The radius of curvature at the point of minimum distance is

\sqrt{\frac{8 R}{3}}

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\sqrt{\frac{5 R}{3}}

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\sqrt{\frac{4 R}{3}}

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\sqrt{\frac{7 R}{3}}

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Solution

The correct option is A $\sqrt{\frac{8 R}{3}}$

Using conservation of angular momentum:

$m_{s a t e l l i t e} v_{1} R_{1} = m_{s a t e l l i t e} v_{2} R_{2}$

$v_{1} R_{1} = v_{2} R_{2}$

$v_{1} 2 R = v_{2} 4 R$

$v_{1} = 2 v_{2}$ ....(i)

Using energy conservation equation:
$- \frac{G M m}{2 R} + \frac{1}{2} m v_{1}^{2} = - \frac{G M m}{4 R} + \frac{1}{2} m v_{2}^{2}$ ....(ii)

substituting $v_{2}$ in equation (ii) from equation (i)

$- \frac{G M m}{4 R} = \frac{3}{2} m v_{1}^{2}$

$v_{1} = \sqrt{\frac{G M}{6 R}}$

$v_{2} = \sqrt{\frac{2 G M}{3 R}}$

Radius of curvature = $\frac{v^{2}}{a_{n}}$ $= \frac{(\sqrt{\frac{G M}{6 R}})^{2}}{\frac{G M}{(2 R)^{2}}}$

$= \frac{2 R}{3}$

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