Using Kepler's law,
Area ABCDA=x.
Area SABCS=4x+2x=43x.
Area SADCS=2x−4x=4x.
Since Arial velocity remains constant. Therefore, area swept will be proportional to time.

4x43x=t2t1
⇒t1=3t2
The figure shows an elliptical path ABCD of a planet around the sun S such that the area of triangle CSA is 41 the area of the ellipse. (see figure) with DB as the major axis, and CA as the minor axis. If t1 is the time taken for the planet to go over path ABC and t2 for path taken over CDA then:

Held on 9 Apr 2016 · Verified 6 Jul 2026.
t1=4t2
t1=2t2
t1=3t2
t1=t2
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