A planet goes around the sun, at A in the figure above, in the Keplerian ellipse

Question

A planet goes around the sun, at A in the figure above, in the Keplerian ellipse c. B is the other focus. Cut the ellipse in half along the semi-MINOR axis EF as indicated. Then one-half of the ellipse is closer to the sun than the other half. Upon going around and around, does the planet spend more time on the far half, or on the near half of the ellipse? Let t_far be the time spent going around the far half of the ellipse, and t_near the time spent going around the near half. Derive a formula for the ratio t_near/t_far Your formula should only involve the eccentricity e of the ellipse and universal constants.

Explanation / Answer

(a)From keplers third law.

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

If the planet is far, it takes more time. so the planet spend more time on far half.

(b) T2 proportonal to R3

Tfar = a+ae

Tnear= a-ae

Tnear2/Tfar2=(ae -a)3/(ae + a)3

                  =[(e-1)/(e+1)]3

Tnear /Tfar =[(e-1)/(e+1)]3/2

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