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

Question

The figure shows the 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 the planet to move from $C$ to $D$ and $t_2$ is the time to move from $A$ to $B$, then:

Accepted Answer

1. **Kepler's Second Law:** The Law of Areas states that the line joining the sun to the planet sweeps out equal areas in equal intervals of time . This implies that the areal velocity ($\frac{\Delta A}{\Delta t}$) is constant. Consequently, the area swept ($A$) is directly proportional to the time taken ($t$).
   $$ A \propto t \quad 	ext{or} \quad \frac{A_1}{t_1} = \frac{A_2}{t_2} $$
2. **Given Data:**
   - Area corresponding to time $t_1$ (path C to D) is $A_{SCD}$.
   - Area corresponding to time $t_2$ (path A to B) is $A_{SAB}$.
   - Condition: $A_{SCD} = 2 	imes A_{SAB}$.
3. **Calculation:**
   Substitute the condition into the proportionality equation:
   $$ \frac{A_{SCD}}{t_1} = \frac{A_{SAB}}{t_2} $$
   $$ \frac{2 A_{SAB}}{t_1} = \frac{A_{SAB}}{t_2} $$
   $$ \frac{2}{t_1} = \frac{1}{t_2} \implies t_1 = 2t_2 $$

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