First slide

Acceleration due to gravity and its variation

Question

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is 0.5 m. If mean density of the Moon is two-thirds that of the Earth and radius is one quarter that of the Earth, the maximum vertical distance through which he can jump on the Moon and the ratio of the time of duration of the jump on the Moon to hold on the Earth are

Moderate

A

3 m,6:1
B

6 m,3:1
C

3 m,1:6
D

6 m, 1:6

Solution

(i) g'= acceleration due to gravity on the Moon
$= G \frac{4}{3} {πR}^{'} ρ^{'} = G \frac{4}{3} π \times \frac{R}{4} \times \frac{2}{3} ρ = \frac{g}{6}$
$Now, h_{m} = \frac{u^{2}}{2 g} or h_{m} \propto \frac{1}{8}$
$Hence h_{m}^{'} g^{'} = h_{m} g$
$or h_{m}^{'} = 6 h_{e} = 6 \times 0 .5 = 3 m$
$(ii) t = \frac{2 u}{g} or t \propto \frac{1}{g}$
$∴ t^{'} g^{'} = tg or \frac{t^{'}}{t} = \frac{g}{g^{'}} = \frac{6}{1}$

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