The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is:

The correct option is 5GmM/6R.

Let us find the minimum energy required to launch a satellite of mass _$m$ from the surface of the Earth to a circular orbit at an altitude of _$2R$ from the surface of the Earth.

Step 1: Find the initial energy of the satellite on the surface of the Earth
Initially, the satellite is at rest on the surface of the Earth.
The distance from the center of the Earth is _{$r_i=R$}.
The initial kinetic energy (_{$K_i$}) is 0 because it is at rest on the surface before launch.
The initial potential energy (_{$U_i$}) is given by:
$U_i=-\frac{GMm}{R}$
Therefore, the total initial energy (_{$E_i$}) is:
$E_i=K_i+U_i=0-\frac{GMm}{R}=-\frac{GMm}{R}$

Step 2: Find the final energy of the satellite in the circular orbit
The satellite is in a circular orbit at an altitude of _$2R$ from the surface of the Earth.
Thus, the orbital radius (_$r$) from the center of the Earth is:
$r=R+2R=3R$
For a satellite in a stable circular orbit of radius _$r$, the total mechanical energy (_{$E_f$}) is:
$E_f=-\frac{GMm}{2r}$
Substituting _$r=3R$ into the equation:
$E_f=-\frac{GMm}{2\left(3R\right)}=-\frac{GMm}{6R}$

Step 3: Calculate the minimum energy required for the launch
The minimum energy required (_$W$) is the difference between the final total energy and the initial total energy:
$W=E_f-E_i$
$W=-\frac{GMm}{6R}-\left(-\frac{GMm}{R}\right)$
$W=\frac{GMm}{R}-\frac{GMm}{6R}$
$W=\left(1-\frac{1}{6}\right)\frac{GMm}{R}$
$W=\frac{5GMm}{6R}$

Thus, the minimum energy required to launch the satellite is 5GmM/6R.