The figure shows the 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 $A$ to $B$ then

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TestHub · Accepted Answer

According to Kepler's second law
Equal areas are swept in equal time.

$t_{1}$ , the time taken to go from C to D = $2 t_{2}$ where $t_{2}$ is the time taken to go from A to B. As it is given that area $S C D = 2 S A B$ .

Physics - Gravitation Question with Solution | TestHub

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