Question

An artificial satellite is revolving round the earth in a circular orbit. Its velocity is half of the escape velocity. Its height from the earth's surface is:

• A. $6400km$
• B. $12800km$
• C. $800km$
• D. $1600km$

Answer 1

The correct option is A $6400km$

As the satellite orbits in a circular path, the centripetal force is provided by the Gravitational force.
$\therefore \frac{m{v}^2}{R+h}=\frac{GMm}{(R+h{)}^2}$
$\Rightarrow v^2=\frac{GM}{R+h}$.........(1)
The escape velocity is given by,
$v_e=\sqrt{\frac{2GM}{R}}$
It is given that the speed of the satellite is half of the escape speed.

From equation (1)

$\therefore {\left[\frac{1}{2}\sqrt{\frac{2GM}{R}}\right]}^2=\frac{GM}{R+h}$
$\Rightarrow \frac{GM}{2R}=\frac{GM}{R+h}$

$\Rightarrow 2R=R+h\Rightarrow h=R\Rightarrow h=6400$ km