Question

A body is released from a point distance$r$ from the centre of 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

$R$ is the radius of the earth

$r$ is point distance from the centre of earth

Step 2. finding the velocity

The initial potential energy of body at distance $r$ is $-\frac{GMm}{r}$

where $G$ is gravitational constant, $M$is mass of Earth and $m$ is mass of body

This energy is converted to kinetic energy and potential energy at height $R$ from centre of Earth $\frac{1}{2}m{v}^2-\frac{GMm}{R}$

Where $v$ is the velocity of the body on reacting earth's surface

Equating the initial and final energy

$$\begin{gathered} -\frac{GMm}{r}=\frac{1}{2}m{v}^2-\frac{GMm}{R} \\ \Rightarrow GMm\left(\frac{1}{R}-\frac{1}{r}\right)=\frac{1}{2}m{v}^2 \\ \Rightarrow v^2=2GM\left(\frac{r-R}{rR}\right) \end{gathered}$$

Since $g=\frac{GM}{R^2}$, we get

$$\begin{gathered} v^2=2gR\left(\frac{r-R}{r}\right) \\ \Rightarrow v=\sqrt{2gR\left(\frac{r-R}{r}\right)} \end{gathered}$$

Hence, option D is correct.