A point mass is orbiting a significant mass M lying at the focus of the elliptical orbit having major and minor axes given by 2a and 2b respectively. Let r be the distance between the mass M and the end point of major axis. Velocity of the particle can be given as

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$\frac{ab}{r} \sqrt{\frac{GM}{a^{3}}}$

$\frac{ab}{r} \sqrt{\frac{GM}{b^{3}}}$

$\frac{a b}{2 r} \sqrt{\frac{G M}{r^{3}}}$

$\frac{2 a b}{r} \sqrt{\frac{G M}{{(\frac{a + b}{2})}^{3}}}$

answer is A.

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Detailed Solution

Gravitational force = Centripetal force
$\Rightarrow \frac{GM}{r^{2}} = \frac{v^{2}}{r}$
here r is the radius of curvature. From Kepler’s law, time period is given by,
$\begin{array}{l} T = 2 π \sqrt{\frac{a^{3}}{GM}} = \frac{2 πab}{rV} \\ (\frac{dA}{dt} = \frac{vr}{2} \Rightarrow dt = T = \frac{2 dA}{vr} = \frac{2}{rv} πab) \\ v = \frac{ab}{r} \sqrt{\frac{GM}{a^{3}}} \end{array}$