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 between the ends of the minor axis and major axis close to the sun is

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

Answer 1

Answer: (D)

As areal velocity o f a planet around the sun is constant. Therefore, the desired time is
$t_{AB}=\left(\frac{\text{area ABS}}{\text{area of ellipse}}\right)\times$ time period
If $a=$ semi-major axis and $b=$ semi-minor axis of ellipse, then, area of ellipse $=\pi ab$
Area $ABS=\frac{1}{4}$ (area of ellipse) - Area of triangle $ASO$
$=\frac{1}{4}\times \pi ab-\frac{1}{2}\left(ea\right)\times \left(b\right)$
$\therefore t_{AB}=\frac{\left[\frac{\pi \left(ab\right)}{4}-\frac{1}{2}eab\right]}{\pi ab}\times T$
$=T\left(\frac{1}{4}-\frac{e}{2\pi }\right)$