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. $T\left(\frac{2\,e}{\pi}-1\right)$
• C. $\frac{T\,e}{2\pi}$
• 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 $ASO$
$=\frac{1}{4} \times \pi ab - \frac{1}{2}\left(ea\right) \times\left(b\right)$
$\therefore t_{AB} = \frac{\left[\frac{\pi\left(ab\right)}{4}-\frac{1}{2}eab\right]}{\pi\,ab}\times T$
$= T\left(\frac{1}{4}-\frac{e}{2\,\pi}\right)$