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  4. The magnitude of binding energy of the satellite is E and kinetic energy is K . The ratio E / K is

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Q. The magnitude of binding energy of the satellite is $E$ and kinetic energy is $K$ . The ratio $E / K$ is

NTA AbhyasNTA Abhyas 2020Gravitation

Solution:

Kinetic energy of satellite $\left(\right.E\left.\right)$
$KE\left(k\right)=\frac{1}{2}mv^{2}$
$=\frac{1}{2}m\left(\sqrt{\frac{G M}{R}}\right)^{2}$ $\left(\because v = \sqrt{\frac{G M}{R}}\right)$
[ $m =$ mass of satellite]
$=\frac{1}{2}m\frac{G M}{R}$
[Symbols have usual meanings]
Potential energy of satellite,
$U=-m\frac{G M}{R}$
$\therefore \, $ Binding energy of the satellite,
$E_{b}=KE+U$
$=\frac{1}{2}m\frac{G M}{R}+\left(- m \frac{G M}{R}\right)$
$=\frac{1}{2}m\frac{G M}{R}-m\frac{G M}{R}$
$=-\frac{G M m}{2 R}$
$\therefore \, \, \, $ Magnitude of the binding energy $E=\left|E_{b}\right|=\frac{G M m}{2 R}$
The ratio of $\frac{E}{K}=\frac{G M m}{2 R}\times \frac{2 R}{G M m}=1$