You've probably heard that the planets in a system orbit their star - but that's not the entire truth. Both the star and the planets orbit a center of mass, sort of the average position of the system, where more massive objects contribute more to the position than less massive objects. That's why we can sort of say that the planets orbit their star - the Earths' sun makes up 99.8 % of the total mass in the system - the center of mass is very close to the sun compared to how far away the planets are.
The center of mass sort of being the 'average' position of the system implies that the center of mass for N bodies can be expressed as:
\(\vec{R} = \frac{1}{M}\sum\limits^N_{i=1}m_i\vec{r}\)
where \(M = \sum_im_i\), which is the sum is over all the masses in the system.
We're looking at the star and a planet. Both of these move, the star has an elliptical orbit around the planet, same as the planet has around the star. How can we say anything about their movement?
We know that all objects that have mass pull at each other, the star exherts a gravitational force on the planet, \(\vec{F}_p\), and the planet exerts a gravitational force on the star, \(\vec{F}_s\). The gravitational force, \(\vec{F} = G\frac{m_sm_p}{\vec{r}^2} \vec{e}_r\), is the only force acting at the moment (\(\vec{e}_r\) is the unit vector pointing in the direction of r).
We have a lot of positions to keep track of here, so instead of standing at the origin, looking at the movement of two bodies, we sit down on the star (auch) and bring the origin with us. Now, we can describe the motion from the rest frame of the star- we are 'standing still'.
When the only force acting on the planet is the gravitational pull from the star, the force on the planet can be described as:
\(\vec{F}_p = -G\frac{m_sm_p}{|\vec{r}|^3} \vec{r} = m_p\ddot{\vec{r}}_p\)
From Newton's third law we know that \(\vec{F}_s = -\vec{F}_p\), and so we have that the force acting on the star from the planet is
\(\vec{F}_s = G\frac{m_sm_p}{|\vec{r}|^3} \vec{r} = m_s\ddot{\vec{r}}_s\)
Recall how \(\ddot{\vec{r}}\) is the derivative of r with respect to time, or \(\vec{a}\)! That should look familiar to you (it's just Newton's second law). We want to eliminate \(\ddot{\vec{r}}_p\) and \(\ddot{\vec{r}}_s\), as this is the position relative to the other body, and since they both move this is hard to keep track of. We therefore subtract \(\vec{F}_p\) from \(\vec{F}_s\)
\(\vec{r} = \ddot{\vec{r}}_p - \ddot{\vec{r}}_s = -G\frac{m_s+m_p}{|\vec{r}|^3} \vec{r} = -m\frac{\vec{r}}{\vec{r}^3}\)
where \(m= G(m_s+m_p)\).
We now have our expression for the motion of two bodies as \(\ddot{\vec{r}} + m\frac{\vec{r}}{r^3} = 0\). We can rewrite this expression using polar coordinates, where we describe the position vector using the angle \(\vec{\theta}\) and \(\vec{e}_r\) , and then solving the differential equation for the position in order to get an equation for the movement of a body:
\(r = \frac{p}{1+e\cos{f}}\)
This should look familiar to you by now!
However, we're interested in how the star and planet orbit around the center of mass. For the clearest results on how the planet affects the stars movement, we want to look at the planet that affects our star the most. This should be the most massive planet, but how far away it is also plays a part. We chose Planet 6, as this planet is very close to the star and has the second largest mass. The more massive planet had a rotational period almost twice as large as the planet we chose, but would probably also be a good candidate due to its large mass.
When we found the planet orbits we considered both the center of mass and the star at the same place, the origin. When considering a star and a planet orbiting a common center of mass, it's better to have them orbit their common center of mass.
Again we can use Leapfrog in order to find our orbits. For this we need initial positions, velocities and accelerations:
We have our planet positionf compared to the stars position, and a center of mass at a position \(\vec{R}\) from the star. We have to change our frame of reference so we know where the star and planets are relative to the center of mass.
From our earlier expression for the position vector of the center of mass we know that the position vector for two bodies can be written as
\(\vec{R} = \frac{m_s\vec{r}_s+m_p\vec{r}_p}{m_s+m_p}\)
When the star is in the origin, this gives the position vector for the center of mass as
\(\vec{R} = \frac{m_p\vec{r}_p}{m_s+m_p}\) be
cause \(\vec{r}_s\) is just 0 (or rather \(\vec{0}\), because we are working with vectors) when it is in the origin. We know that the center of mass lies somewhere between the two bodies we are looking at, and therefore we move our coordinate system to the right so that the origin lie in the center of mass.
We can consider \(\vec{R}\) the displacement along the x-axis when changing coordinate systems. This makes \(\vec{r}_s = -\vec{R}\) and , \(\vec{r}_p = \vec{r}_p-\vec{R}\). At this point both the star and planet orbit a common center of mass at the origin.
Let's take a moment to use these positions and the expression for elliptical movement, \(r = \frac{a(1-e^2)}{1+e\cos{f}}\), where e is the eccentricity to the planet and a is the semi-major axis, or the furthest point from the center of the ellipse, in order to plot the orbits not dependent on time.
We already know the semi-major axis of the planet with respect to the star, but what of the semi major axis with respect to the center of mass? The star is moving around the center of mass too, maybe we should find the semi-major axis of the star in our new coordinate system?
We can describe the movement of the star compared to the center of mass as \(|\vec{r}_s| = \frac{\mu}{m_s}r=\frac{\mu}{m_s}\frac{a(1-e^2)}{1+e\cos{f}}\) (recall that there are no external forces on the system and so the acceleration of the center of mass is zero), and the same for the planet, meaning we can set
\(a_s = \frac{\mu a_{p,s}}{m_s}\) and \(a_p = \frac{\mu a_{p,s}}{m_p}\). Here, we've set \(\mu = \frac{m_pm_s}{m_p + m_p}\), this is called the reduced mass.
Word of warning: This is the semi-major axis, and should not be confused with the acceleration! The alphabet has only so many letters, meaning that we sometimes have to recycle them. From here on, \(a_s\) and \(a_p\) will be the acceleration of the star and the planet.
And finally, we get this plot of the analytic orbits.
The eccentricity is still small, and we are therefore still getting nice, almost circular orbits. Yes, orbits in plural form! If you squint your eyes, you can see a tiny, blue dot. This is the orbit of the star around the center of mass. Let's zoom in:
The star orbits the center of mass, but the orbit is tiny compared to the orbit of the planet. As previously mentioned, a star tends to have way more mass than the other planets in a system combined (all of the planets make up 1.05% of all the mass in this solar system), and so the center of mass will lie very close to it. So close that we can almost consider the center of mass to be the star! If that were the case, the star would be at a standstill. Here we can see that the star is in fact moving, although very little. (It should be mentioned here that 0.005 AU is 747 989 kilometers, so not exactly a walking distance- everything is relative, as they say).
Another noticalbe quirk, is that since our stars radius is 1 746 710 km, the center of mass is inside the star! This is perfectly possible, imagine you're using a hula hoop- you're the center of mass, and the hoop is 'orbiting' around you, but you're still inside the hoop!
These orbits are useful because we can compare them with our time dependent orbits!
Now we want to find the position of the star and the planet by as a function of time! This way, we will know exactly where we can find them at any given time. The technique is much the same as before. We utilize the leapfrog method, and are therefore dependent on finding the acceleration of our two bodies.
We already found the initial positions but still need the initial acceleration and velocities. We have our expression for the gravitational force between the star and the planet
\(\vec{F} = -G\frac{m_sm_p}{|\vec{r}|^3} \vec{r}\)
where \(\vec{r}\) is the position vector from the source of the force to the body the force is acting upon.
The position vectors we have been working with so far have been very easy to find, as the star was lying in the origin. We still want to use the position vector between the star and the planet, but we only have the position vectors from the center of mass to the star and planet respectively. So in order to find the position vectors between the two bodies, we do the following vector additions:
\(\vec{r}_{s,p} = \vec{r_s} - \vec{r_p} \\ \vec{r}_{p,s} = \vec{r_p} - \vec{r_s}\)
where \(\vec{r}_{s,p}\) is the vector from the star to the planet, and \(\vec{r}_{p,s}\) is the vector from the planet to the star. Note that \(\vec{r}_{s,p}\) is the exact same vector as \(\vec{r}_{p,s}\), only directed the other way. We will therefore work with only one vector, and then change the sign according to which body we are working with.
The acceleration of the planet will be on the same form as previously, and the acceleration of the star will be the same, but opposite. Sound familiar? It should. According to Newton's third law, the force acting upon the star is equal to the force acting upon the planet, but oppositely directed. This is because the force is dependent on the position vectors, which are equal with opposite directions, as we have already shown.
We simulate the orbits of the star and planet over 20 years and get a circle and yet another blue dot. This means that the planet we have chosen to study has a period shorter (or equal to) than 20 years.
What about the motion of the star? We take a closer look at the blue dot:
Zooming in, we see this weirdly shaped line. Have we made some numerical error, perhaps? We have reason not to think so! The star orbits the center of mass in an elliptical pattern, but when viewed over time from the center of mass it will look like a corkscrew. When the star moves, the center of mass will shift. Seeing as we are looking at this movement from the center of mass' point of view, the star's elliptical orbit will be moving through space, and the circle becomes a spiral!
Why did we get a circle when we previously plotted our stars movement? Because the orbit of the star is circular around the center of mass when we ignore the movement over time! This prediction looks very close to our numerical calculation, and we can therefore conclude that the results from our integration so far looks satisfying.
As usual we wonder, can we be sure that we are correct?
We can do another test to see if our system holds by checking whether the energy in our integrated orbits are conserved. We know that the total energy in a system where the energy is conserved is the sum of the potential and kinetic energy. If the planets have not lost or gained energy between the first and last calculation, we know that the energy is conserved.
We can express the total energy in a system of two moving bodies as:
\(E = \frac{1}{2}m_s \mid\vec{v}_s \mid^2 + \frac{1}{2}m_p\mid \vec{v}_p \mid^2 - \frac{Gm_sm_p}{r}\) (why?)
Where we've used the velocities and positions of the star and planet relativt to the center of mass, therefore \(r = \mid \vec{r_s} -\vec{ r_m} \mid\), or \(\mid r_{s,p} \mid\), which is the vector between the star and the planet that we used in our integration. We also take the absolute values of the velocities, as energy is a scalar, not a vector. If we set the reduced mass, \(\hat{\mu}\) \(=\frac{m_1 m_2}{m_1 + m_2}\) and \(M = m_1 m_2\), we can rewrite this as:
\(E = \frac{1}{2} \hat{\mu} v^2 - \frac{GM \hat{\mu}}{r}\)
Once again, we look at the absolute value of difference, allowing for a small margin of error:
\(\mid E_2 - E_1 \mid < \epsilon\)
Our calculations prove to be very precise, as this holds for \(\epsilon = 0.00001\), or an error of 0.001%! At \(\epsilon = 0.000001\) we get a difference of 0.000004396 Joule. This is a negligible difference which we can write off as a numerical error. In other words, an overall satisfying result!
Why are we taking all this time obsessing over how our star moves? Did we simply want to show you more accurate results?
Well yes, that too. But did you know that watching for a certain 'wobble' in a stars movement is also a way to search for planets orbiting them? Orbiting stars cause planets to wobble in space, changing the color of the light astronomers observe!
Keep reading as we totally don't procastinate our space trip, by looking for aliens. Or at least a planet orbiting a distant star (fingers crossed there are aliens).