Worksheet 13: Problem 1

a) We often say that planets orbit stars. But planets and their stars actually orbit their mutual
řř center of mass, which in general is given by xcom=∑mixi∑mi. 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 mp. Also, set xcom=0. How do ap and a∗ depend on the masses of the star and planet? 

From the definition of the center of mass, we know that \(x_{com}\:=\:\frac{m_{p}a_{p}-M_{*}a_{*}}{M_{*}+m_{p}}\). Since the center of mass is defined on our timeline as 0, we know that this expression is equal to zero. Thus we can set the numerator equal to zero:

\[m_{p}a_{p}-M_{*}a_{*}\:=\:0\]

\[m_{p}a_{p}\:=\:M_{*}a_{*}\]

\[\frac{a_{p}}{a_{*}}\:=\:\frac{M_{*}}{m_{p}}\]

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=ap+a∗. Label this on your diagram. Now derive the relationship between the total mass M∗+mp≈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). 

Starting with the Virial Theorem, we can expand and find our desired relationship (one of Kepler's laws).

\[K\:=\:-\frac{1}{2}U\]

\[\frac{1}{2}m_{p}v^{2}\:=\:\frac{1}{2}\frac{GM_{*}m_{p}}{r}\]

\[m_{p}(\frac{2\pi a_{p}}{P})^{2}\:=\:\frac{GM_{*}m_{p}}{r}\]

Canceling out the mass of the planet and solving for \(P^{2}\), we find that:
\[P^{2}\:=\:\frac{4 \pi^{2} a_{p}^{3}}{GM_{*}}\:\approx\:\frac{4 \pi^{2} a^{3}}{GM_{*}}\]

This approximation holds true when the mass of the star is substantially greater than the mass of the planet.

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. (Useful numbers: M⊙≈1000MJup, PJup≈12 years, R⊙=7×1010cm ). 

From the first part of the problem we know that the absolute value of the Sun's displacement from the Solar System's center of mass, \(a_{*}\), is:

 

\[a_{*}\:=\:\frac{m_{p}a_{p}}{M_{*}}\]

 

We know that \(\frac{m_{p}}{M_{*}}\) is equal to \(\frac{1}{1000}\). Thus,

\[a_{*}\:=\:\frac{a_{p}}{1000}\]

From our period equation we know that:

\[a_{p}\:=\:(\frac{GM_{*}P^{2}}{4\pi^{2}})^{\frac{1}{3}}\]

Plugging in the for G, the mass of the Sun and Jupiter's period in seconds, we get:

\[a_{j}\:=\:(\frac{6.67\:\times\:10^{-8}\frac{cm^{3}}{gs^{2}}\:\times\:2\:\times\:10^{33}g\:\times\:(3.8\:\times\:10^{8}s)^{2}}{36})^{\frac{1}{3}}\:\times\:10^{-3}]

This comes out to roughly \(8\:\times\:10^{10}\) centimeters, which is about 1 solar radius.