A planet revolves around the Sun in an ellipt | vedclass

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(C) According to Kepler's second law,the areal velocity of a planet is constant. The time taken to sweep an area is proportional to the area swept.Let

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$804$ minutes

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(C) According to Kepler's second law,the areal velocity of a planet is constant. The time taken to sweep an area is proportional to the area swept.Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here $a = 2b$.The Sun is at the focus $S(ae, 0)$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.So,$S = (2 \cdot \frac{\sqrt{3}}{2}, 0) = (\sqrt{3}, 0)$.The path $bed$ corresponds to the area swept by the radius vector from $S$ to the arc $bed$. The area of the ellipse is $A = \pi ab$.The area swept by the radius vector in path $bed$ is $A_{bed} = 	ext{Area of rectangle} + 	ext{Area of sector}$.Using the property of areal velocity,the time taken is proportional to the area swept. After calculating the areas for path $bed$ and $dab$,the ratio of times is found. Given $T_{bed} = 24$ hours = $1440$ minutes,the calculation yields $T_{dab} = 804$ minutes.

Similar Questions

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