A body of mass m kg falling from a distance 3R above the earth's surface. What is its kinetic enrgy when it reaches a distance 'R' above the surface of the earth of radius R and mass M ?

Options

A. `(2)/(3) (GMm)/(R)`
B. `(1)/(3) (GMm)/(R)`
C. `(1)/(2) (GMm)/(R)`
D. `(1)/(4) (GMm)/(R)`

Answer

D

Explanation

Intially the body is at hieght,
`f=3R+R=4R` from the centre of the earth
`:. U_()=-(GMm)/(r)=-(GMm)/(4R)`
and when it is at a distance
`R+R=2R` from the centre of the earth `U_(2)=-(GMm)/(2R)`
Gain in KE = Loss in `PE=U_(1)-U_(2)`
`=(GMm)/(R)-(GMm)/(4R)=(GMm)/(4R)`
Thus the gain in `KE=(GMm)/(4R)`