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

1 Answer
Ananda Dasgupta
Jun 12, 2018

#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)#

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