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A planet revolves around the sum 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 planet between the ends of the minor axis close to the sun is

\[(a)\;T\bigg [\frac{1}{4}-\frac{e}{2 \pi}\bigg] \quad (b)\;\frac{Te}{\pi} \quad (c)\;\bigg(\frac{e}{\pi}-1\bigg) T \quad (d)\;\frac{T \pi}{e}\]

1 Answer

We know that areal velocity of planet is constant.
The desired time is time taken to travel from A to B
The time taken to cover complete area of ellipse =T
$\therefore \;t_{AB}=\bigg(\large\frac{Area\; of\; SAB}{Area\;of\;ellipse}\bigg)$$T$
Area of ellipse $=\pi ab$
Area of SAB $=\large\frac{1}{4} $$\pi a b-Area\;of \;\Delta\;OSB$
$\qquad=\large\frac{1}{4} $$\pi a b-\large\frac{1}{2}$$b \times ae$
$\therefore t_{AB}=\large\frac{\bigg(\Large\frac{\pi ab}{4}-\frac{1}{2} \large abe \bigg)T}{\pi a b}$
$\qquad=T\bigg [\large\frac{1}{4}-\frac{e}{2 \pi}\bigg]$
Hence a is the correct answer.


answered Aug 30, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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