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?

Short answer

The period of revolution of the binary-star system is $0.71\mathrm{year}$

Step-by-step solution

Step 1: Given

The mass of each star of the binary-star system is$\text{M}=1.99\times {10}^{30}\text{\hspace{0.17em}kg}$

The distance between two stars of the binary-star system is $\text{d}=\text{2r}=1.5\times {10}^{11}\text{\hspace{0.17em}m}$

Determining the concept

Find the period by equating the gravitational force of attraction and centripetal force in the binary star system. Gravitational force is the force exerted by the Earth towards it.

Formulae are as follows:

The Centripetal force,$\mathrm{F}=\frac{{\mathrm{Mv}}^2}{\mathrm{r}}$.

The Gravitational force of attraction between two bodies of masses M and m separated by distance d is$\mathrm{F}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$.

where F is the centripetal force, G is the gravitational constant, M, and m are masses, v is velocity and r is the radius.

Determining the period of revolution of the binary-star system

Now,

The gravitational force between stars = centripetal force between two stars

$\mathrm{F}=\frac{{\mathrm{GM}}^2}{{\mathrm{d}}^2}=\frac{{\mathrm{Mv}}^2}{\mathrm{r}}$

$\begin{array}{c}\frac{{\mathrm{GM}}^2}{{\mathrm{d}}^2}=\frac{\mathrm{M}{\left(\frac{2\mathrm{\pi r}}{\mathrm{T}}\right)}^2}{\mathrm{r}}\\ \frac{{\mathrm{GM}}^2}{4{\mathrm{r}}^2}=\frac{\mathrm{M}{\left(\frac{2\mathrm{\pi r}}{\mathrm{T}}\right)}^2}{\mathrm{r}}\\ \frac{{\mathrm{GM}}^2}{4{\mathrm{r}}^2}=\frac{4{\mathrm{M\pi }}^2{\mathrm{r}}^2}{{\mathrm{T}}^2\mathrm{r}}\\ {\mathrm{T}}^2=\frac{16{\mathrm{\pi }}^2{\mathrm{r}}^3}{\mathrm{GM}}\\ \mathrm{T}=\sqrt{\frac{16{\mathrm{\pi }}^2{\mathrm{r}}^3}{\mathrm{GM}}}\\ \mathrm{T}=\sqrt{\frac{16\times {\left(3.142\right)}^2{(0.75\times {10}^{11}\mathrm{m})}^3}{6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2(1.99\times {10}^{30}\mathrm{kg})}}\\ \mathrm{T}=2.2\times {10}^7\mathrm{s}\end{array}$

But,

$1\text{\hspace{0.17em}\hspace{0.17em}year}=3.156\times {10}^7\text{s}$

So,

$\begin{array}{c}\mathrm{T}=\frac{2.241\times {10}^7\mathrm{s}}{3.156\times {10}^7\mathrm{s}}\times 1\mathrm{year}\\ =0.71\mathrm{year}\end{array}$

Therefore,the period of revolution of the binary-star system$0.71\mathrm{year}$.

Therefore, using the gravitational force of attraction between two objects and the centripetal force, the period of revolution of a system can be found.