The energy required to take a satellite to a height ' $h$ ' above Earth surface (radius of Earth $=6.4 \times 10^3 \mathrm{~km}$ ) 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
Select the correct option:
A
$3.2 \times 10^3 \mathrm{~km}$
B
$1.6 \times 10^3 \mathrm{~km}$
C
$1.28 \times 10^4 \mathrm{~km}$
D
$64 \times 10^3 \mathrm{~km}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & E_1=\frac{G M m}{R}=-\frac{G M m}{(R+h)} \\ & E_1=\frac{G M m}{R}-\frac{G M m}{(R+h)} \\ & E_1=\frac{G M m h}{R(R+h)} \\ & E_2=\frac{1}{2} \frac{G M m}{R(R+h)} \\ & \text { Given } E_1=E_2 \\ & \frac{h}{R}=\frac{1}{2}, h=\frac{R}{2}\end{aligned}$
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