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If A Body Falls Freely In The Earth’s Gravitational Field From Infinity, Show That It Attains The Same Velocity On Reaching The Earth’s Surface As That Attained By A Freefall From A Height Above The Earth Equal To Its Radius (R), Under A Constant Acceleration Due To Gravity (g), Where g Refers To The Value On Earth’s Surface.

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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 .

Fig. 1

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}+C

where 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’ .

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