Question

A particle is projected from a point A that is at a distance of $4R$ from the centre of the earth with speed $V_1$ in a direction making ${30}^{\circ }$ with the line joining the centre of the earth and point A as shown in figure. Consider the gravitational interaction only between these two bodies (use $\frac{GM}{R}=6.4\times {10}^7{m}^2/s^2$). The speed $V_1$ if particle passes grazing the surface of the earth is:

• A. $2\sqrt{2}\times {10}^{3}m/s$
• B. $4\sqrt{2}\times {10}^{3}m/s$
• C. $4\times {10}^{3}m/s$
• D. $4\sqrt{3}\times {10}^{3}m/s$

Answer 1

The correct option is B $4\sqrt{2}\times {10}^{3}m/s$

Conserving angular momentum
$m{V}_{1}cos{60}^{\circ }\times 4R=m_{}{V}_{2}R$
$\frac{V_2}{V_1}=2$
Conserving energy of the system
$\frac{-GMm}{4R}+\frac{1}{2}m{V}_1^2=\frac{-GMm}{R}+\frac{1}{2}m{V}_2^2$
$\frac{1}{2}{V}_2^2-\frac{1}{2}{V}_1^2=\frac{3GM}{4R}$
or
$V_1^2=\frac{1}{2}\frac{GM}{R}$
$V_1=\frac{1}{\sqrt{2}}\sqrt{64\times {10}^6}=\frac{8000}{\sqrt{2}}m/s$