Question

Find the maximum height reached from the centre of the earth if a body is projected up with a velocity equal to $\frac{3}{4}$ th of the escape velocity from the surface of the earth? $\left(R a d i u s \, o f \, t h e \, e a r t h = R\right)$ :

• A. $\frac{10 \, R}{9}$
• B. $\frac{16 \, R}{7}$
• C. $\frac{9 \, R}{8}$
• D. $\frac{10 \, R}{3}$

Answer 1

Answer: (B)

$v=\frac{3}{4} \, v_{e}$
$K.E.=\frac{1}{2} \, \left(m v\right)^{2}=\frac{1}{2}m \, \left(\frac{3}{4} v_{e}\right)^{2}$
$=\frac{9}{32} \, mv_{e}^{2}$
$=\frac{9}{32}m \, \left(\frac{2 G M}{R}\right)$
$K.E=\frac{9}{16} \, \frac{G M m}{R} \, $
$P.E.=-\frac{G M m}{R}$
$Total \, energy=K.E.+P.E.=-\frac{7}{16} \, \frac{G M m}{R}$
Let the distance from the centre of earth be $h$ ; then $P.E.=-\frac{G M m}{h}$
$Total \, energy=P.E.$ above earth's surface
$-\frac{7}{16}\frac{G M m}{R}=- \, \frac{G M m}{h}$
$\therefore h=\frac{16 R}{7}$