Question

The figure shows the elliptical orbit of a planet P about the sun S. The shaded area SCD is a twice shaded area SAB. if $t_1$ is the 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. $t_1={2t}_2$
• C. $t_1={4t}_2$
• D. $t_1>t_2$

Answer 1

The correct option is B $t_1={2t}_2$

We know,

According to Kepler's law, time taken to complete a path in an elliptical orbit is directly proportional to the area swept from the focus as the centre.

$A\propto t$

$A_{SAB}\propto t_2$

$A_{SCD}\propto t_1$

Given,

$\frac{A_{SAB}}{A_{SCD}}=\frac{1}{2}$

$\frac{t_2}{t_1}=\frac{1}{2}$

$t_1=2t_2$

Option $\text{B}$ is the correct answer