The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area

Question

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

Accepted Answer

According to Kepler's Second Law of planetary motion (Law of Areas), the line joining the Sun and the planet sweeps out equal areas in equal intervals of time. This implies that the areal velocity is constant ($rac{dA}{dt} = 	ext{constant}$), meaning the area swept ($A$) is directly proportional to the time taken ($t$).
Mathematically, $rac{A_1}{t_1} = rac{A_2}{t_2}$.
Given that the Area SCD ($A_1$) is twice the Area SAB ($A_2$), i.e., $A_1 = 2A_2$, we have:
$rac{2A_2}{t_1} = rac{A_2}{t_2}$
$\Rightarrow t_1 = 2t_2$.

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