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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