \color{blue}{\frac{4\pi^{2}m(R + h)^{3}}{T^{2}R^{2}}}\\ &= \color{blue}{\frac{4\pi^{2}\times 6135\times \left(3.85\times 10^{6} + 3.80\times 10^{5}\right)^{3}}{7488^{2}\times \left(3.85\times 10^{6}\right)^{2}}}\\ &\approx \color{red}{22,056.79\, \rm N}\\ \end{align*}} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from Physics: High School

Chapter 8 / Lesson 16