Question

a body falls freely towards the earth from a height 2R, above the surface of the earth where intially it was rest .If R is the radius of the then the velocity on reaching surface of the earth is
$\sqrt{4 \div 3 \times gr}$


options:-
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Answer 1

The gravitational potential energy of a body at a distance of r from the centre of the earth is given by:

U∴GMm=−GMmr=universal gravitational constant=mass of the earth=mass of the body

In this case r=3R

So, the gravitational potential energy at the initial position is:

U1=−GMm3R

The velocity of the body is zero. So, the initial kinetic energy is also zero.

This means the total mechanical energy initially is:

E1=U1+K1=−GMm3R

On reaching the ground r=R

So, the gravitational potential energy at this position is:

U2=−GMmR

Let the velocity on reaching the ground be v

So, the kinetic energy is:

K2=12mv2

Total mechanical energy on the ground is now:

E2=U2+K2=−GMmR+12mv2

These two energies should be equal:

−GMmR+12mv2v22v=−GMm3R=2GM3R=4GM3R−−−−−√

Now:

gGM=GMR2=gR2

That gives finally:

v=4gR23R−−−−−√=4gR3−−−−√