 # This question has to do with binary star systems, where 'i' is the angle of incl Nann 2021-09-15 Answered
This question has to do with binary star systems, where i is the angle of inclination of the system.
Calculate the mean expectation value of the factor ${\mathrm{sin}}^{3}$ i, i.e., the mean value it would have among an ensemble of binaries with random inclinations. Find the masses of the two stars, if ${\mathrm{sin}}^{3}$ i has its mean value.
Hint: In spherical coordinates, $\left(\theta ,\varphi \right)$, integrate over the solid angle of a sphere where the observer is in the direction of the z-axis, with each solid angle element weighted by $\mathrm{sin}\left\{3\right\}\left(\theta \right)$.
${v}_{1}=100k\frac{m}{s}$
${v}_{2}=200k\frac{m}{s}$
Orbital period $=2$ days
${M}_{1}=5.74e33g$
${M}_{2}=2.87e33g$
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Step 1
For a binary star system, the ratio of their velocities and their masses are related as,
$\frac{{m}_{1}}{{m}_{2}}=\frac{{v}_{2}}{{v}_{1}}$
$=\frac{200k\frac{m}{s}}{100k\frac{m}{s}}$
$=2$

The Kepler equation for a binary star system is,
${m}_{1}+{m}_{2}=\frac{P}{2\pi G}\frac{\left({v}_{1}+{v}_{2}{\right)}^{3}}{{\mathrm{sin}}^{3}\left(i\right)}$
$=\frac{\left(2×24×3600s\right)}{2\pi \left(6.67×{10}^{-11}N{m}^{2}k{g}^{-2}\right\}}$ $\frac{{\left({10}^{5}m{s}^{-1}+2×{10}^{5}m{s}^{-1}\right)}^{3}}{{\mathrm{sin}}^{3}\left(i\right)}$
$=\frac{\left(1.11×{10}^{31}kg\right)}{{\mathrm{sin}}^{3}\left(i\right)}\left(\frac{{M}_{\circ \le d\circ }}{2×{10}^{30}kg}\right)$
$=\frac{5.556{M}_{\circ \le d\circ }}{{\mathrm{sin}}^{3}\left(i\right)}$
Step 3
The mean expectation value of ${\mathrm{sin}}^{3}$ i be given as,
$⟨{\mathrm{sin}}^{3}\left(i\right)⟩={\int }_{0}^{\pi /2}{\mathrm{sin}}^{3}\left(i\right)\left\{\mathrm{sin}idi\right\}$
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{4}\left(i\right)di$
$={\left[\frac{3i}{8}-\frac{\mathrm{sin}\left(2i\right)}{4}+\frac{\mathrm{sin}\left(4i\right)}{32}\right]}_{0}^{\frac{\pi }{2}}$
$=\frac{3\pi }{16}$
$=0.59$
Step 4
From equation (1), ${m}_{1}=2{m}_{2}$
Hence,
$3{m}_{2}=\frac{5.556{M}_{\circ \le d\circ }}{{\mathrm{sin}}^{3}\left(i\right)}$
${m}_{2}=\frac{5.556{M}_{\circ \le d\circ }}{3\left(0.59\right)}$
$=3.1389{M}_{\circ \le d\circ }$
Similarly,
${m}_{1}=6.3{M}_{\circ \le d\circ }$
Here ${M}_{\circ \le d\circ }$ defined as the solar mass. Solar mass equal to approximately