Question

A particle falls on earth: (1)from infinity (2)from a height 10 times the radius of earth. The ratio of the velocities gained on reaching at the earth's surface is?

Answer 1

Dear student

1) By the conservation of energy

(K.E.+P.E.) at the surface of earth=(K.E.+P.E.) at the infinity
but as the body fall the velocity is zero at infinity and also the potential energy of the particle is zero at infinity so
$$\begin{gathered} \frac{1}{2}m{v}^2-\frac{GMm}{R}=0 \\ \frac{1}{2}m{v}^2=\frac{GMm}{R} \\ v=\sqrt{\frac{2GM}{R}} \\ v=\sqrt{2gR} \\ \left(2\right)forthiscase \\ \frac{1}{2}mv{\text{'}}^2-\frac{GMm}{R}=0-\frac{GMm}{10R} \\ \frac{1}{2}mv{\text{'}}^2=\frac{GMm}{R}-\frac{GMm}{10R} \\ \frac{1}{2}mv{\text{'}}^2=\frac{9GMm}{10R} \\ v\text{'}=\sqrt{\frac{18GM}{10R}} \\ v\text{'}=\sqrt{\frac{18}{10}gR} \\ so \\ \frac{v}{v\text{'}}=\frac{\sqrt{2gR}}{\sqrt{\frac{18}{10}gR}} \\ \frac{v}{v\text{'}}=\sqrt{\frac{20}{18}} \\ \frac{v}{v\text{'}}=1.054 \end{gathered}$$