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

Answer 1

The correct option is C $\frac{5GmM}{6R}$

The energy required is the difference in the energies in the orbit and the energy on the surface of the earth,
$KE=\frac{m{v}^2}{2}$
We know that for orbital velocity
$v^2=\frac{GM}{r_{orbit}}$
$\Rightarrow v^2=\frac{GM}{3R}$
Energy $=\Delta U+KE$
$=[\frac{GMm}{R}-\frac{GMm}{3R}]+\frac{1}{2}m\left(\frac{GM}{3R}\right)=\frac{5}{6}\frac{GMm}{R}$