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

Answer 1

Answer: (A)

Given that,
Mass of satellite $= m$
Mass of planet $= M$
Radius $= R$
Altitude $h =2 R$
Now,
The gravitational potential energy
P.E $=\frac{- Gm }{ r }$
Potential energy at altitude $=\frac{G m M}{3 R}$
Orbital velocity $v _{0}=\frac{ GmM }{ R + h }$
Now, the total energy is
$E_{f}=\frac{1}{2} m v_{0}^{2}-\frac{G m M}{3 R}$
$E_{f}=\frac{1}{2} \frac{G m M}{3 R}-\frac{G M m}{3 R}$
$E_{f}=\frac{G m M}{3 R}\left[\frac{1}{2}-1\right]$
$E_{f}=\frac{-G m M}{6 R}$
Now, $E_{i}=E_{f}$
Now, the minimum required energy
$K \cdot E =\frac{ Gmm }{ R }-\frac{ GmM }{6 R }$
$K \cdot E =\frac{5 GmM }{6 R }$
Hence, the minimum required energy is $\frac{5 GmM }{6 R }$