Question

What is minimum energy required to launch a satellite of mass m from the surface of the earth(radius = R) in a circular orbit at an altitude of 2R from the centre?

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

Answer 1

The correct option is A

$\frac{5GmM}{6R}$

Consider a satellite of mass m moving with a speed v at altitude r (measured from the centre of the earth).

Then Kinetic energy (KE) = $\frac{1}{2}m{v}^2$

Gravitational potential energy (PE) = - $\frac{GmM}{r}$

Where M is the mass of the earth.

For a satellite in circular orbit, we have

$\frac{m{v}^2}{r}=\frac{GmM}{r^2}or{v}^2=\frac{GM}{r}$

Or $\frac{1}{2}m{v}^2=\frac{GmM}{2r}$

i.e. $KE=\frac{GmM}{2r}$

Thus the KE of a satellite in a circular orbit is numerically half its PE but opposite in sign,

The total energy of the satellite in orbit is

$E=KE+PE=\frac{GmM}{2r}-\frac{GmM}{r}$

$=-\frac{GmM}{2r}$
It is given that r = 2R + R = 3R, where R is the radius of the earth.

$\therefore E=-\frac{GmM}{6R}$

Now PE on the surface of the earth = - $\frac{GmM}{R}$

$\therefore$ MInimum energy required $\left(E_{min}\right)$

=-$\frac{GmM}{6R}-(-\frac{GmM}{R})$

= $\frac{5GmM}{6R}$