Question

What is the minimum eneregy required to launch a satellite of mass $m$ from the surface of earth of radius $R$ in a circular orbit at an altitude of $2R$ from surface of earth?

• A. $\frac{GMm}{R}$
• B. $\frac{GMm}{2R}$
• C. $\frac{3}{4}\frac{GMm}{R}$
• D. $\frac{5}{6}\frac{GMm}{R}$

Answer 1

The correct option is D $\frac{5}{6}\frac{GMm}{R}$

Potential energy in the orbit is
$U_h=-\frac{GMm}{R+h}\text{for}h=2R,$
$U_h=-\frac{GMm}{3R}$
Kinetic energy of the satellite moving with velocity v at a distance $h=2R$ from the surface of earth is
$K_h=\frac{1}{2}m{v}^2=\frac{GMm}{2r}(\because \frac{m{v}^2}{r}=\frac{GMm}{r^2})$
$K_h=\frac{GMm}{2(R+h)}=\frac{GMm}{2(R+2R)}=\frac{GMm}{6R}$

$E_h=U_h+K_h=-\frac{GMm}{3R}+\frac{GMm}{6R}=-\frac{GMm}{6R}$
Since total energy remains conserved,
$K_R+U_R=K_h+U_h$
$\Rightarrow K_R-\frac{GMm}{R}=-\frac{GMm}{6R}$
$\Rightarrow K_R=\frac{GMm}{R}-\frac{GMm}{6R}=\frac{5}{6}\frac{GMm}{R}$
Hence, the correct answer is option (d).