Question

Taking the earth revolves around the sun in a circular orbit of $15\times {{10}^{10}}m$, with a time period of $1yr$, the time taken by another planet, which is at distance of $540\times {{10}^{10}}m$ to revolve around the sun in circular orbit once, will be given as,
$\begin{align}
  & A.216yr \\
 & B.144yr \\
 & C.72yr \\
 & D.36yr \\
\end{align}$

Answer 1

Hint: As we all know the time period is related to the radius of the orbit as, the square of time period is directly proportional to cube of the radius of the orbit. Use this relation while solving this problem by comparing the situation. These all may help you to solve this problem.

Complete step by step answer:
As we all know the relation between the time period and the radius of the orbit is given as,
${{T}^{2}}\propto {{R}^{3}}$
Where $T$ be the time period of the orbit and $R$ be the radius of the orbit.
The square of the time period is given as directly proportional to the cube of the radius of the orbit. Using this relation, we can compare the two conditions given in the question. That is we can write that,
${{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}={{\left( \dfrac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}$
Where ${{T}_{1}}$ be the time period of the orbital in first situation, which is given as,
${{T}_{1}}=1yr$
${{R}_{1}}$ be the radius of the orbital in first situation which is given as,
${{R}_{1}}=15\times {{10}^{10}}m$
The radius of the orbit in the second situation is given as,
${{R}_{2}}=540\times {{10}^{10}}m$
We have to find the time period of the orbit in the second situation. That is we can substitute the values in it,
${{\left( \dfrac{1}{{{T}_{2}}} \right)}^{2}}={{\left( \dfrac{15\times {{10}^{10}}}{540\times {{10}^{10}}} \right)}^{3}}$
Rearranging this equation will give,
${{T}_{2}}={{\left( 36 \right)}^{\dfrac{3}{2}}}={{6}^{3}}=216yr$

So, the correct answer is “Option A”.

Note: Kepler’s laws of planetary motion are basically three in numbers. First one states that every planet’s orbit is an ellipse in which the sun is kept at one of the focus. Second one says that a line joining the Sun and a planet is taking out equal areas at the same times, and the final law states that the square of the time period of the orbital is proportional to the cube of the radius of its orbit.