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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
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B

2v0
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C
v2ev20
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D
v2e2v20
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Solution

The correct option is D v2e2v20
The 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=12mv2emv20

v2=v2e2v20

v=(v2e2v20)

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.


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