
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. \[{y^{\dfrac{3}{2}}}\]
Answer
474.9k+ views
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.
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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
