Question

A planet revolves around the sun in an elliptical orbit of eccentricity $e$. If $T$ is the time period of the planet then the time spent by the planet while moving from closest point to the sun to the end points of minor axis is

[For ellipse eccentricity , $0<e<1$]

• A. $T(\frac{1}{4}-\frac{e}{2\pi })$
• B. $\frac{Te}{\pi }$
• C. $(\frac{e}{\pi }-1)$
• D. $\frac{\pi T}{e}$

Answer 1

The correct option is A $T(\frac{1}{4}-\frac{e}{2\pi })$


Let $t$ be the time taken by the planet to go from point ${\text{P}}_1$ to ${\text{P}}_2$.

Since, the areal velocity is constant.

$\frac{\text{Area of ellipse}}{\text{T}}=\frac{\text{Area}{\text{P}}_1{\text{P}}_2\text{S}}{\text{t}}...\left(1\right)$

Where, $T$ is the time period of revolution.

Now, area of ellipse $=\pi ab$

${\text{Area of P}}_1{\text{P}}_2{\text{S= Area of P}}_1{\text{P}}_2\text{O}-{\text{Area of P}}_1\text{S}\text{O}...\left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$, we get

$\frac{\pi ab}{T}=\frac{\pi ab}{4t}-\frac{1}{2t}\left(ae\right)b$

$\Rightarrow \frac{\pi ab}{T}=\frac{\pi ab-2abe}{4t}$

$\Rightarrow t=T[\frac{1}{4}-\frac{e}{2\pi }]$

Hence, option (a) is the correct answer.

Why this question? To make students familiar with the concept of time calculation in an elliptical orbit.