Question

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

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

Answer 1

The correct option is B $t_1=2t_2$

According to Kepler's law of Areas:
The line that joins any planet to the sun sweeps out equal areas in equal intervals of time

i.e. $\frac{\Delta A}{\Delta t}$ is constant.

Given that:

$\text{Area}\text{SCD}=2\times \text{Area}\text{SAB}$

Using above Kepler's law

$\frac{\Delta A_{SCD}}{\Delta A_{SAB}}=\frac{t_1}{t_2}=2$

$\Rightarrow t_1=2t_2$

Hence, option $\left(b\right)$ is correct.

Key Concept: Kepler’s 2nd Law of planetary motion - A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.