Question

A body is released from height equal to the radius of earth the velocity of the object when it strikes the surface is? 1) ✓gr 2) ✓2gr 3) 2✓gr

Answer 1

`sqrt(gR)`

Explanation:

The potential energy of a mass `m` at a distance `r` from the center of the earth (assumed to be a spherically symmetric object of mass `M` and radius `R` ) is

`V(r) = -(GMm)/r`

Initially the mass is at rest at `r=2R`. So, its total energy is

`V(2R) = -(GMm)/(2R)`

Thus, from the law of conservation of energy, its speed `v` when it hits the earth (at `r=R`) is given by

`1/2 mv^2 -(GMm)/R = -(GMm)/(2R) implies`

`1/2 mv^2 = GMm(1/R-1/(2R)) = (GMm)/(2R) implies`

`v^2 = GM/R`

Now, the force on the mass at the surface of the earth is given by

`mg = (GMm)/R^2 implies g = (GM)/R^2`

Thus

`v^2 = gR implies v = sqrt(gR)`