Question

The kinetic energy of a revolving satellite (mass m) at a height equal to thrice the radius of the earth (R) is ______.

Options

• ${\displaystyle \frac{\text{mgR}}{8}}$

• ${\displaystyle \frac{\text{mgR}}{16}}$

• ${\displaystyle \frac{\text{mgR}}{2}}$

• ${\displaystyle \frac{\text{mgR}}{4}}$

Answer 1

The kinetic energy of a revolving satellite (mass m) at a height equal to thrice the radius of the earth (R) is ${\displaystyle \underline{\frac{\text{mgR}}{8}}}$.

Explanation:

As we know that the kinetic energy of a revolving satellite at a height h above the earth's surface is given as

KE = ${\displaystyle \frac{1}{2}{\text{mv}}^2=\frac{\text{GMm}}{\text{2r}} ...(\because \text{v}=\frac{\sqrt{\text{GM}}}{\text{r}}\text{for circular orbit})}$

${\displaystyle =\frac{\text{GMm}}{2\left(\text{R + h}\right)}}$

where, R is the radius of the earth.

Given, height of a satellite from the earth surface h = 3R and mass of the satellite = m

So, KE = ${\displaystyle \frac{\text{GMm}}{2(\text{R}+3\text{R})}=\frac{\text{GMm}}{2\left(\text{4R}\right)}}$

KE = ${\displaystyle \frac{\text{GMm}}{8\text{R}}\times \frac{\text{R}}{\text{R}}}$

∴ GMm${\displaystyle \frac{\text{R}}{8{\text{R}}^2}=\frac{\text{gmR}}{8} ...(\therefore \text{g}=\frac{\text{GM}}{{\text{R}}^2})}$

So, the kinetic energy, KE = ${\displaystyle \frac{\text{gmR}}{8}}$