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Question

A particle of mass m is projected with a velocity v=kve(k<1) from the surface of the earth. (ve= escape velocity). The maximum height above the surface reached by the particle is -

R is the radius of earth.

A
Rk21k2
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B
R(k1k)2
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C
R(k1+k)2
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D
R2k1+k
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Solution

The correct option is A Rk21k2
From conservation of mechanical energy,

12mv2GmMR=GmM(R+h)

v22=GMRGM(R+h)=GMhR(R+h)

12k2v2e=GMhR(R+h)

We know that, ve=2GMR

12k2(2GMR)=GMhR(R+h)

k2=hR+h

Rk2+hk2=h

Rk2=h(1k2)

h=Rk21k2

Hence, option (A) is the correct answer.

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