Question

A satellite is revolving around earth with orbital speed $v_0$. If it stops suddenly, find the speed with which it will strike the surface of earth. $(v_e=$ escape velocity).

• A. $\frac{v_e^2}{v_0}$
• B. $v_0$
• C. $\sqrt{v_e^2-v_0^2}$
• D. $\sqrt{v_e^2}-2v_0^2$

Answer 1

The correct option is D $\sqrt{v_e^2}-2v_0^2$

As the satellite is stopped suddenly we have is velocity as zero.
Thus we get the initial energy of the satellite as

$E_1=-\frac{GMm}{R}$

Now when the satellite was in its circular orbit, the centripetal force is given as

$\frac{m{v}_o^2}{R}=\frac{GMm}{R^2}$

or

$m{v}_o^2=\frac{GMm}{R}$

or

$E_1=-\frac{GMm}{R}=-m{v}_o^2$

Now at the earth's surface, the total energy is given as

$E_2=\frac{1}{2}m{v}^2-\frac{GM}{R_e}$

Equating the two energies we get

$-m{v}_o^2=\frac{1}{2}m{v}^2-\frac{GM}{R_e}$

or

$v^2=-2v_o^2+\frac{2GM}{R_e}$

or

$v^2=v_e^2-2v_o^2$

or

$v=\sqrt{v_e^2-2v_o^2}$