Question

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about their center of mass. The distance between them is the 6.2 times the distance between Earth and the Sun. What is their period of revolution in years?

Answer 1

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

`T^2 = (4\pi^2 d^3)/(2GM)`

The given data are given with respect to known constants, for example the mass of the sun is

`m_s = 1.989*10^(30)`

The radius between the earth and the sun is given by

`r = 149.6*10^9m`

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

`m = 8.2*1.989*10^(30)`

`d = 6.2*149.6*10^6`

Substituting in Kepler's third law:

`T^2 = (4\pi^2 d^3)/(2)`

`T^2 = (4\pi^2(6.2*149.6*10^9)^3)/(2(6.674*10^(-11) )(8.2*1.989*10^30 ))`

`T=\sqrt{(4\pi^2(6.2*149.6*10^9)^3)/(2(6.674*10^(-11) )(8.2*1.989*10^30))}`

`T = 120290789.7s`

`T = 120290789.7s((1year)/(31536000s))`

`T \approx 3.8143 years`

Therefore the period of this star is 3.8years