Question

# A satellite is revolving round the earth with orbital speed v0. If it stops suddenly, the speed with which it will strike the surface of earth would be (ve= escape velocity of a particle on earth's surface.)

A

v2ev0
No worries! Weāve got your back. Try BYJUāS free classes today!
B

2v0
No worries! Weāve got your back. Try BYJUāS free classes today!
C
v2ev20
No worries! Weāve got your back. Try BYJUāS free classes today!
D
v2e2v20
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D √v2e−2v20The orbital velocity of a satellite in circular orbit is given by v0=√GMrWhere r is the distance of satellite from the center of the earth. ⇒r=GMv20...(1) Now, when the satellite stops , its velocity becomes zero and it is at distance, r=GMv20 from the center of the earth. Since, here external and non conservative forces are absent, so the mechanical energy of the system will be conserved. Conserving mechanical energy between the points when the satellite is in its orbit and when it hits the surface of the earth : P.Ei+K.Ei=P.Ef+K.Ef −GMmr+0=−GMmR+12mv2 Where, v is the speed with which the satellite strike the Earth. Using, GMR=v2e2 and equation (1), we get ⇒−mv20+0=−12mv2e+12mv2 ⇒12mv2=12mv2e−mv20 ⇒v2=v2e−2v20 ∴v=√(v2e−2v20) Hence option (d) is the correct answer. Key Concept: The escape velocity for a planet is given ​​​​​byve=√2GMR Why this question: To make students apply conservation of mechanical energy in a planet-satellite system.

Suggest Corrections
19
Join BYJU'S Learning Program
Join BYJU'S Learning Program