Turns out stars rotate within their solar system

We are bounded in a nutshell of Infinite Space: Week 8: Worksheet #13: Problem #1: Turns out stars rotate within their solar system 

1. (a) We often say that planets orbit stars. But planets and their stars actually orbit their mutual center of mass, which in general is given by $x_{com} = \Sigma_i m_i x_i / \Sigma_i m_i$ . Set up the problem by drawing an x-axis with the star at $x = -a_\star$ with mass $M_\star$, and the planet at $x = a_P$ and $m_P$. Also, set $x_{com} = 0$. How do $a_P$ and $a_\star$ depend on the masses of the star and planet?

(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 (assume circular orbits from here on).

(c) By how much is the Sun displaced from the Solar System’s center of mass (a.k.a. the Solar System “barycenter”) as a result of Jupiter’s orbit? Express this displacement in a useful unit such as Solar radii. (Potentially useful numbers $M_\odot \approx 1000 M_{Jup}$ and $a_{Jup} \approx 5.2 AU$.)

a. Taking this problem as a simple center of mass problem, we can use the illustration below:

 and now we can better understand the original sum equation: $$x_{com} = \frac{\sum_i m_i x_i }{\sum_i m_i} ,$$ and identifying all the masses and their position relative to the center: $$x_{com} = \frac{M_\star (a_\star) + m_P a_P}{M_\star + m_P},$$ and setting $x_{com}$ to 0:   $$0 = \frac{M_\star (- a_\star) + m_P a_P}{M_\star + m_P},$$ we continue simplifying   $$\frac{-m_P a_P}{M_\star + m_P} = \frac{-M_\star a_\star}{M_\star + m_P},$$ and a simpler equality begins to emerge:   $$m_P a_P = M_\star a_\star,$$ meaning we find a direct correlation of mass of an entangled system between the objects’ masses and their semi-major axis of rotation: $$\frac{m_P}{M_\star} = \frac{a_\star}{a_P}$$

b. 

Using the virial theorem (similar to a problem directly involving the Orbital period earlier in the semester) we can find the correlation of mass to Period, and semi-major axis. Starting off with the base equation: $$K = -\frac{1}{2} U ,$$ we begin representing these energy values with the objects in question, mainly the gravitational potential energy and the kinetic energy (the movement of the star at the center of the system is negligible here):   $$\frac{1}{2} m_P v^2 = -\frac{1}{2} \frac{-GM_\star m_P}{a_\star + a_P},$$ and we take into account the definitions discussed by the problem: $$M_\star + m_P \approx M_\star,$$ $$a_\star + a_P = a,$$   which allow the previous equation to simplify down to: $$v^2 = \frac{GM_\star}{a},$$ and now we define the velocity as a rotational system: $$v = \frac{2\pi a}{P} ,$$ and plug this in: $$\frac{4\pi^2 a^2}{P^2} = \frac{GM_\star}{a},$$ which is then placed simplified into a version of Kepler’s third law: $$P = \left(\frac{4\pi^2 a^3}{GM_\star}\right)^{\frac{1}{3}}$$

c. Using the equations we just derived, we know the proportion is $$\frac{m_P}{M_\star} = \frac{a_\star}{a_P},$$ and we substitute in the values for the mass of Jupiter and the axis of the orbit of Jupiter: $$\frac{1/1000 M_\odot}{1 M_\odot} = \frac{a_\star}{5.2 AU},$$ which solves to: $$\frac{5.2 AU}{1000} = a_\star$$ $$a_\star = 5.2 \times 10^{-3} AU,$$ and knowing a particular equivalence of: $$1 AU =    214.94 R_\odot,$$ we can see how the Sun’s orbit is shifted by the mass of Jupiter by the following distance: $$a_\star = 1.1 R_\odot.$$