Q 11-07-049JEE MainJEE Main 2025 (23 Jan, Shift 2)Medium
A satellite of mass $\dfrac{M}{2}$ is revolving around the Earth in a circular orbit at a height of $\dfrac{R}{3}$ from the Earth's surface. The angular momentum of the satellite is $M\sqrt{\dfrac{GMR}{x}}$. The value of $x$ is ______, where $M$ and $R$ are the mass and radius of the Earth, respectively. ($G$ is the gravitational constant)
Numerical value type. Enter your answer.
Answer: 3
Orbit radius $r = R + \dfrac{R}{3} = \dfrac{4R}{3}$. Orbital speed $v = \sqrt{\dfrac{GM}{r}}$, so
$$L = mvr = m\sqrt{GMr}$$
With $m = \dfrac{M}{2}$:
$$L = \frac{M}{2}\sqrt{GM\cdot\frac{4R}{3}} = M\sqrt{\frac{1}{4}\cdot\frac{4GMR}{3}} = M\sqrt{\frac{GMR}{3}}$$
$x = 3$
Solution by Sreeraj P, M.Sc Physics