2021 GravitationEscape Speed and Energy Conservation Hard Question 20 of 66 in Gravitation

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:

A $R\left(\frac{k}{1+k}\right)^2$
B $\frac{R^2k}{1+k}$
C $\frac{Rk^2}{1-k^2}$
D $R\left(\frac{k}{1-k}\right)^2$

The answer is C.

Solution

By energy conservation: $\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R + h}$ This gives $h = \frac{R}{\frac{2gR}{v^2} - 1}$ With $v = kV_e$ and $V_e^2 = 2gR$: $h = \frac{R}{\frac{V_e^2}{k^2V_e^2} - 1} = \frac{Rk^2}{1 - k^2}$

Concept maximum height for sub-escape velocity

Gravitation in past papers

66 questions from this chapter have appeared across 26 exam years.