Question

The minimum energy required to launch a $m$ kg satellite from the earth's surface into a circular orbit at an altitude $2\text{R}$, where $R$ is the radius of earth is

• A. $\frac{5}{3}mgR$
  
• B. $\frac{4}{3}mgR$
  
• C. $\frac{5}{6}mgR$
• D. $\frac{5}{4}mgR$

Answer 1

The correct option is C $\frac{5}{6}mgR$

Since, the kinetic energy near the surface of the Earth is zero, so total mechanical energy of the satellite near the surface of the earth $(r=R)$, $$E_1=U_1=-\frac{GMm}{R}$$ Where, $U_1$ is the potential energy.

The total mechanical energy of a satellite in a circular orbit of radius $r$ is given by, $$E=-\frac{GMm}{2r}$$So, total mechanical energy of the satellite at an altitude $2R$i.e.,at $r=R+2R=3R$,

$E_2=-\frac{GMm}{2\left(3R\right)}=-\frac{GMm}{6R}$

Let $K$ be the energy required to launch the satellite into the orbit of altitude $2R$. From energy conservation,

$E_1+K=E_2$

$\Rightarrow -\frac{GMm}{R}+K=\frac{-GMm}{6R}$

$\Rightarrow K=\frac{5GMm}{6R}$

As, $g=\frac{GM}{R^2}$

$\Rightarrow K=\frac{5g{R}^{2}m}{6R}=\frac{5mgR}{6}$

Hence, option (c) is the correct answer.

Key Concept: The energy required to shift a satellite from one configuration to the other is equal to the difference between the total mechanical energy of the two configurations.