Question

If velocity of a satellite is half of escape velocity, then distance of the satellite from earth surface will be.

• A. $6400km$
• B. $12800km$
• C. $6400\sqrt{2}km$
• D. $\frac{6400}{\sqrt{2}}km$

Answer 1

The correct option is A $6400km$

Orbital velocity , $v=\sqrt{\frac{GM}{r}}$ where $r$ is the radius of orbit

For $v=\frac{v_e}{2}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

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

As $r=h+R$, thus $h=R=6400km$