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Gravitation
Escape Speed and Energy Conservation
1 question from 1 paper, each with its answer.
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Concepts tested
- maximum height for sub-escape velocity
All Questions
2021 1 question
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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: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 $V_e^2 = 2gR$: $h = \frac{R}{\frac{V_e^2}{k^2V_e^2} - 1} = \frac{Rk^2}{1 - k^2}$
