Question

The differential equation of a body fired vertically from the earth is given by,$v\frac{dv}{dt}=-\frac{g{r}^2}{x^2}$.The initial velocity of a body supposed to escape is?

• A. $\sqrt{3gr}$
• B. $\sqrt{2gr}$
• C. $\sqrt{gr}$
• D. $\sqrt{5gr}$

Answer 1

The correct option is B $\sqrt{2gr}$

We have $v\frac{dv}{dx}=-\frac{g{r}^2}{x^2}$.

This is a differential equation of variable separable type.

$\therefore vdv=g{r}^2\cdot \frac{dx}{x^2}$

By intergration, $\frac{v^2}{2}=\frac{g{r}^2}{x}+c$

If $u$ is the required velocity on the surface of the earth where $x=r$ then

$\frac{u^2}{2}=gr+c$ $\therefore c=\frac{u^2}{2}-gr$

$\therefore \frac{v^2}{2}=\frac{g{r}^2}{x}+\frac{u^2}{2}-gr$

$\therefore v^2=\frac{2g{r}^2}{x}+u^2-2gr$

This is the equation of motion of a body projected from the surface of the earth with initial velocity $u$.

If the body is not to return to the earth its velocity $v$ must be always positive. (If the velocity $v$ becomes zero the body will to rest and then will start to descend.) As $x$ increases. $2g{r}^2/x$ decreases. Hence, $v$ will be positive it

$u^2-2gr\ge 0$ $\therefore u^2\ge 2gr$ i.e $u\ge \sqrt{2gr}$.

$\therefore$ The least velocity of projection $=\sqrt{2gr}$

A particle projected with this velocity will never return to the earth. This is called the escape velocity from the earth.