A satellite of mass (M)/(2) is revolving around earth in a circular orbit at a height of (R)/(3) from earth surface. The angular momentum of the satellite is M (G M R)/(x) . The value of x is ______ , where M and R are

Correct answer: 0 aligned & orbital velocity v_0= GM 4 R / 3 = 3 GM 4 R & Angular momentum of satellite = M 2 v_0 (4 R)/(3) & =(M)/(2) · 3 GM 4 R · 4 R 3 & =M GMR 3 & x=3 aligned (ii) Since mass of satellite is comparable to the mass of earth. aligned & G.M. M 2 ( 4 R 3 )^2 = M 2 ω^2 · 8 R 9 & ω= 81 GM 128 R^3 aligned Angular momentum of satellite about common centre of mass, aligned & L=(M)/(2) ·((8 R)/(9))^2 · ω & L=M GMR((8)/(81)) & x=(81)/(8) 10 aligned

JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Physics
Chapter: Gravitation

JEE Main 2025 · 23 January · Evening Shift · Physics · Gravitation

A satellite of mass (M)/(2) is revolving around earth in a circular...

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{G M R}{x}}

. The value of

x

is ______ , where

M

and

R

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

Correct answer

Step-by-step solution

\begin{aligned} & \text{ orbital velocity } v_0=\sqrt{\frac{\mathrm{GM}}{4 \mathrm{R} / 3}}=\sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}} \\ & \text{ Angular momentum\text{ of } satellite }=\frac{\mathrm{M}}{2} \mathrm{v}_0 \frac{4 R}{3} \\ & =\frac{M}{2} \cdot \sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}} \cdot \frac{4 \mathrm{R}}{3} \\ & =M \sqrt{\frac{\mathrm{GMR}}{3}} \\ & x=3\end{aligned}

(ii) Since mass of satellite is comparable to the mass of earth.

\begin{aligned} & \frac{\text{ G.M. } \frac{\mathrm{M}}{2}}{\left(\frac{4 \mathrm{R}}{3}\right)^2}=\frac{\mathrm{M}}{2} \omega^2 \cdot \frac{8 \mathrm{R}}{9} \\ & \omega=\sqrt{\frac{81 \mathrm{GM}}{128 \mathrm{R}^3}}\end{aligned}

Angular momentum of satellite about common centre of mass,

\begin{aligned} & L=\frac{M}{2} \cdot\left(\frac{8 R}{9}\right)^2 \cdot \omega \\ & L=M \sqrt{\operatorname{GMR}\left(\frac{8}{81}\right)} \\ & x=\frac{81}{8} \simeq 10 \end{aligned}

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