Question

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 $30^{∘}$ 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 the figure. Consider gravitational interaction only between these two. (use $\frac{GM}{R}=6.4×10^7{m}^2{s}^{−2}$ )

• A. $4\sqrt{2}km{s}^{−1}$
• B. $3\sqrt{2}km{s}^{−1}$
• C. $6\sqrt{2}km{s}^{−1}$
• D. $5\sqrt{2}km{s}^{−1}$

Answer 1

Answer: (A)

Conserving angular momentum w.r.t centre of earth,
$(V_{1}cos60^{∘})4R={\left(mV\right)}_{2}R$
$⇒\frac{V_2}{V_1}=2$
Conserving mechanical energy of the system,
$−\frac{GMm}{4R}+\frac{1}{2}m{V}_1^2=−\frac{GMm}{R}+\frac{1}{2}m{V}_2^2$
$\frac{1}{2}{V}_2^2−\frac{1}{2}{V}_1^2=\frac{3}{4}\frac{GM}{R}$
$V_1^2=\frac{1}{2}\frac{GM}{R}$
$V_1=\frac{1}{\sqrt{2}}\sqrt{64×10^6}=\frac{8000}{\sqrt{2}}m{s}^{−1}=4\sqrt{2}km{s}^{−1}$