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Q.
A potential energy of satellite, having mass m and rotating at height of $ 6.4\times {{10}^{6}}m $ from the earth surface will be:
ManipalManipal 2000
Solution:
Here : Mass of the satellite $=m$
Height of the satellite from the earth
$h=6.4 \times 10^{6} m$
Gravitation potential energy of the satellite at a height $h$ is given by,
$U=-\frac{G M_{e} m}{R_{e}+h}=-\frac{g R_{e}^{2} m}{2 R_{e}}$
$=-\frac{g R_{e} m}{2}=-0.5\, m g R_{c}$
(where $G M_{e}=g R_{e}^{2}$ and $h=R_{e}$ )