Question 221 of 223

The figure shows an elliptical path $ABCD$ of a planet around the sun $S$ such that the area of triangle $CSA$ is $\dfrac{1}{4}$ the area of the ellipse. (see figure) with $DB$ as the major axis, and $CA$ as the minor axis. If ${t}_{1}$ is the time taken for the planet to go over path $ABC$ and ${t}_{2}$ for path taken over $CDA$ then:

A${t}_{1}=4{t}_{2}$B${t}_{1}=2{t}_{2}$C${t}_{1}={3t}_{2}$D${t}_{1}={t}_{2}$

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Answer: (C) ${t}_{1}={3t}_{2}$

Using Kepler's law,

Area $ABCDA$$=x$.

Area $SABCS=\dfrac{x}{4}+\dfrac{x}{2}=\dfrac{3x}{4}$.

Area $SADCS=\dfrac{x}{2}-\dfrac{x}{4}=\dfrac{x}{4}$.
Since Arial velocity remains constant. Therefore, area swept will be proportional to time.


$\dfrac{\dfrac{3x}{4}}{\dfrac{x}{4}}=\dfrac{{t}_{1}}{{t}_{2}}$
$\Rightarrow {t}_{1}=3{t}_{2}$