Figure shows elliptical path abcd of a planet around the sun S such… | JEE Main 2016 (9 Apr, Shift 1)

Q23JEE Main 2016 · 9 APR · SHIFT 1Medium

Kepler's Laws & Planetary Motion

Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is $\frac{1}{4}$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then:

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Answer: (c)

Kepler's second law: time $\propto$ area swept by the line from S.
Path abc: half-ellipse on the b side of $ca$ plus triangle $csa$ : $\tfrac12 A + \tfrac14 A = \tfrac34 A$ .
Path cda: other half-ellipse minus triangle $csa$ : $\tfrac12 A - \tfrac14 A = \tfrac14 A$ .
$\dfrac{t_1}{t_2} = \dfrac{3/4}{1/4} = 3$ , so the answer is (c) $t_1 = 3t_2$ .

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