First slide

Motion of planets and satellites

Question

A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is

Moderate

A

R/3
B

R/2
C

R/4
D

R/5

Solution

If body is projected with velocity v $(v < v_{e})$ then height up to which it will rise, $h = \frac{R}{\frac{v_{e}^{2}}{v^{2}} - 1}$
$v = \frac{v_{e}}{2}$ $∴$ (given) $h = \frac{R}{{(\frac{v_{e}}{v_{e} / 2})}^{2} - 1} = \frac{R}{4 - 1} = \frac{R}{3}$
II method:
Total energy on surface=Total energy at a height
$\frac{- G m M}{R} + \frac{1}{2} m v^{2} = \frac{- G m M}{R + h} \Rightarrow \frac{1}{2} m v^{2} = \frac{- G m M}{R + h} + \frac{G m M}{R} \Rightarrow v^{2} = 2 G M (\frac{1}{R} - \frac{1}{R + h}) \Rightarrow v^{2} = 2 g R^{2} (\frac{R + h - R}{R (R + h)}) (∵ g = \frac{G M}{R^{2}}) \Rightarrow v^{2} = 2 g R (\frac{h}{(R + h)}) = \frac{2 g h}{1 + \frac{h}{R}}$
$N o w, {(\frac{v_{e}}{n})}^{2} = \frac{2 g h}{1 + \frac{h}{R}} \Rightarrow {(\frac{\sqrt{2 g R}}{n})}^{2} = \frac{2 g h}{1 + \frac{h}{R}} \Rightarrow \frac{2 g R}{n^{2}} = \frac{2 g h}{1 + \frac{h}{R}} \Rightarrow \frac{R}{n^{2}} = \frac{h}{1 + \frac{h}{R}} \Rightarrow R (1 + \frac{h}{R}) = n^{2} h \Rightarrow (R + \frac{R h}{R}) = n^{2} h \Rightarrow (R + h) = n^{2} h \Rightarrow h (n^{2} - 1) = R \Rightarrow h = \frac{R}{(n^{2} - 1)} = \frac{R}{(2^{2} - 1)} = \frac{R}{3}$

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