JEE Main 2025 Jan 23 Shift 2, Physics Q22: Angular Momentum, Its Conservation and Angular Impulse

JEE Main2025Jan 23, Shift 2Physics

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

Your answer

Solution

Orbital radius (centre of earth to satellite): $r = R + \frac{R}{3} = \frac{4 R}{3} .$

Treating earth as stationary (mass of satellite $M /2$ is small compared to $M$ ), the orbital speed satisfies

$\frac{( M /2 ) v ^{2}}{r} = \frac{G M ( M /2 )}{r ^{2}} \Rightarrow v = \frac{GM}{r} = \frac{3 GM}{4 R} .$

Angular momentum about earth:

$L = \frac{M}{2} v r = \frac{M}{2} \frac{3 GM}{4 R} \cdot \frac{4 R}{3} .$

Combine: $L = \frac{M}{2} \cdot \frac{4 R}{3} \cdot \frac{3 GM}{4 R} = \frac{2 M R}{3} \frac{3 GM}{4 R} = M \frac{4 R ^{2}}{9} \cdot \frac{3 GM}{4 R} = M \frac{GM R}{3} .$

So $x = 3$ .

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JEE Main 2025 Jan 23 Shift 2, Physics Q22: Angular Momentum, Its Conservation and Angular Impulse

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