Question

What is the period of revolution of a planet in circular orbit around a star of 4.0 solar mass if it is at a distance from the star equal to that of the Earth from the sun?

Answer 1

Hint: The time period of revolution of a planet around its star is given by the formula, the circumference of the planet's revolution around the star upon the speed by which the planet is travelling around the star. In this problem, since the circumference of revolution of the planet is the same initially and finally, we need to find the change in velocity of the planet.

Complete step-by-step solution:
We will first find the relation between the velocity of Earth and the mass of the Sun and then use it to find the velocity by which the other planet is orbiting its star.
Since, the Earth is revolving around the sun in a circular orbit, the force of attraction is balanced by the centrifugal force acting on the Earth.
Let the mass of the sun be ‘M’ and that of the Earth be ‘m’. Also, let the distance between the Earth and the Sun be R, then we can write the balanced force equation on Earth as:
$\begin{align}
  & \Rightarrow \dfrac{GMm}{{{R}^{2}}}=\dfrac{m{{v}^{2}}}{R} \\
 & \therefore v=\sqrt{\dfrac{GM}{R}} \\
\end{align}$
Thus, the time period of revolution of Earth is:
$\begin{align}
  & \Rightarrow {{T}_{E}}=\dfrac{2\pi R}{v} \\
 & \Rightarrow {{T}_{E}}=\dfrac{2\pi R}{\sqrt{\dfrac{GM}{R}}} \\
 & \therefore {{T}_{E}}=\dfrac{2\pi {{R}^{\dfrac{3}{2}}}}{\sqrt{GM}} \\
\end{align}$
Now, in our problem the mass of the star is four times the mass of the sun, that is, ‘4M’ with the same radius of revolution ‘R’. Thus, the time period of the planet revolving its star is:
$\Rightarrow {{T}_{P}}=\dfrac{2\pi {{R}^{\dfrac{3}{2}}}}{\sqrt{G{{M}_{S}}}}$
Where, ${{M}_{S}}$ is the mass of the star which is equal to ‘4M’. Thus, we can write that:
$\begin{align}
  & \Rightarrow {{T}_{P}}=\dfrac{2\pi {{R}^{\dfrac{3}{2}}}}{\sqrt{G{{M}_{S}}}} \\
 & \Rightarrow {{T}_{P}}=\dfrac{2\pi {{R}^{\dfrac{3}{2}}}}{\sqrt{G(4M)}} \\
 & \Rightarrow {{T}_{P}}=\dfrac{2\pi {{R}^{\dfrac{3}{2}}}}{2\sqrt{GM}} \\
 & \Rightarrow {{T}_{P}}=\dfrac{1}{2}{{T}_{E}} \\
\end{align}$
Since, we know that the time period of revolution of Earth is 365 days. Therefore, we can write that:
$\begin{align}
  & \Rightarrow {{T}_{P}}=\dfrac{1}{2}\times 365 \\
 & \therefore {{T}_{P}}=182.5days \\
\end{align}$
Hence, the time period of the planet revolving around its star comes out to be 182.5 days.

Note: In this problem, we first derived the formula for the time period of revolution of Earth and then generalized it for any planetary system. Also, we work under the assumption that the mass of the Earth is very small in comparison to the mass of Earth or else the Earth wouldn’t be moving in a perfectly circular orbit.