Question 364: Gravitation

Question

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

Accepted Answer

As we know that the kinetic energy of a revolving satellite at a height $h$ above the earth's surface is given as
$$\begin{aligned}
K E=\frac{1}{2} m v^2 & =\frac{G M m}{2 r}\left(\because v=\frac{\sqrt{G M}}{r} 	ext { for circular orbit }ight) \
& =\frac{G M m}{2(R+h)}
\end{aligned}$$
where, R is the radius of the earth.
Given, height of a satellite from the earth surface $h=3 R$ and mass of the satellite $=m$
So,
$$\begin{aligned}
& K E=\frac{G M m}{2(R+3 R)}=\frac{G M m}{2(4 R)} \
& K E=\frac{G M m}{8 R} 	imes \frac{R}{R} \
	herefore\quad & G M m \frac{R}{8 R^2}=\frac{g m R}{8} \quad\left(	herefore g=\frac{G M}{R^2}ight)
\end{aligned}$$
So, the kinetic energy, $K E=\frac{g m R}{8}$.

Answer

$\frac{m g R}{8}$

Answer

$\frac{m g R}{16}$

Answer

$\frac{m g R}{2}$

Answer

$\frac{m g R}{4}$

Gravitation

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

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