Question

A tunnel is dug across the diameter of the earth. A ball is released from the surface of Earth into the tunnel. The velocity of the ball when it is at a distance $\frac{R}{2}$ from the centre of the earth is (where R = radius of Earth and M = mass of Earth)

• A. $\sqrt{\frac{3GM}{4R}}$
• B. $\sqrt{\frac{2GM}{3R}}$
• C. $\sqrt{\frac{GM}{2R}}$
• D. $\sqrt{\frac{2GM}{R}}$

Answer 1

The correct option is A $\sqrt{\frac{3GM}{4R}}$

$\begin{array}{c}workdonebythegravity=changeinkineticenergy\\ g^{\prime }=g\left(1-\frac{h}{R}\right)\\ Force=m{g}^{\prime }=mg\left(1-\frac{h}{R}\right)\\ mv\frac{dv}{dh}=mg\left(1-\frac{h}{R}\right)\\ \underset{0}{\overset{V}{\int }}vdv=\underset{0}{\overset{\frac{R}{2}}{\int }}g\left(1-\frac{h}{R}\right)dh\\ {\left(\frac{V^2}{2}\right)}_0^V=g\left(\frac{R}{2}-\frac{R^2}{4\times 2R}\right)\\ \frac{V^2}{2}=g\left(\frac{R}{2}-\frac{R}{8}\right)\\ \frac{V^2}{2}=\frac{3gR}{8}\\ v=\sqrt{\frac{3gR}{4}}\\ F_g=\frac{G{M}^2}{d^2}:F_e=\frac{K{Q}^2}{d^2}\\ \frac{G{M}^2}{d^2}=\frac{K{Q}^2}{d^2}\\ \frac{G}{K}={\left(\frac{Q}{M}\right)}^2\\ \left(\frac{Q}{M}\right)={\left(\frac{G}{K}\right)}^{\frac{1}{2}}=\sqrt{\frac{6.67\times {10}^{-11}}{9\times {10}^9}}\\ \frac{Q}{M}=0.86\times {10}^{-10}\frac{C}{kg}\end{array}$