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

Answer 1

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

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 $v_0=\sqrt{\frac{2GmM}{R+h}}$

Now, the total energy is

$E_f=\frac{1}{2}m{v}_0^2-\frac{GmM}{3R}$

$E_f=\frac{1}{2}\frac{GmM}{3R}-\frac{GMm}{3R}$

$E_f=\frac{GmM}{3R}[\frac{1}{2}-1]$

$E_f=\frac{-GmM}{6R}$

Now, $E_i=E_f$

Now, the minimum required energy

$K.E=\frac{Gmm}{R}-\frac{GmM}{6R}$

$K.E=\frac{5GmM}{6R}$

Hence, the minimum required energy is $\frac{5GmM}{6R}$