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)/(6)(GMm)/( R)`

Answer

D

Explanation

Initially the body is at a height, `r=2R+R=3R` from the centre of the earth.
`therefore U_(1)=-(GMm)/(r )=-(GMm)/(3R)`
and when it is at a distance `R+R=2R` from the centre of the earth `U_(2)=-(GMm)/(2R)`
Gain in K.E. = Loss in P.E. `=U_(1)-U_(2)`
`=(GMm)/(2R)-(GMm)/(3R)=(GMm)/(6R)`
Thus the gain in `K.E. =(GMm)/(6R)`