Question

A binary star is a system of two stars moving around the centre of inertia of the system due to mutual gravitational force. If the total mass of two stars = $M_0$ and the period of revolution is T. Find the distance in km between stars? (Take $GM_0 = 8 \times 10^6$ and $T = 2\pi \times 10^3$ seconds).

Answer 1

20

Answer 2

The distance between the two stars in a binary system is related to the total mass $M_0$ and the period of revolution T by the formula derived from Kepler's third law for binary systems: $r^3 = \frac{G M_0 T^2}{(2\pi)^2}$ or $r = \left( \frac{G M_0 T^2}{4\pi^2} \right)^{1/3} = \left( G M_0 \left( \frac{T}{2\pi} \right)^2 \right)^{1/3}$

We are given the values: $GM_0 = 8 \times 10^6$ $T = 2\pi \times 10^3$ seconds

Substitute these values into the formula: $r = \left( (8 \times 10^6) \left( \frac{2\pi \times 10^3}{2\pi} \right)^2 \right)^{1/3}$ $r = \left( 8 \times 10^6 \times (10^3)^2 \right)^{1/3}$ $r = \left( 8 \times 10^6 \times 10^6 \right)^{1/3}$ $r = \left( 8 \times 10^{12} \right)^{1/3}$ $r = (8)^{1/3} \times (10^{12})^{1/3}$ $r = 2 \times 10^{12/3}$ $r = 2 \times 10^4$ meters

The question asks for the distance in kilometers. Convert meters to kilometers: $r_{km} = r_m / 1000$ $r_{km} = \frac{2 \times 10^4 \text{ m}}{10^3 \text{ m/km}}$ $r_{km} = 2 \times 10^{4-3} \text{ km}$ $r_{km} = 2 \times 10^1 \text{ km}$ $r_{km} = 20 \text{ km}$

The distance between the stars is 20 km.