# A satellite has a mass of 5823 kg and is in a circular orbit 4.45 10 5 m above the surface of...

## Question:

A satellite has a mass of 5823 kg and is in a circular orbit {eq}4.45\times10^{5} m {/eq}above the surface of a planet. The period of the orbit is 1.9 hours. The radius of the planet is {eq}4.57\times10^{6} m {/eq}. What would be the true weight of the satellite if it were at rest on the planet s surface?

## Time Period & True Weight:

Let a satellite of mass {eq}m {/eq} be orbiting at a height {eq}h {/eq} from the surface of a planet of mass {eq}M {/eq} and radius {eq}R {/eq}. Then we can derive time period of the orbit of the satellite as:

{eq}\displaystyle{T = 2\pi\, \sqrt{\frac{(R + h)^3}{GM}}} {/eq}.

{eq}\displaystyle{\therefore GM = \frac{4\pi^2}{T^2}(R + h)^3} {/eq}.

Then, weight of the satellite on the surface of the planet can be given by:

{eq}\displaystyle{\begin{align*} \color{green}{w} &= \color{green}{\frac{GMm}{R^2}}\\ &= \color{green}{\frac{4\pi^2(R + h)^3m}{T^2R^2}}\\ \end{align*}} {/eq}

## Answer and Explanation:

Mass of the satellite: {eq}\color{blue}{m = 5823\, \rm kg} {/eq}.

Height of the satellite from the surface of the planet: {eq}\color{blue}{h = 4.55\times 10^5\, \rm m} {/eq}.

Time period of the orbit: {eq}\color{blue}{T = 1.9\, \rm h = 1.9\times 3600\, \rm s} {/eq}

Radius of the planet: {eq}\color{blue}{R = 4.57\times 10^6\, \rm m} {/eq}.

From the data given above, we can easily give true weight of the satellite on the surface of the planet as:

{eq}\displaystyle{\begin{align*} \color{red}{w} &= \color{blue}{\frac{4\pi^2(R + h)^3m}{T^2R^2}}\\ &= \color{blue}{\frac{4\pi^2\left(4.57\times 10^6\, \rm m + 4.45\times 10^5\, \rm m\right)^3\times 5823\, \rm kg}{(1.9\times 3600\, \rm s)^2\times \left(4.57\times 10^6\, \rm m\right)^2}}\\ &\approx \color{red}{29,673.91\, \rm N}\\ \end{align*}} {/eq}

#### Learn more about this topic:

Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
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