Worksheet 13.1 Problem 1: A Subtle Wobble

(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 \( X_{com} = \frac{ \sum{m_ix_i}}{\sum{m_i}} \) . Set up the problem by drawing an x-axis with the star at \( x = -a_* \) with mass \(M_*\), and the planet at \( x = a_p \, and \, m_p \). Also, set \( x_{com} = 0 \). How do \( a_p \) and \( a_* \) depend on the masses of the star and planet?

To demonstrate the actual orbiting behavior of stars and planets, we began by sketching out the system below:

In this problem, we examine the orbital relationship of a star (left), a planet (right), and their center of mass (Xcom). This position, declared as \(X_{com} \) is derived from the following: \[ X_{com} = \frac{\sum{m_i x_i}}{\sum{m_i}} \] For this system described above, this center of mass would be: \[  X_{com} = \frac{M_* a_* + m_p a_p }{ M_* + m_p} \] Since we're setting our center of mass to the origin of our system, we can set this equation to zero, which allows us to relate our distances and masses. \[ M_* a_* = m_p a_p \] \[ \frac{a_*}{a_p} = \frac{m_p}{M_*} \]

(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: \( a = a_p + a_* \). Label this on your diagram. Now derive the relationship between the total mass \(M_* + m_p \approx M_* \), 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).

First, we recall the virial theorem: \[K= - \frac{1}{2}U \] To use this, we can insert the value of K for the total mass of the system: \[\frac{1}{2}M_* v^2 = - \frac{1}{2} U \] Furthermore, we can insert the gravitational potential energy in the place of U: \[\frac{1}{2}M_* v^2 = - \frac{1}{2} ( \frac{-G {M_*}^2}{a}) \] \[ v^2 = \frac{GM}{a} \] Additionally, since v is equal to angular velocity, we can further expand this to: \[ (\frac{2 \pi a}{P})^2 = \frac{GM}{a} \] Finally, solving for P, we get: \[ P^2 = \frac{4 \pi^2 a^3}{GM} \] Should this look familiar, that's because this is none other than Kepler's Third Law.

(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.

Knowns:
\( M_{sun} \approx 1000 M_{Jup}\)
\( P_{Jup} \approx 12 \, years \)
\( R_{sun} = 7 \times 10^{10} \, cm \)

Knowing that Jupiter's period is 12 years, we can determine the distance between jupiter and the sun. Since we are trying to maintain simple units, such as the AU, we can actually calculate the jupiter-sun distance but setting up a proportion to the earth-sun distance: \[ \frac{{a_{jupiter-sun}}^3}{12^2 \, years} = \frac{{a_{earth-sun}}^3}{1^2} \] \[a_{jupiter-sun} = 12^{\frac{2}{3}} a_{earth-sun} = 12^{\frac{2}{3}} AU \] We can now use our equation from part a that relates distances and masses: \[a_p = a_* \frac{M_*}{m_p} \] The total distance, a, then equals: \[ a = a_* + a_* \frac{M_*}{m_p} \] Since we are approximating the sun to be 1000x heavier than jupiter, we can solve for \(a_* \): \[ a_* = \frac{ a } { 1+1000}\] \[ a_* = \frac{ 12^{\frac{2}{3}}AU} { 1+1000}\] \[ 1 \, AU = 214 \, R_{sun} \] \[ a_* = \frac{ 10^{3}R_{sun}} { 1+1000}\]

Acknowledgements: Team EE