Question

If 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, the height of the satellite above the surface of the earth is (where $R=$ Radius of the earth)

• A. $2R$
• B. $\frac{R}{2}$
• C. $R$
• D. $\frac{R}{4}$

Answer 1

Answer: (C)

Escape velocity, $v_e=\sqrt{\frac{2GM}{R}}$
Orbital velocity, $v_o=\sqrt{\frac{GM}{R+h}}$
Given that $v_o=\frac{v_e}{2}$
$\therefore \sqrt{\frac{GM}{R+h}}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$
or $\frac{GM}{R+h}=\frac{1}{4}\times \frac{2GM}{R}$
On solving, we get $h=R$