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

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 of radius $R$ and mass $M$ into a circular orbit at an altitude of $2R$ step-by-step.

Step 1: Calculate the initial energy of the satellite on the surface of the Earth.
Initially, the satellite is at rest on the surface of the Earth. Its distance from the center of the Earth is $r_i=R$.
Since it is at rest before launching, its initial kinetic energy is zero:
$K_i=0$
Its initial gravitational potential energy is:
$U_i=-\frac{GMm}{R}$
Therefore, the total initial energy ($E_i$) of the satellite is:
$E_i=K_i+U_i=-\frac{GMm}{R}$

Step 2: Calculate the final energy of the satellite in the circular orbit.
The satellite is placed in a circular orbit at an altitude of $h=2R$ above the Earth's surface.
The orbital radius ($r_f$) from the center of the Earth is:
$r_f=R+h=R+2R=3R$
For a satellite in a stable circular orbit of radius $r$, its total mechanical energy ($E_f$) is given by the formula:
$E_f=-\frac{GMm}{2r}$
Substituting $r=3R$ into the expression, we get the total final energy:
$E_f=-\frac{GMm}{2\left(3R\right)}=-\frac{GMm}{6R}$

Step 3: Calculate the minimum energy required for the launch.
The minimum energy required ($E$) to launch the satellite is the difference between its total final energy and its total initial energy:
$E=E_f-E_i$
Substituting the values of $E_i$ and $E_f$:
$E=-\frac{GMm}{6R}-\left(-\frac{GMm}{R}\right)$
$E=\frac{GMm}{R}-\frac{GMm}{6R}$
Taking the common denominator:
$E=\frac{6GMm-GMm}{6R}=\frac{5GMm}{6R}$

Thus, the minimum energy required to launch the satellite is $\frac{5GMm}{6R}$.