Question 39: A particle of mass ‘ m ‘ is projected with a velocity v=kV_e(k<1) from the surface of the earth.
( Ve = escape velocity).
The maximum height above the surface reached by the particle is:
(1) \mathrm{R}\left(\frac{\mathrm{k}}{1-\mathrm{k}}\right)^{2}
(2) \mathrm{R}\left(\frac{\mathrm{k}}{1+\mathrm{k}}\right)^{2}
(3) \frac{\mathrm{R}^{2} \mathrm{k}}{1+\mathrm{k}}
(4) \frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}}
Answer: Option (4)
Explanation:
The escape velocity from the surface of the earth is given by
\mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2GM}{R}}The particle is projected with velocity
v=k\mathrm{V}_{\mathrm{e}}Initial kinetic energy of the particle at the surface of the earth is
\frac{1}{2}mv^{2}=\frac{1}{2}m k^{2}\mathrm{V}_{\mathrm{e}}^{2}Substituting \mathrm{V}_{\mathrm{e}}^{2}=\frac{2GM}{R},
\text{Initial kinetic energy}=m k^{2}\frac{GM}{R}The gravitational potential energy at the surface of the earth is
-\frac{GMm}{R}Let the maximum height reached above the surface be h. At this point, the velocity becomes zero, so the total mechanical energy is only potential energy:
-\frac{GMm}{R+h}Using conservation of mechanical energy,
m k^{2}\frac{GM}{R}-\frac{GMm}{R}=-\frac{GMm}{R+h}Dividing throughout by GMm,
\frac{k^{2}}{R}-\frac{1}{R}=-\frac{1}{R+h} \frac{k^{2}-1}{R}=-\frac{1}{R+h} \frac{1-k^{2}}{R}=\frac{1}{R+h}Taking reciprocal on both sides,
R+h=\frac{R}{1-k^{2}} h=\frac{R}{1-k^{2}}-R h=\frac{Rk^{2}}{1-k^{2}}Hence, the maximum height reached above the surface of the earth is
\frac{\mathrm{Rk}^{2}}{1-\mathrm{k}^{2}},
which corresponds to option (4).