Question

Let $E$ be the energy required to raise a satellite to height $h$ above earth's surface and $E^{\prime }$ be the energy required to put the same satellite into orbit at that height. Then $\frac{E}{E^{{}^{\prime }}}$ is equal to

• A. $\frac{2h}{(R+2h)}$
  
• B. $\frac{2h}{(2R+3h)}$
  
• C. $\frac{R}{(R+h)}$
  
• D. $\frac{R}{(2h+R)}$
  

Answer 1

The correct option is A $\frac{2h}{(R+2h)}$


As the velocity of the satellite on the earth is zero so, the total mechanical energy of the satellite at the surface of the earth will be equal to the potential energy,

$E_1=U_1=-\frac{GMm}{R}$

Total mechanical energy of the satellite at altitude $h$,

$E_2=U_2=-\frac{GMm}{(R+h)}$

The energy required to raise the satellite by height $h$ will be:

$E=E_2-E_1=-\frac{GMm}{(R+h)}+\frac{GMm}{R}$

$\Rightarrow E=\frac{GMmh}{R(R+h)}...\left(1\right)$

The total mechanical energy of a satellite in a circular orbit of radius $r$ will be equal to the sum of kinetic and potential energy of the satellite, which is given by, $$E=-\frac{GMm}{2r}$$ Total mechanical energy of the satellite at $\text{r = R + h,}$

$E_3=-\frac{GMm}{2(R+h)}$

The energy required to put the satellite in orbit will be:

$E^{\prime }=E_3-E_1=-\frac{GMm}{2(R+h)}+\frac{GMm}{R}$

$\Rightarrow E^{\prime }=GMm\left[\frac{(R+2h)}{2R(R+h)}\right]...\left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$,

$\frac{E}{E^{\prime }}=\frac{\frac{GMmh}{R(R+h)}}{\frac{GMm(R+2h)}{2R(R+h)}}$

$\Rightarrow \frac{E}{E^{\prime }}=\frac{2h}{(R+2h)}$

Hence, option (a) is the correct answer.

Key Concept: The work done (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.