Question

A satellite of mass $m$ is revolving around the earth at height $R$ (radius of the earth) from the earth's surface. Its potential energy will be:

• A. $mgR$
• B. $0.67mgR$
• C. $-\frac{mgR}{2}$
• D. $0.33mgR$

Answer 1

The correct option is C $-\frac{mgR}{2}$

At a height h, acceleration due to gravity be $g^{\prime }$

As, $g^{\prime }=g(1+\frac{h}{R}{)}^{-2}$

Now, for $h=R,⟹g^{\prime }=\frac{g}{4}$

potential energy, $P.E=-\frac{GMm}{(R+h)}$ , and $g^{\prime }=\frac{GM}{(R+h{)}^2}$

$⟹P.E=-m{g}^{\prime }(R+h)$

For $h=R,P.E=-m\times \frac{g}{4}\times 2R$

Thus, $P.E=-\frac{mgR}{2}$