JEE Main 2022 · Physics · Gravitation · medium

A satellite is moving in a circular orbit of radius 3R about the Earth (R is the Earth's radius). It is suddenly brought to rest and then falls…

Single correct option

A satellite is moving in a circular orbit of radius

3R

about the Earth (

R

is the Earth's radius). It is suddenly brought to rest and then falls freely along a radius towards the Earth. Neglecting the atmosphere, the speed with which it reaches the Earth's surface is (

g

is the surface gravity)

A. $\sqrt{gR}$
B. $\sqrt{\dfrac{2gR}{3}}$
C. $2\sqrt{\dfrac{gR}{3}}$
D. $\sqrt{2gR}$

Show answer and solution

Answer: (C) $2\sqrt{\dfrac{gR}{3}}$

Starting from rest at

r=3R

, energy conservation with

GM=gR^{2}

gives

\dfrac12v^{2}=GM\left(\dfrac1R-\dfrac1{3R}\right)=\dfrac{2GM}{3R}=\dfrac{2gR}{3}

. So

v^{2}=\dfrac{4gR}{3}

and

v=2\sqrt{\dfrac{gR}{3}}

. The orbital speed of the satellite plays no role once it is stopped. Using

v^{2}=2gh

with

h=2R

would give

\sqrt{4gR}

, which wrongly assumes constant

g

Report a mistake

← Previous Gravitation question Next Gravitation question →

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake