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Q. An artificial satellite is moving in a circular orbit around the earth, with a speed equal to half the magnitude of escape velocity from the Earth. The height of satellite above the surface of earth is (where $R$ is the radius of the Earth)

NTA AbhyasNTA Abhyas 2022

Solution:

$\frac{1}{2}\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{G M}{R \, + \, h}}$
$\frac{1}{4}\times \, \frac{2 G M}{R}= \, \frac{G M}{R \, + \, h}$
$2R \, = \, R \, + \, h$
$h \, = \, R$