The figure shows the elliptical path abcd of a planet around the sun S, such that the area of the triangle csa is 1/4 of the area of the ellipse, with db as the semi-major axis and ca as the semi-minor axis (see figure). If t₁ is the time taken by the planet over the path abc and t₂ the time taken…

Question

The figure shows the elliptical path abcd of a planet around the sun S, such that the area of the triangle csa is 1/4 of the area of the ellipse, with db as the semi-major axis and ca as the semi-minor axis (see figure). If t₁ is the time taken by the planet over the path abc and t₂ the time taken over the path cda, then

Accepted Answer

(3) t₁=3t₂. By Kepler's second law the time spent on a stretch of orbit is proportional to the area swept by the radius from S. The minor axis ca cuts the ellipse into two halves of area A/2 each, and S sits on the d side of it, as the figure shows. Along a→ b→ c the radius sweeps that half plus the triangle csa: A/2+A/4=(3A)/4. Along c→ d→ a it sweeps what is left, A/4. So (t₁)/(t₂)=(3A/4)/(A/4)=3, that is t₁=3t₂.

Step-by-step solution

Why the other options are wrong