Question

A tunnel is dug across diameter of earth.A ball is released from surface of earth into the tunnel. The velocity of ball when it is at a distance R/2 from centre of earth is
answer in terms of G,M and R
R is radii of earth and M is mass of earth

Answer 1

Answer on Question #40824 – Physics – Mechanics

A tunnel is dug across diameter of earth. A ball is released from surface of earth into the tunnel. The velocity of ball when it is at a distance $R/2$ from centre of earth is answer in terms of $G, M$ and $R$.

R is radii of earth and M is mass of earth.

Solution:

Gravitational potential at a point on the surface of the Earth

$$U_1 = - \frac{G M}{R}$$

If Earth is assumed to be a solid sphere, then the gravitational potential at the distance $x = \frac{R}{2}$ from the centre of Earth is

$$\begin{aligned} U_2 &= - \int_{\infty}^{x} \vec{g} \cdot \overrightarrow{dx} = - \left[ \int_{\infty}^{R} \vec{g}_{outside} \cdot \overrightarrow{dx} + \int_{R}^{x} \vec{g}_{inside} \cdot \overrightarrow{dx} \right] = \\ &= \int_{\infty}^{R} \frac{G M}{x^2} dx + \int_{R}^{x} \frac{G M x}{R^3} dx = - \frac{G M}{R} + \frac{G M}{2 R^3} (x^2 - R^2) = \\ &= - \frac{G M}{2 R^3} (3 R^2 - x^2) \\ &\quad x = \frac{R}{2} \Longrightarrow \end{aligned}$$ $$U_2 = - \frac{G M}{2 R^3} \left(3 R^2 - \left(\frac{R}{2}\right)^2\right) = - \frac{G M}{2 R^3} \left(3 R^2 - \frac{R^2}{4}\right) = - \frac{11 G M R^2}{8 R^3} = - \frac{11 G M}{8 R}$$

Loss in the potential energy:

$$m(U_1 - U_2) = m\left(- \frac{G M}{R} + \frac{11 G M}{8 R}\right) = \frac{3}{8} \frac{G M m}{R}$$

Now, gain in the kinetic energy = loss in potential energy:

$$\begin{aligned} \frac{1}{2} m v^2 &= \frac{3}{8} \frac{G M m}{R} \\ v^2 &= \frac{3}{4} \frac{G M}{R} \\ v &= \sqrt{\frac{3 G M}{4 R}} \end{aligned}$$

**Answer:** The velocity of ball when it is at a distance $R/2$ from centre of earth is

$$v = \sqrt{\frac{3 G M}{4 R}}.$$