Since we know that \(x_{com}=\sum_im_ix_i\sum_im_i\), we can say that \(x_{com}=\frac{-a_\star M_\star+a_pm_p}{M_\star+m_p}\). Setting \(x_{com}=0\) allows us to determine that \(M_\star a_\star=m_pa_p\).
b) In a two-body orbital system the variable a is the mean semimajor axis, or the sum of the planet's and star's distances away from their mutual center of mass: \(a=a_p+a_\star\). Label this on your diagram. Now derive the relationship between the total mass \(M_\star+m_p\approx M_\star\), orbital period P, and the mean semimajor axis a, starting with the Virial Theorem for a two-body orbit.
The Virial Theorem states that \(-\frac{1}{2}U=K\) or \(-\frac{1}{2}\frac{GMm}{r}=\frac{1}{2}mv^2\). Since the planet is much smaller than the star, the center of mass will be much closer to the star and therefore \(a\approx a_p\). This allows us to find the velocity based on the planet's period: \(v=\frac{2\pi a}{P}\).
\(m_p(\frac{2\pi a}{P})^2=\frac{GM_\star m_p}{a}\)
\(\frac{4\pi a^2}{P^2}=\frac{GM}{a}\)
\(P^2=\frac{4\pi^2a^3}{GM}\)
This is Kepler's third law!
c) By how much is the Sun displaced from the Solar System's center of mass as a result of Jupiter's orbit? Express this displacement in a useful unit such as Solar radii. (Potentially useful numbers: \(M_\odot\approx1000M_{Jup}\) and \(a_{jup}\approx5.2AU\).)
\(M_\star a_\star=m_pa_p\)
\(1000M_{Jup}a_\odot=M_{Jup}5.2AU\)
\(a_\odot=0.0052AU\)
\(\frac{0.0052AU}{1}\frac{1.5\times10^8km}{1AU}\frac{1R_\odot}{7.0\times10^5km}\approx1R_\odot\): Jupiter displaces the Sun by about one Solar radius.
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