Question

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 surface of earth. (Radius of earth $=6400km)$. Determine the height of the satellite above the earth's surface.

Answer 1

From Newton's second law,

$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$

$v=\sqrt{\frac{GM}{r}}$

Thus, as per the question,

$\sqrt{\frac{GM}{r}}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

$r=2R=12800km$

Answer is $12800km$