Question

A body is released from a point distance $r$ from the center of the earth. If $R$ is the radius of the earth and $r>R$, then the velocity of the body at the time of striking the earth will be

• A. $\sqrt{gR}$
• B. $\sqrt{2gR}$
• C. $\sqrt{\frac{2gR}{r-1}}$
• D. $\sqrt{\frac{2gR\left(r-R\right)}{r}}$

Answer 1

The correct option is D

$\sqrt{\frac{2gR\left(r-R\right)}{r}}$

Step 1. Given Data:

Distance of the body from the earth's center is $r$

Radius of the earth is $R$

Where, $r>R$

Step 2. Velocity of the body at time of striking the earth is:

The formula used: Total initial energy = Total final energy

Total initial energy $=$ $-\frac{GmM}{R}$…….$\left(i\right)$

Total final energy $=$ $\frac{1}{2}m{v}^2$$-\frac{GmM}{R}$…….$\left(ii\right)$

By Comparing Equation $\left(i\right)$ and $\left(ii\right)$, we get:-

$$\begin{gathered} \Rightarrow -\frac{GmM}{R}=\frac{1}{2}m{v}^2-\frac{GmM}{R} \\ \Rightarrow \frac{v^2}{2}=\frac{GM}{R}-\frac{GM}{r} \\ \Rightarrow \frac{v^2}{2}=GM\left(\frac{r-R}{rR}\right) \\ \Rightarrow \frac{v^2}{2}=gR\left(\frac{r-R}{r}\right)\left[\sin ce:g=\frac{GM}{R^2}\right] \\ \Rightarrow v=\sqrt{2gR\frac{\left(r-R\right)}{r}} \end{gathered}$$

Hence, Option D is correct.