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

Answer 1

- Hint – In order to solve this question, firstly we will use the equation of gravitational potential energy i.e. $P.E = \dfrac{{ - Gm}}{r}$. Then we will find the kinetic energy from the total energy calculated to get the required result.

Formula used-

1. $P.E = \dfrac{{ - Gm}}{r}$
2. ${V_{orbit}} = \sqrt {\dfrac{{GM}}{R}}$
   We are given that-
   Mass of satellite = m
   Mass of planet = M
   Radius = R
   Altitude h = 2R

Complete step-by-step solution -

Now, gravitational potential energy is the energy stored in an object as the result of its vertical position or height.
Using the equation of gravitational potential energy,
We get-
$P.E = \dfrac{{ - Gm}}{r}$
Potential energy at altitude$= \dfrac{{GmM}}{{3R}}$
Orbital velocity- Orbital velocity is the velocity at which a body revolves around the other body. The velocity of orbit depends on the distance between the object and the center of the earth.
${V_{orbit}} = \sqrt {\dfrac{{GM}}{R}}$
Where mass of the body at center = M
Gravitational constant = G
Radius of the orbit = R
Here,${V_0} = \dfrac{{GmM}}{{R + h}}$
Now the total energy is-
${E_f} = \dfrac{1}{2}m{v_0}^2 - \dfrac{{GmM}}{{3R}}$
$\Rightarrow {E_f} = \dfrac{1}{2}\dfrac{{GmM}}{{3R}} - \dfrac{{GMm}}{{3R}}$
$\Rightarrow {E_f} = \dfrac{{GmM}}{{3R}}\left[ {\dfrac{1}{2} - 1} \right]$
$\therefore {E_f} = \dfrac{{ - GmM}}{{6R}}$
Now,${E_i} = {E_f}$ (total energy of satellite is equal to total energy of planet)
Also, the minimum required energy will be-
$K.E = \dfrac{{GmM}}{R} - \dfrac{{GmM}}{{6R}}$
$\therefore K.E = \dfrac{{5GmM}}{{6R}}$
Hence, the minimum required energy is $\dfrac{{5GmM}}{{6R}}$.

Therefore, option A is correct.

Note- While addressing this issue, we must realize that an object's gravitational potential energy is directly proportional to its height above the zero location, a doubling of the height would result in the gravitational potential energy being doubled.