Gravitation 3 Question 2
5. The energy required to take a satellite to a height ’ $h$ ’ above earth surface (where, radius of earth $=6.4 \times 10^{3} km$ ) is $E _1$ and kinetic energy required for the satellite to be in a circular orbit at this height is $E _2$. The value of $h$ for which $E _1$ and $E _2$ are equal is
(2019 Main, 9 Jan II)
(a) $3.2 \times 10^{3} km$
(b) $1.28 \times 10^{4} km$
(c) $6.4 \times 10^{3} km$
(d) $1.6 \times 10^{3} km$
Show Answer
Answer:
Correct Answer: 5. (a)
Solution:
- The energy required for taking a satellite upto a height $h$ from earth’s surface is the difference between the energy at $h$ height and energy at surface, then
$\Rightarrow$
$$ E _1=U _f-U _i $$
$$ E _1=-\frac{G M _e m}{R _e+h}+\frac{G M _e m}{R _e} $$
(where, $U$ =potential energy)
$\therefore$ Orbital velocity of satellite,
$$ v _o=\sqrt{\frac{G M _e}{\left(R _e+h\right)}} \quad\left(\text { where, } M _e=\right.\text { mass of earth) } $$
So energy required to perform circular motion
$$ \begin{aligned} \Rightarrow \quad E _2 & =\frac{1}{2} m v _o^{2}=\frac{G M _e m}{2\left(R _e+h\right)} \\ E _2 & =\frac{G M _e m}{2\left(R _e+h\right)} \end{aligned} $$
According to the question,
$$ \begin{aligned} E _1 & =E _2 \\ \therefore \quad \frac{-G M _e m}{R _e+h}+\frac{G M _e m}{R _e} & =\frac{G M _e m}{2\left(R _e+h\right)} \\ \Rightarrow \quad 3 R _e & =2 R _e+2 h \\ h & =\frac{R _e}{2} \end{aligned} $$
As radius of earth, $R _e \approx 6.4 \times 10^{3} km$
Hence, $\quad h=\frac{6.4 \times 10^{3}}{2} km$ or $=3.2 \times 10^{3} km$
