JEE Main 2025 — Gravitation Question with Solution

The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following formula:

$\mathrm{KE}=\frac{1}{2}m{v}^2=\frac{1}{2}m\frac{G{M}_e}{r}=\frac{G{M}_{e}m}{2r}$

For this satellite, the radius $r$ is the sum of the Earth's radius $R_E$ and the height $h$ above the Earth's surface:

$r=R_E+h$

Given:

Mass of the satellite, $m=1000\text{kg}$

Mass of the Earth, $M_e=6\times 10^{24}\text{kg}$

Radius of the Earth, $R_E=6.4\times 10^6\text{m}$

Height of orbit above the Earth's surface, $h=270\text{km}=2.7\times 10^5\text{m}$

Gravitational constant, $G=6.67\times 10^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$

Substitute these values into the equation:

$\mathrm{KE}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 1000}{2(6.4\times 10^6+2.7\times 10^5)}$

$=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 1000}{2\times 6.67\times 10^6}$

$=3\times 10^{10}\text{J}$

Hence, the kinetic energy of the satellite in its orbit is $3\times 10^{10}\text{J}$.