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 \(x_{com}=\sum _i m_i x_i / \sum_i m_i\). Set up the problem by
drawing an x-axis with the star at \(x=-a_*\) with mass \(M_*\), and the planet at \(x=a_p \text{ and } m_p\).
Also, set \(x_com =0\). How do ap and a* 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: \(a=a_p+a_*\) 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).
(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: \(M_* \approx 1000 M_{Jup}, P_{Jup} \approx 12 \text { years}, R_*=7\times 10^{10} cm\))
Let's begin.
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 \(x_{com}=\sum _i m_i x_i / \sum_i m_i\). Set up the problem by drawing an x-axis with the star at \(x=-a_*\) with mass \(M_*\), and the planet at \(x=a_p \text{ and } m_p\). Also, set \(x_com =0\). How do ap and a* depend on the masses of the star and planet?
Since we only have 2 objects, the summation is easy.
\[\sum_i m_i x_i /\sum_i m_i=\frac{-M_*a_*+m_pa_p}{M_*+m_p}=0\]
We can ignore the denominator, and add one tern to the other side to get:
\[\boxed{M_*a_*=m_pa_p}\]
(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: \(a=a_p+a_*\) 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).
We've done this before:
(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).
(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: \(M_* \approx 1000 M_{Jup}, P_{Jup} \approx 12 \text { years}, R_*=7\times 10^{10} cm\))
Let's begin.
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 \(x_{com}=\sum _i m_i x_i / \sum_i m_i\). Set up the problem by drawing an x-axis with the star at \(x=-a_*\) with mass \(M_*\), and the planet at \(x=a_p \text{ and } m_p\). Also, set \(x_com =0\). How do ap and a* depend on the masses of the star and planet?
\[\sum_i m_i x_i /\sum_i m_i=\frac{-M_*a_*+m_pa_p}{M_*+m_p}=0\]
We can ignore the denominator, and add one tern to the other side to get:
\[\boxed{M_*a_*=m_pa_p}\]
(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: \(a=a_p+a_*\) 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).
\(M\) is the Mass of the Star
\(m\) is the Mass of the Planet
\(m\) is the Mass of the Planet
\(a\) is the distance between the Planet and the Star
\(G=6.7\times 10^8 \frac{cm^3}{gs^2}\) is Newton's gravitational constant
\(P\) is the period of the Planet
\(K\) is Kinetic Energy
\(U\) is Gravity Potential Energy
\(v\) is the planet's velocity
We've done this before:
First let's start with the virial theorem.
\[K=-\frac{1}{2}U\]
We also know that \(K=\frac{1}{2}mv^2\) and that \(U=-\frac{GmM}{a}\).
Therefore:
\[\frac{1}{2}mv^2=\frac{GmM}{2a}\]
We can cancel the 1/2 and m.
\[v^2=\frac{GM}{a}\]
We can cancel the 1/2 and m.
\[v^2=\frac{GM}{a}\]
A planet's period is the time it takes to travel all the way around the planet so, \(P=\frac{2\pi a}{v}\). We can rearrange this to say that \(v=\frac{2\pi a}{P}\).
Substituting this in for v we get:
\[\frac{4\pi^2 a^2}{P^2}=\frac{GM}{a}\]
Simplifying and solving for P, we get:
\[\frac{4\pi^2 a^3}{P^2}=GM\]
\[\boxed{P^2=\frac{4\pi^2 a^3}{GM}}\]
(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: \(M_* \approx 1000 M_{Jup}, P_{Jup} \approx 12 \text { years}, R_*=7\times 10^{10} cm\)
We can start with Kepler's Third Law of Planetary Motion:
\[P^2=\frac{4\pi^2 a^3}{GM}\]
Simplifying and solving for P, we get:
\[\frac{4\pi^2 a^3}{P^2}=GM\]
\[\boxed{P^2=\frac{4\pi^2 a^3}{GM}}\]
(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: \(M_* \approx 1000 M_{Jup}, P_{Jup} \approx 12 \text { years}, R_*=7\times 10^{10} cm\)
We can start with Kepler's Third Law of Planetary Motion:
\[P^2=\frac{4\pi^2 a^3}{GM}\]
Next, let's rearrange to solve for a
\[a^3=\frac{P^2GM_*}{4\pi^2}\]
We want our final answer to be in terms of \(R_*\), so we now need to use a neat trick.
\[M_*=\frac{4}{3}\pi R_*^3 \rho \]
The density of the sun is almost \(1g/cm^3\) so we can "cancel" the term, giving us:
\[M_*=\frac{4}{3}\pi R_*^3\]
We can now substitute this into our prior equation:
\[a^3=\frac{P^2GR_*^3}{3\pi}\]
We want our final answer to be in terms of \(R_*\), so we now need to use a neat trick.
\[M_*=\frac{4}{3}\pi R_*^3 \rho \]
The density of the sun is almost \(1g/cm^3\) so we can "cancel" the term, giving us:
\[M_*=\frac{4}{3}\pi R_*^3\]
We can now substitute this into our prior equation:
\[a^3=\frac{P^2GR_*^3}{3\pi}\]
The next equation we need is the relation that we derived in part (a).
\[M_*a_*=m_pa_p\]
We know that \(M_* \approx 1000 M_{Jup}\), so:
\[1000m_pa_*=m_pa_p\]
\[1000a_*=a_p\]
We also know that \(a=a_p+a_*\) so:
\[a=1001a_*\]
Reverting to our original equation,
\[(1000a_*)^3=\frac{P^2GR_*^3}{3\pi}\]
\[a_*^3=\frac{P^2GR_*^3}{3\pi10^9}\]
Let's plug in some numbers. (12 years is about \(3.8\times 10^8\) seconds)
\[a_*^3=\frac{(3.8\times 10^8sec)^2 (6.7\times 10^{-8}c^3g^{-1}s^{-2})R_*^3}{3\pi10^9}\]
\[a_*^3=1.03R_8^3\]
\[\boxed{a_*\approx R_*}\]
This effectively means that the sun orbits a spot on it's equator due to the gravitational pull of Jupiter.
I worked with G. Grell and S. Morrison to solve this problem.
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