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Question

A satellite is orbiting around earth in a circular orbit of radius r. A particle of mass m is projected from satellite in forward direction with velocity v=23 times orbital velocity (this velocity is given with respect to earth). During subsequent motion of the particle, its minimum distance from the centre of earth is

A
r2
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B
r3
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C
2r3
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D
4r5
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Solution

The correct option is A r2

As the velocity of particle is less than orbital velocity of satellite, the particle goes in elliptical orbit of semi-major axis less than r.

Let r1 be the minimum distance and v1 be velocity of particle at this position.
Then from the conservation of angular momentum,

m0×23v0×r= m0v1r1

where m0 is the mass of particle and v0 is orbital speed equal to GMr

v1r1=23v0r...(1)

From energy conservation, we have

12m0(23v0)2GMm0r=m0v212GMm0r1

Using equation (1) and v0=GMr,

13m0v20v20rm0r=m0(23v0rr1)22rv20m0r1

131=r23r21rr1

r23r21rr1+23=0

r23r1r+2r21=0

r(r2r1)r1(r2r1)=0

Solving above equations, we get

r1=r2 and r1=0

So, the acceptable value is, r1=r2

Hence, option (a) is the correct answer.

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