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 satellite of mass $m$ is initially on the Earth’s surface. The Earth has mass $M$ and radius $R$. We need the minimum mechanical energy that must be supplied to place the satellite into a circular orbit whose altitude is 2 R above the surface.

**1. Initial energy on the surface** When the satellite is at rest on the surface, its kinetic energy is zero. Its gravitational potential energy (taking zero at infinity) is

$-\frac{GMm}{R}$

Thus,

$\mathrm{E\_i}=\; -\frac{G}{M}/R$

**2. Radius of the desired orbit** Altitude = 2 R, so the orbital radius measured from the Earth’s centre is

$\mathrm{r\_f}=R+\; 2R=\; 3R$

**3. Energy of a circular orbit** For a circular orbit of radius $r$, the kinetic energy is half the magnitude of the (negative) potential energy:

$K=\; \backslash frac\{1\}\{2\}\backslash left(\backslash frac\{G\; M\; m\}\{r\}\backslash right)$

The gravitational potential energy is

$U=\; -\backslash frac\{G\; M\; m\}\{r\}$

Hence the total mechanical energy of the orbit is

$\mathrm{E\_f}=\; K\; +\; U\; =\; -\backslash frac\{G\; M\; m\}\{2\; r\}$

Substituting the final radius $r\_f\; =\; 3R$ gives

$\mathrm{E\_f}=\; -\backslash frac\{G\; M\; m\}\{2\; \backslash cdot\; 3R\}\; =\; -\backslash frac\{G\; M\; m\}\{6R\}$

**4. Minimum energy that must be supplied** The required energy is the increase in mechanical energy:

$\mathrm{\Delta E}=\; E\_f\; -\; E\_i$

Insert the expressions for $E\_f$ and $E\_i$:

$\mathrm{\Delta E}=\; -\backslash frac\{G\; M\; m\}\{6R\}\; -\; \backslash left(-\backslash frac\{G\; M\; m\}\{R\}\backslash right)$

Simplify:

$\mathrm{\Delta E}=\; -\backslash frac\{G\; M\; m\}\{6R\}\; +\; \backslash frac\{G\; M\; m\}\{R\}\; =\; G\; M\; m\backslash !\backslash left(\backslash frac\{1\}\{R\}\; -\; \backslash frac\{1\}\{6R\}\backslash right)\; =\; G\; M\; m\backslash !\backslash left(\backslash frac\{6-1\}\{6R\}\backslash right)\; =\; \backslash frac\{5\; G\; M\; m\}\{6R\}$

**5. Result** The minimum energy required to launch the satellite into the specified orbit is

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

which matches the given correct option.