Question

A particle falls towards earth from infinity. Its velocity on reaching the earth would be

• A. Infinity
• B. $\surd 2gR$
• C. $2\surd gR$
• D. Zero

Answer 1

The correct option is B

$\surd 2gR$

Explanation for correct answer(s):

Let

Mass of earth $=M$

Mass of particle$=m$

Velocity with which particle move $=v$

Radius of earth $=R$

Force acting on particle $=F$

Since there is not any external force acting on the particle, we can apply the conservation of mechanical energy

${(Kineticenergy)}_{Infinity}+{(PotentialEnergy)}_{Infinity}={(Kineticenergy)}_{surface}+{(PotentialEnergy)}_{surface}$

Since the object falls from infinity, the kinetic energy and potential energy will be zero at infinity.

The force acting on unit mass on a point at distance $x$ is,

$F=\frac{GMm}{x^2}$

The work done is given by

$$\begin{gathered} dW=F.dx \\ dW=\frac{GMm}{x^2}.dx \\ W={\int }_{\infty }^R\frac{GMm}{x^2}dx \\ W=GMm{\int }_{\infty }^R\frac{1}{x^2}dx \\ W=GMm{\left[\frac{x^{-1}}{-1}\right]}_{\infty }^R \\ W=-\frac{GMm}{R} \end{gathered}$$

The work is stored as potential energy in the body. Thus,

${\left(Potentialenergy\right)}_{surface}=-\frac{GMm}{R}$

Therefore,

$0={(Kineticenergy)}_{surface}+{(PotentialEnergy)}_{surface}$

$$\begin{gathered} 0=(1/2)m{v}^2+(-GMm/R) \\ (1/2)m{v}^2=(GMm/R) \\ {v}^2=(2GM/R) \\ {v}^2=2GMR/R^2 \\ \sin ceGM/R^2=g \\ {v}^2=2gR \\ v=\surd 2gR \end{gathered}$$

Hence, option B is correct.