The figure shows elliptical orbit of a planet $m$ about the sun $S$ . The shaded area $S C D$ is twice the shaded area $S A B$ . If $t_{1}$ is the time for the planet to move from $C$ to $D$ and $t_{2}$ is the time to move from $A$ to $B$ then :-

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$t_{1} = t_{2}$

$t_{1} = 2 t_{2}$

$t_{1} < t_{2}$

$t_{1} = 4 t_{2}$

answer is D.

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Detailed Solution

according to Kepler's law,

$\Rightarrow \frac{A_{S C D}}{t_{1}} = \frac{A_{S A B}}{t_{2}} \Rightarrow \frac{2 A_{S A B}}{t_{1}} = \frac{A_{S A B}}{t_{2}} \Rightarrow t_{1} = 2 t_{2}$

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