Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed (in $km/sec$) with which it hits the surface of the earth ($g=10m/s^2$ and $R=6400km$).

Answer 1

Given $v_0=v_0/2$
${\left(\frac{GM}{R+h}\right)}^{1/2}=\frac{1}{2}{\left(\frac{2GM}{R}\right)}^{1/2}$
On solving $h=R$
From law of conservation of energy,
$-\frac{GMm}{(R+h)}=\frac{1}{2}m{v}^2-\frac{GMm}{R}$
or $\frac{1}{2}m{v}^2=\frac{GMm}{R}-\frac{GMm}{2R}=\frac{GMm}{2R}$
or $v=\sqrt{\frac{GM}{R}}=\sqrt{gR}$