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 $M \sqrt{\frac{\text{GMR}}{x}}$. The value of $x$ is $\_\_\_\_$, where M and $R$ are the mass and radius of earth, respectively. ( G is the gravitational constant)

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Gravitation

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 \frac{GMR}{x}

. The value of

x

____

, where M and

R

are the mass and radius of earth, respectively. ( G is the gravitational constant)

Answer: 3

Solution

The satellite, of mass

M /2

, orbits at a radius

r = R + \frac{R}{3} = \frac{4 R}{3}

For a circular orbit around Earth (mass

M

), the orbital speed is

v_{0} = \frac{GM}{r} = \frac{3 GM}{4 R}

The angular momentum of the satellite about Earth's centre is

L = (\frac{M}{2}) v_{0} r = \frac{M}{2} \frac{3 GM}{4 R} \cdot \frac{4 R}{3}

L = \frac{2 M R}{3} \frac{3 GM}{4 R} = M \frac{GM R}{3}

Comparing with

M \frac{GM R}{x}

gives

x = 3

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Integer answer

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