An object of mass m is released from a point at a height H above the surface of the earth .show that it's velocity v when it strikes the earths surface is given by
v =[2GM(1/R-1/r)] race to 1/2 where M = mass of earth , R = radius of earth and r = R+H

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Solution

Gravitational potential energy at a height r=R+H is,
$U_{1} = - \frac{G M m}{r}$
Gravitational potential energy on the surface of the earth is,
$U_{2} = - \frac{G M m}{R}$
Therefore,
$F i n a l K E o f t h e o b j e c t = U_{1} - U_{2} \Rightarrow \frac{1}{2} m v^{2} = - \frac{G M m}{r} + \frac{G M m}{R} \Rightarrow v = \sqrt{2 G M (\frac{1}{R} - \frac{1}{r})}$

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An object of mass m is released from a point at a height H above the surface of the earth .show that it's velocity v when it strikes the earths surface is given byv =[2GM(1/R-1/r)] race to 1/2 where M = mass of earth , R = radius of earth and r = R+H

Gravitational potential energy at a height r=R+H is, U1=-GMmr Gravitational potential energy on the surface of the earth is, U2=-GMmR Therefore, Final KE of the object=U1-U2⇒12mv2=-GMmr+GMmR⇒v=2GM1R-1r

An object of mass m is released from a point at a height H above the surface of the earth .show that it's velocity v when it strikes the earths surface is given by
v =[2GM(1/R-1/r)] race to 1/2 where M = mass of earth , R = radius of earth and r = R+H