Question

In a certain binary-star system, each star has the same mass as our Sun, and they revolve about their center of mass. The distance between them is the same as the distance between Earth and the Sun. What is their period of revolution in years?

A. 0.31 years

B. 0.71 years

C. 0.82 years

D. 0.94 years

Answer 1

Hint: The binary star system is a system that consists of two stars that orbit around their barycenter. The gravitational force is defined as the force which attracts any two objects with mass. It is attractive because it always tries to pull the masses together and it never pushes apart.

Complete step by step solution:

In the binary-star system with the two identical stars, each having the mass equals the mass of the sun, \[M = 1.99 \times {10^{30}}kg\]. The distance between them equals the mass between the Earth and the Sun, \[r = 1.5 \times {10^{11}}m\]. Since the two stars are having the same mass and are rotating about the center of the mass, the gravitational force between them is, 

$F = \dfrac{{G{M^2}}}{{{r^2}}}$

This force must equal the centrifugal force and this centrifugal force is equal to the centrifugal acceleration multiplied by the mass of the star. The centrifugal acceleration is equal to the square of the speed divided by the distance from one of the stars to the center of mass. Thus,

\[{a_c} = \dfrac{{{v^2}}}{{r/2}}\]

\[ \Rightarrow {a_c} = \dfrac{{2{v^2}}}{r}\]

Hence we can write \[\dfrac{{G{M^2}}}{{{r^2}}} = \dfrac{{M2{v^2}}}{r}\_\_\_\_\_\_\_\_\_\_(1)\]

In terms of period, the velocity can be written as, 

$v = \dfrac{{2\pi \left( {r/2} \right)}}{T} = \dfrac{{\pi r}}{T}$

Substituting the value of v in equation 1, we get

\[\dfrac{{G{M}}}{{{r^2}}} = \dfrac{{2{\pi ^2}r}}{{{T^2}}}\]

Thus $T = \sqrt {\dfrac{{2{\pi ^2}{r^3}}}{{GM}}} $

Substituting the values of radius, r, and mass, M, we get

\[T = \sqrt {\dfrac{{2 \times \pi \times \pi \times {{(1.5 \times {{10}^{11}}\;\;)}^3}}}{{(6.67 \times {{10}^{ - 11}})(1.99 \times {{10}^{30}})}}} \]

\[ \Rightarrow T = 2.24 \times {10^7}s\]

But we know that for one year \[T = 3.154 \times {10^7}s\]

Thus $T = \dfrac{{2.24 \times {{10}^7}}}{{3.154 \times {{10}^7}}} \times 1y$

$ \therefore T = 0.71\,yr$

Thus the period of revolutions in years, T=0.71 yr

Hence the correct option is (B).

Note: Gravitation or just gravity is the force of attraction between any two bodies. All the objects in the universe attract each other with a certain amount of force, but in most of the cases, the force is too weak to be observed due to the very large distance of separation. Besides, gravity’s range is infinite but the effect becomes weaker as objects move away.