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Question

A body is projected up with a velocity equal to (34)th to the escape velocity from the surface of the earth. Find the height above earth's surface up to which it reaches. Given R is the radius of the earth.

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Solution

The velocity of projection, v=34×ve
=34×2GMR ------ (i)
Total energy of the body when it is projected
E=KE+PE=v=12×mv2GMmR ----(ii)
From (i) ,v2=916×2GMR
Putting it in (ii)
E=12m×916×2GMRGMmR
=716GMmR
Suppose, h= maximum height attained by the body.
So, distance of the body from the center of earth, r = R+h
At this height, KE = 0
So from E=0GMmr=GMm(R+h)
Using the principle of conservation of energy.
716GMmR=GMm(R+h)
7h=16R7R
h=97R

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