• # question_answer A particle is projected from point A, that is at a distance 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. Find the speed ${{V}_{1}}$ if particle grazes the surface of the earth as shown in figure. Consider gravitational interaction only between these two. $(use\,\,\frac{GM}{R}=6.4\times {{10}^{7}}{{m}^{2}}/{{s}^{2}})$ A) $4\sqrt{2}\,km/s$ B) $3\sqrt{2}\,km/s$ C) $6\sqrt{2}\,km/s$ D) $5\sqrt{2}\,km/s$

 Conserving angular momentum w.r.t. centre of earth $m.({{V}_{1}}\cos 60{}^\circ ).4R=m.{{V}_{2}}.R\,\,;\,\,\frac{{{V}_{2}}}{{{V}_{1}}}=2.$ Conserving mechanical energy of the system $m.({{V}_{1}}\cos 60{}^\circ ).4R=m.{{V}_{2}}.R$;   $\frac{{{V}_{2}}}{{{V}_{1}}}=2.$
 $-\frac{G\,\,Mm}{4R}+\frac{1}{2}mV_{1}^{2}=-\frac{GMm}{R}+\frac{1}{2}mV_{2}^{2}$ $\frac{1}{2}{{V}_{2}}^{2}-\frac{1}{2}{{V}_{1}}^{2}=\frac{3}{4}\frac{GM}{R}$or ${{V}_{1}}^{2}=\frac{1}{2}\frac{GM}{R}$ ${{V}_{1}}=\frac{1}{\sqrt{2}}\sqrt{64\times {{10}^{6}}}=\frac{8000}{\sqrt{2}}m/s=4\sqrt{2}km/s$