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Expert-verified Found in: Page 413 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Derive an expression for the work required to move an Earth satellite of mass m from a circular orbit of radius to one of radius

The derivation will be $\Delta E=\frac{G{M}_{E}m}{12{R}_{E}}$

See the step by step solution

## Step 1: Given

$\text{Mass of the satellite}=m\phantom{\rule{0ex}{0ex}}\text{circular}\text{orbit}\text{radius from}{\text{2R}}_{E}\text{to}3{R}_{E}$

## Step 2: Concept

Earth is a planet that revolves around the sun, and the moon is the natural satellite of Earth. Apart from this, the artificial satellite of the earth can be set in orbit around the earth.

For placing any kind of satellite in the earth’s orbit, the earth satellite energy requirement is as follows:

$E=-\frac{G{M}_{E}}{2r}$

## Step 3: Derive an expression

When the circular orbit radius changes from the radius $r=2{R}_{E}\text{}to\text{}r=3{R}_{E}$ required to work, W will be required to follow.

$W=\Delta E\phantom{\rule{0ex}{0ex}}\Delta E=\frac{G{M}_{E}m}{2{r}_{f}}+\frac{G{M}_{E}m}{2{r}_{i}}\phantom{\rule{0ex}{0ex}}\Delta E=G{M}_{E}m\left[\frac{1}{4{R}_{E}}-\frac{1}{6{R}_{E}}\right]\phantom{\rule{0ex}{0ex}}\Delta E=\frac{G{M}_{E}m}{12{R}_{E}}$

Here,

$\text{W = workdone}\phantom{\rule{0ex}{0ex}}\Delta E=\text{Energy}\phantom{\rule{0ex}{0ex}}{M}_{E}=\text{mass of the earth}\phantom{\rule{0ex}{0ex}}G=\text{gravitational constant}\phantom{\rule{0ex}{0ex}}{r}_{f}=\text{final radius}\phantom{\rule{0ex}{0ex}}{r}_{i}=\text{initial radius}\phantom{\rule{0ex}{0ex}}m=\text{mass of the satellite}$

Hence, the derivation will be $\Delta E=\frac{G{M}_{E}m}{12{R}_{E}}$ . ### Want to see more solutions like these? 