The velocity on reaching the earth’s surface of a body which falls freely:
Let us consider the mass of the earth is M and the radius is R . There is a body of mass m at a the point P at a distance r from the centre O of the earth.
Now the differential equation of motion of the body is,
m\frac{d^2{r}}{dt^2}=-G\frac{Mm}{r^2} \\or,\ \frac{d^2{r}}{dt^2}=-\frac{GM}{r^2}=-\frac{\mu}{r^2}where G is the gravitational constant and \mu=GM .
Multiplying both sides by 2\frac{dr}{dt} , we get
2\frac{dr}{dt}\frac{d^2{r}}{dt^2}=-2\frac{\mu}{r^2}\frac{dr}{dt}Integrating both side we get,
\left(\frac{dr}{dt}\right)^2=\frac{2\mu}{r}+Cwhere C is the integration constant.
Since the body starts from infinity so the initial velocity of the body is zero, i.e, at x=\infty , \frac{dr}{dt}=0 .
Therefore C=0 .
now, \left(\frac{dr}{dt}\right)^2= \frac{2\mu}{r} .
Again when r=R , \frac{dr}{dt}=v , the velocity at the earth’s surface.
v^2=\frac{2\mu}{R}or, \displaystyle{v=\sqrt{\frac{2GM}{R}}}
A body falls freely under g from a height R :
A body falls freely under constant acceleration due to gravity g from a height R which is the radius of the earth. v’ is the velocity attains by the body then we can write,
{v’}^2=0^2+2gR (here initial velocity is zero)
or, v’=\sqrt{2gR}=\sqrt{2\frac{GM}{R^2}R}
or, v’=\displaystyle{\sqrt{\frac{2GM}{R}}}
Therefore v=v’ .