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

Answer 1

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

As areal velocity of a planet around the sun is constant. Therefore, the desired time is

$t_{AB}=\left(\frac{areaABS}{areaofellipse}\right)\times timeperiod$

If a = semi-major axis and b= semi-minor axis of ellipse then, area of ellipse = $\pi$ab

Area $ABS=\frac{1}{4}\left(areaofellipse\right)-AreaoftriangleASO$

$=\frac{1}{4}\times \pi ab-\frac{1}{2}\left(ea\right)\times \left(b\right)$

$\therefore t_{AB}=\frac{[\frac{\pi \left(ab\right)}{4}-\frac{1}{2}eab]}{\pi ab}\times T=T(\frac{1}{4}-\frac{e}{2\pi })$