An object is released from a height twice the radius of the earth on the surface of earth. Find the speed with which it will collide with group by neglecting effect of air. (Take, R radius of earth and mass of earth as M)

Options

A. `2sqrt((GM)/(3R))`
B. `3sqrt((GM)/(2R))`
C. `2 sqrt((GM)/(R))`
D. `3 sqrt((GM)/(R))`

Answer

A

Explanation

The initial potential energy `(U_(i))` of object is `U_(1)=-(GMm)/(3R)`
Final potential energy `U_(f)=-(GMm)/(R)`
By law of conservation of energy, `DeltaKE=-DeltaPE`
`rArr (1)/(2)mv^(2)=-[U_(f)-U_(i)]=U_(i)-U_(f)`
`rArr (1)/(2)mv^(2)=-(GMm)/(3R)+(GMm)/(R)`
`rArr (1)/(2)v^(2)=(2GM)/(3R)rArr v=sqrt((4GM)/(3R))=2sqrt((GM)/(3R))`.