A piece of matter of mass m is thrown up vertically from the earth's surface and it rises up to a height R.
(Radius of the earth=R.) What is the initial velocity of the piece of matter ? Show that the increase in potential energy of the piece of matter `=1/2 mgR.`

Explanation

Potential energy of a body of mass m on the surface of the earth , i.e., at a distance R from the centre of the earth `=-(GMm)/R`. If the initial upward velocity is u,its kinetic energy `=1/2 "mu"^2`. Thus, the total energy of the body on the surface of the earth =`1/2 "mu"^2 -(GMm)/R`. At a height R from the earth.s surface , which is at a distance 2R from the centre of the earth,potential energy =`(GMm)/(2R)`. As per the given condition , kinetic energy at that height =0 . Hence , the total energy `=-(GMm)/(2R)`
From the law of conservation of energy
`1/2 "mu"^2-(GMm)/R =-(GMm)/(2R)`
or, `1/2 "mu"^2=(GMm)/(2R)`
or, initial velocity `u=sqrt((GM)/R)`
Also, as per the law of conservation of energy , increase in potential energy = decrease in kinetic energy
`=1/2 "mu"^2=(GMm)/(2R)=1/2m(GM)/(R^2)*R=1/2 mgR`.