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. $\frac{GmM}{2R}$
• D. $\frac{GmM}{2R}$

Answer 1

Answer: (A)

Given that,
Mass of satellite $=m$
Mass of planet $=M$
Radius $=R$
Altitude $h=2R$
Now,
The gravitational potential energy
P.E $=\frac{-Gm}{r}$
Potential energy at altitude $=\frac{GmM}{3R}$
Orbital velocity $v_0=\frac{GmM}{R+h}$
Now, the total energy is
$E_f=\frac{1}{2}m{v}_0^2-\frac{GmM}{3R}$
$E_f=\frac{1}{2}\frac{GmM}{3R}-\frac{GMm}{3R}$
$E_f=\frac{GmM}{3R}\left[\frac{1}{2}-1\right]$
$E_f=\frac{-GmM}{6R}$
Now, $E_i=E_f$
Now, the minimum required energy
$K\cdot E=\frac{Gmm}{R}-\frac{GmM}{6R}$
$K\cdot E=\frac{5GmM}{6R}$
Hence, the minimum required energy is $\frac{5GmM}{6R}$