Question

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

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

Answer 1

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


Energy of the system when the satellite is at the surface of the planet,
$E_i=\frac{-GMm}{R}$

Energy of the system when the satellite is at the altitude of $2R$ from the surface of the planet,
$E_f=\frac{-GMm}{3R}+\frac{1}{2}m{v}^2$

Also, required centripetal force $=$ gravitational force
$\Rightarrow \frac{m{v}^2}{3R}=\frac{GMm}{(3R{)}^2}$
$\Rightarrow v^2=\frac{GM}{3R}$

$\therefore E_f=\frac{-GMm}{3R}+\frac{1}{2}m\times \frac{GM}{3R}=\frac{-GMm}{6R}$

Now, required energy $=$ change in energy
$\Rightarrow E=E_f-E_i=\frac{-GMm}{6R}-\left(\frac{-GMm}{R}\right)$
$\Rightarrow E=\frac{5GMm}{6R}$