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A planet revolves around the sun in an elliptical orbit of semi minor and major axes x and y respectively. Then the time period of revolution is proportional to:
A. \[{(x + y)^{\dfrac{3}{2}}}\]
B. \[{(y - x)^{\dfrac{3}{2}}}\]
C. \[{x^{\dfrac{3}{2}}}\]
D. \[{y^{\dfrac{3}{2}}}\]

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Last updated date: 13th Jun 2024
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Answer
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Hint: We can solve this question using Kepler’s law. Time period of revolution depends upon the axes of the elliptical orbit. Knowledge about ellipse helps to solve this problem easily.

Complete step by step answer:
Let us know the Kepler’s law:
According to Kepler’s law, the square of the time is directly proportional to the cube of semi-major axis of the ellipse.
\[{T^2} \propto {R^3}\], Where T is the time period of revolution and R is semi-major axis.
Now in this question we have semi minor axis as x and semi minor axis y.
\[{T^2} \propto {Y^3}\]
.\[{T^{}} \propto {Y^{\dfrac{3}{2}}}\].

So our correct option is D.

Additional Information:
Kepler gave us three laws of planetary motion:
All planets move around the sun in an elliptical orbit having the sun as foci of the ellipse.
A radius vector joining any planet to the sun sweeps out equal areas in equal lengths of time.
The square of the time is directly proportional to the cube of semi-major axis of ellipse.

Note:Mostly students forgot that square of time is proportional to cube of semi-major axis and they write time is proportional to cube of semi-major axis so remember carefully.