Question

A satellite of mass $\frac{M}{2}$ is revolving around earth in a circular orbit at a height of $\frac{R}{3}$ from earth surface. The angular momentum of the satellite is $M\sqrt{\frac{GMR}{x}}$. The value of $x$ is _______, where $M$ and $R$ are the mass and radius of earth, respectively. (G is the gravitational constant)

Answer 1

3

Answer 2

Solution Explanation:

1. Orbital radius:
     $r = R + \frac{R}{3} = \frac{4R}{3}$.

2. Orbital speed (using $v = \sqrt{\frac{GM}{r}}$):
     $v = \sqrt{\frac{GM}{\frac{4R}{3}}}$.

3. Angular momentum of the satellite (mass $m = \frac{M}{2}$):
     $L = mrv = \frac{M}{2} \times \frac{4R}{3} \times \sqrt{\frac{GM}{\frac{4R}{3}}} = \frac{M}{2} \times \frac{4R}{3} \times \sqrt{\frac{3GM}{4R}} = \frac{M}{2} \times \frac{4R}{3} \times \frac{\sqrt{3GM}}{2\sqrt{R}} = \frac{M}{2} \times \frac{2}{3} \sqrt{3GMR} = M\sqrt{\frac{GM R}{3}}$.

4. Given expression:
     $L = M\sqrt{\frac{GM R}{x}}$.
     Equate:
     $M\sqrt{\frac{GM R}{x}} = M\sqrt{\frac{GM R}{3}}$
     Thus, $x = 3$.