Question

The figure shows elliptical orbit of a planet m about the sun 'S'. The shaded area SCD is twice the shaded are SAB. If $t_1$ is the time for the planet to move from C to D and $t_2$ is the time to move from A to B then:

• A. $t_1=t_2$
• B. $t_1>t_2$
• C. $t_1=4t_2$
• D. $t_1=2t_2$

Answer 1

The correct option is D $t_1=2t_2$

Planets sweep an equal area in equal intervals of time. Area under SCD is twice the area under SAB so it must take double the time to sweep the area SCD than SAB.
So $t_1=2t_2$.