Q.10. A body is projected vertically upwards from earth’s surface of radius ‘ R’ with velocity equal to \frac{1^{\text {rd }}}{3} of escape velocity. The maximum height reached by the body is
A. \frac{R}{8}
B. \frac{\mathrm{R}}{6}
C. \frac{\mathrm{R}}{4}
D. \frac{R}{9}
Answer :- A
Explanation :-
\Delta \text{K.E.} = \Delta \text{U} \text{Let mass of the particle be } M \text{ and that of the Earth be } M_e \therefore \quad \frac{1}{2} Mv^2 = GM_e M\left(\frac{1}{R} - \frac{1}{R+h}\right). \text{Also, } g = \frac{GM_e}{R^2}. \text{Equation (i) can be written as,} \frac{1}{2} v^2 = G_e\left[\frac{R+h-R}{R(R+h)}\right] = \frac{G_e}{R^2}\left[\frac{R h}{(R+h)}\right] \therefore \quad \frac{1}{2}\left(\frac{1}{3} v_e\right)^2 = \frac{g R h}{R+h} \therefore \quad \frac{1}{2}\left(\frac{1}{3} \sqrt{2 g R}\right)^2 = \frac{g R h}{R+h} \therefore \quad \frac{1}{2} \times \frac{1}{9}(2 g R) = \frac{g R h}{R+h} \therefore \quad \frac{h}{R+h} = \frac{1}{9} \therefore \quad 9 h = R+h8h=R
Therefore, h= \frac{R}{8}
Given V]=\frac{V_{e}}{8}
h=\frac{R}{\left [ \frac{V_{e}}{V_{e}/3} \right ]^{2}-1}=\frac{R}{9-1} =\frac{R}{8}