[QUE/ME-12003] ME-PROBLEM

J.S.Plaskett's star is one of the most massive stars known at present. It is a double or a binary star, that is, it consists of two stars bound together by gravity. From spectroscopic studies it is known that the period of revolution of each component is 14.4 days; the velocity of each component is about 220 km/s; The orbit is nearly circular

1. [(a)] Argue that the masses of two stars are nearly equal and that they are nearly equidistant from the centre of mass of the system. 
2. [(b)] Compute the reduced mass and and the separation of the two components.

• Assume that the masses, velocities, radii of the circular orbits of the two stars are \(m_{1,2} v_{1,2}, r_{1,2}\). Let \(F_1, F_2\) be the gravitational pulls on a star due to the other star. Since the orbits are given to be circular \begin{equation}\label{EQ901} \frac{m_1v_1^2}{r_1}= F_1, \qquad \frac{m_2v_2^2}{r_2}= F_2, \end{equation} Now use \(F_!=F_2\) to get \begin{equation}\label{EQ902} \frac{m_1v_1^2}{r_1} =\frac{m_2v_2^2}{r_2} \Longrightarrow \frac{m_1}{r_1} =\frac{m_2}{r_2} \end{equation} the last step follows because it is given that \(v_1=v_2\). For a circular orbit, the speed is constant and time period is equal to (perimeter/ velocity): \begin{equation}\label{EQ903} T_1 = \frac{2\pi r_1}{v_1 }, \qquad T_2 = \frac{2\pi r_2}{v_2 } \end{equation} We are given \(T_1=T_2\) and \(v_1=v_2\), hence we get \(r_1=r_2\), and \EqRef{EQ902} then then gives \(m_1=m_2\).
• The radius of the orbits can be obtained from \eqref{\labelPrefix;EQ903}: \begin{eqnarray} r_1 &=& \frac{T_1 v_1}{2\pi} \\ &=& \frac{\HighLight{\( 14.4 \times 24 \times 60\)} \times 60 \ \text{\,sec})(220\times 10^{3} \text{m/s})}{2\pi}, \text{ use }\HighLight{\((14.4 \times 24)\times 60 \sim (14 \times (25 \times 60)\))} \nonumber\\ &\approx& (\HighLight{\(14\times 1500\)})(10)(220)\times 10^{3} \quad \HighLight{used \((60/2\pi) \sim 10\)} \\ &\approx& (200,00) (22)\times 10^5, \HighLight{Recall \(15^2=225\)}, \, \text{so used } \HighLight{\((14)(15)\sim 200\)}\\ &\approx& 4.4 \times 10^{10} \text{m}.\\ \end{eqnarray} Thus the distance between stars is \(2r_1 \sim 8.8 \times 10^{10}\) \, m. The mass can be computed by using the second law \begin{equation} \frac{mv^2}{r_1}= \frac{G m m }{(2r_1)^2} \Rightarrow m = \frac{4 v^2 r_1}{G} \end{equation} Substituting numerical values \begin{eqnarray} m &=& \frac{4 \times 220^2 \times 10^6 \times 4.4 \times 10^{10}}{6.67\times 10^{-11}} = \frac{4\times 484 \times 4.4 \times 10^{29}}{6.67}\\ &=& 4\times 484 \times \Big(\frac{2}{3}\Big) \times 10^{29} = 4 \times (1.61 \times 2) \times 10^{31} \\ &=& 4 (3.2)\times 10^{31} = 1.28 \times 10^{32} \text{ kg}. \end{eqnarray}