First slide

Variation of g

Question

A rocket is launched vertically from the surface of the earth of radius fi with an initial speed v. If atmospheric resistance is neglected, the maximum height h attained by the rocket is given by

Moderate

A

$\frac{R}{(\frac{2 gR}{v^{2}} - 1)}$
B

$\frac{R}{[\frac{2 gR}{v^{2}} + 1]}$
C

$R [\frac{2 gR}{v^{2}} - 1]$
D

$R [\frac{2 gR}{v^{2}} + 1]$

Solution

The total energy on the surface of the earth is given by
$T = K.E. + P.E. = \frac{1}{2} {mv}^{2} - (GmM / R)$
where m is the mass of rocket. At highest point, y =0. Thus at a distance {R+h) i.e .highest point, the energy is entirely potential. So, energy at highest point is given by
$P.E. = - [GMm / (R + h)]$
According to law of conservation of energy
$\frac{1}{2} {mv}^{2} - \frac{GmM}{R} = - \frac{GMm}{(R + h)}$
or $\frac{1}{2} {mv}^{2} = GmM [\frac{1}{R} - \frac{1}{R + h}]$
or $\frac{v^{2}}{2} = GM \frac{h}{R (R + h)}$
or $\frac{R + h}{R} = \frac{2 gR}{v^{2}} (∵ g = \frac{GM}{R^{2}})$
$h = \frac{R}{(\frac{2 gR}{v^{2}} - 1)}$

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