A space vehicle approaching a planet has a speed $v$ , when it is very far from the planet. At that moment tangent of its trajectory would miss the centre of the planet by distance $R$ . If the planet has mass $M$ and radius $r$ , what is the smallest value of $R$ in order that the resulting orbit of the space vehicle will just miss the surface of the planet?

\frac{r}{v} {[v^{2} + \frac{2 G M}{r}]}^{\frac{1}{2}}

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v r [1 + \frac{2 G M}{r}]

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\frac{r}{v} [v^{2} + \frac{2 G M}{r}]

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\frac{2 G M v}{r}

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Solution

The correct option is B $\frac{r}{v} {[v^{2} + \frac{2 G M}{r}]}^{\frac{1}{2}}$
Consider conservation of angular momentum,
$m v R = m v_{0} r$
$⟹ v_{0} = \frac{R}{r} v$
From conservation of energy,
$\frac{1}{2} m v^{2} - 0 = \frac{1}{2} m v_{0}^{2} - \frac{G M m}{r}$

$\frac{1}{2} m v^{2} - 0 = \frac{1}{2} m v^{2} (\frac{R}{r})^{2} - \frac{G M m}{r}$
Rearranging to get the value of R gives,
$\frac{R^{2}}{r^{2}} = \frac{2 G M}{r v^{2}} + 1$

$⟹ R = \frac{r}{v} [v^{2} + \frac{2 G M}{r}]^{1 / 2}$

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The correct option is B rv[v2+2GMr]12Consider conservation of angular momentum,mvR=mv0r⟹v0=RrvFrom conservation of energy,12mv2−0=12mv20−GMmr12mv2−0=12mv2(Rr)2−GMmrRearranging to get the value of R gives,R2r2=2GMrv2+1⟹R=rv[v2+2GMr]1/2