Question

The figure shows the elliptical orbit of a planet $m$ about the sun $S$. The area $SCD$ is twice the area $SAB$. If $t_1$ be 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. $t_1=4t_2$
• B. $t_1=8t_2$
• C. $t_1=6t_2$
• D. $t_1=2t_2$

Answer 1

The correct option is D $t_1=2t_2$

From Kepler's law, areal velocity is constant.

$\frac{\text{Area SCD}}{{\text{t}}_1}=\frac{\text{Area SAB}}{{\text{t}}_2}$

$\frac{\text{2 Area SAB}}{{\text{t}}_1}=\frac{\text{Area SAB}}{{\text{t}}_2}$

$\Rightarrow {\text{t}}_1=2{\text{t}}_2$