Q 11-07-011NEETNEET 2021Top questionMedium
A particle of mass 'm' is projected with a velocity $v = kV_e$ ($k < 1$) from the surface of the earth. ($V_e$ = escape velocity) The maximum height above the surface reached by the particle is :
Answer: (A) $\dfrac{Rk^2}{1 - k^2}$
Energy conservation, with $V_e^2 = \dfrac{2GM}{R}$:
$$\frac{1}{2}mk^2V_e^2 - \frac{GMm}{R} = -\frac{GMm}{R + h}$$
$$k^2\frac{GM}{R} = \frac{GM}{R} - \frac{GM}{R + h} = \frac{GMh}{R(R + h)}$$
$$k^2(R + h) = h \;\Rightarrow\; h = \frac{Rk^2}{1 - k^2}$$
Solution by Sreeraj P, M.Sc Physics