WS 10.2, Problem 1: Center of Mass

In the first part of this problem, we were asked to draw a picture illustrating the relationship between the masses of a star and planet and their center of mass.

The relationship between the center of mass and and masses is

$x_{COM}=\frac{(-a_*m_*)+(a_Pm_P)}{m_*+m_P}$

If we set the center of mass to 0, like it is in the picture, the equation becomes 

$a_*m_*=a_Pm_P$

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.  Label this on your diagram. Now derive the relationship between the total mass, , 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).

So we start with the Virial Theorem like the problem told us to.  

$K=-\frac{1}{2}U$

$\frac{1}{2}(m_*v_*^2+m_Pv_P^2)=\frac{Gm_*m_P}{a}$

In order to conserve momentum, the periods of the star and planet have to be the same.  

$v_*=\frac{2\pi a_*}{P}$

$v_P=\frac{2\pi a_P}{P}$

The star is so much more massive than the planet, so it moves a lot more slowly.  When the two are compared, the velocity of the star is practically 0.  

The distance from the center of mass to the planet, $a_P$ is also a lot larger than the distance to the star, $a_*$.  Since $a=a_*+a_P$, $a=a_P$.  

We can now rewrite the Virial Theorem as 

$m_P\left ( \frac{4\pi^2a_P^2}{P^2} \right )=\frac{Gm_*m_P}{a_P}$

and we can rearrange it to say

$m_*=\frac{4\pi^2a_P^3}{GP^2}$

which is basically just Kepler's Third Law of Motion, so yay!

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.

This should be a pretty simple plug and chug problem.  

$a_\odot M_\odot=a_{Jup}M_{Jup}$

$a_\odot=\frac{a_{Jup}M_{Jup}}{M_\odot}$

Plugging in all of those values, we find that the sun is displaced $5\times 10^4$cm, or $7\times 10^{-7}$ solar radii.