# Solar System barycenter - Orbit of planets

1. Jun 5, 2015

### Andrew1955

Hi

As far as I know the Earth orbits around the Sun Earth barycenter while the Sun orbits the Solar System BaryCenter formed by the changing center of mass of the Solar system. So even while the Sun orbits the SSBC and Earth orbits the Sun-Earth BC it would not be true to say the Earth orbits the SSBC.

So for example ISS orbits earth rather than earth moon BC.

I have though got myself into an almighty argument about this topic and I feel I need a fresh pair of scientifically minded eyes on the subject.

All thoughts welcomed!

Andrew

Last edited: Jun 6, 2015
2. Jun 5, 2015

### marcus

I think the positions and motions of all the bodies affect the shape of the gravitational field.
they contribute to the dynamic changing geometry encompassing the solar system,
a web of distances,angles, "straight" lines in the sense that great circles on the earth surface are the shortest distances between points, so-called geodesics.
I think that the path of a planet in orbit is a geodesic defined by whatever the geometry is, and all the bodies contribute to forming that.

3. Jun 5, 2015

### marcus

Talking about the Earth and Sun each orbiting the barycenter of the Earth-Sun system is an approximation. Often just studying a two-body subsystem of the whole is an extremely useful approximation but it is probably not the best way to think of it in reality. Might be good for calculating stuff though. Let's see what other people say in response to your question. You said "All thoughts welcome" and those are my thoughts, not in any sense authoritative or conclusive.

4. Jun 5, 2015

### Staff: Mentor

My understanding was that the Earth orbits the Earth-Moon barycenter, which itself orbits the solar system's barycenter.

5. Jun 6, 2015

### Andrew1955

>>My understanding was that the Earth orbits the Earth-Moon barycenter,

Yes, I should have mentioned that.

So for example the ISS orbits earth rather than the barycenter of the earth and moon

>> which itself orbits the solar system's barycenter.

I dont think so. The barycenter created by a theoretical object orbiting the ISS would still orbit Earth rather than the earth moon bc

6. Jun 6, 2015

### Andrew1955

Actually it appears to be precise about the topic, we need to say that at each instant of time the Moon orbits the center of the Earth while simultaneously the Earth is accelerated towards the centre of the moon. Therefore when we look at the results of this behaviour after it has happened we find the Earth and Moon appear to orbit the center of mass of the two objects.

7. Jun 7, 2015

### Bandersnatch

When asking whether A orbits B you need to be precise about what frame of reference are you using. It's also good to stick to one meaning of orbit.

ISS orbits the Earth (or rather the ISS-Earth barycentre) in the frame of reference centred on the barycentre of the Earth-ISS system. Note, that it means that it is easy in this reference frame to describe the motion of ISS using 2-body solutions (i.e. Keplerian orbits).

If you were to switch to the reference frame of the Earth-Moon (and, implicitly, ISS as well) barycentre, you'd find out that ISS follows a spiralling path around it that can no longer be described in terms of Keplerian conic sections. But it can be treated as a combination of ISS orbiting Earth-ISS barycentre, plus the Earth-ISS barycentre orbiting the Earth(with ISS)-Moon barycentre.
As long as when you say 'orbit', you're thinking of Keplerian orbits and not just of the fact that something goes around some point in whatever fashion, you can't say that in this FoR ISS orbits just the Earth, and you can't say it orbits just the E-M barycentre.

To visualise this, imagine attaching thrusters to ISS and raising its orbit. When you begin, you're likely to say it's orbiting Earth (you use the Earth-centric FoR), and if you raise it e.g. far beyond the orbit of the Moon you'd be inclined to say that it now orbits the E-M barycentre. Notice how it was a smooth process. While magnitudes of forces acting on ISS varied, there was no sudden qualitative jump. There was never a moment when the Moon 'turned on' its influence on ISS. The only thing that changed is your choice of a FoR to describe motion in a more convenient way.

Keep in mind, though, even these piecemeal Keplerian orbits are going to be just approximations of the actual paths of the objects, due to perturbations from the objects you disregard at any given stage. Since there are always more than 2 objects outside idealised thought experiments, perfect Keplerian orbits don't exist.

Back to the initial question. When you say that the Sun orbits the CoM of the solar system, you obviously don't mean Keplerian orbits. The path our star follows in this reference frame is an irregular, looping pattern that doesn't admit analytical solutions.
There's no difference if you were to say that Earth 'orbits' the CoM, as it also follows an irregular path around it, affected by all the bodies in the system.

When you say that Earth orbits the Earth-Sun barycentre, you choose a different reference frame and decide to treat all the other influences as perturbations in the hope of drastically simplifying the calculations (to a 2-body orbit) and allowing for at least an approximate solution without having to resort to numerical simulations.

In the same way you could say that the Sun orbits Sun-Jupiter barycentre, as this planet's gravitational pull on our star is the strongest, and treat all other planets as perturbers.

8. Jun 7, 2015

### rootone

I would assume that the the solar system does not have a fixed center of mass.
The center of mass will vary depending on the positions of planets in their orbit at any given time, with the gas giants contributing mostly to shifting it.
It won't correspond with the center of the Sun's core, at least only rarely would it do so, although it is probably always at some point within the body of the Sun.

9. Jun 7, 2015

### Bandersnatch

When you say that you're using some unspecified reference frame not coincident with the CoM. In the reference frame of CoM it is fixed by definition.

10. Jun 7, 2015

### rootone

Hmm yes I guess so.
The Solar system's center of mass clearly cannot be moving in relation to itself.
However the center of mass will always be changing it's position in relation to every other gravitationally significant solar system object, including the geometric center of the Sun core.
Gets difficult to visualise considering that center of mass at a given time is (probably) always going to be *somewhere* between the Sun's core and the surface.

11. Jun 7, 2015

### Bandersnatch

There's always this picture from the Wikipedia article on the Sun:

12. Jun 9, 2015

### Andrew1955

Thanks for the answers so far.

It seems to me that the force of Jupiter's gravity at the surface of the Sun is absolutely tiny, and if we calculate where the gravitational center of the Solar system is it is very near the center of the Sun. The barycenter therefore misleads a person into thinking Jupiter has some great ability to cause changes upon the Sun, and also causes a person to think that the Earth must orbit the center of mass of the solar system.

Can somebody guide me towards working out:

1. What the pull of Jupiter's gravity is at the Suns surface and

2. How i can calculate a 'center of gravity' of the two object system.

13. Jun 9, 2015

### Staff: Mentor

If by gravitational center you mean the center of mass, then I don't see how you're coming to that conclusion since the barycenter is the center of mass of the system.

From wiki: http://en.wikipedia.org/wiki/Barycenter

The distance from the center of a body (thought of as a point-mass) to the barycenter in a simple two-body case can be calculated as follows:

where :

r1 is the distance from body 1 to the barycenter
rtot is the distance between the two bodies
m1 and m2 are the masses of the two bodies.

14. Jun 9, 2015

### Bandersnatch

How comfortable are you with algebra and around equations in general? Do you know how to use Newton's law of gravity? $F_g=GMm/R^2$
It's just a matter of plugging in the numbers for masses and distance you can find on the wikipedia.
Note: while you're at it, calculate the same for the Sun on Earth or the Sun on Jupiter, or Jupiter on Earth (at their closest) for comparison.
It might be more meaningful to calculate acceleration than force, though. Take $F_g=ma$ and go from there.
Notice the meaning of the equation Drakkith supplied: the heavier body is as much closer to the CoM than the lighter one as it is heavier.

edit: jesus, forgot the gravitational constant there

Last edited: Jun 9, 2015
15. Jun 9, 2015

### Andrew1955

Thanks i have learnt about that equation and could use it. But if i want to calculate the suns gravity at jupiter it appears you are asking me to use the same equation?

16. Jun 9, 2015

### Bandersnatch

Yes, and it means that the force the Sun exerts on Jupiter is the same as Jupiter exerts on the Sun. That's why it'd be more meaningful to calculate the acceleration (just divide the force by the mass of whichever body).

17. Jun 9, 2015

### Andrew1955

It could be I am totally mixed up here but here is the nuts and bolts of the situation

Leif svalgaard a famous solar scientist has said the Earth orbits around the Sun and this is known to great precision using cm accurate results from the JPL lab where ephemerides are available to 7 decimal places.

Another group say he is lying! They say the Earth must be orbiting the SSBC.

If we calculate the BC for the Sun jupiter system we get a result of 742,723km from the center of the sun as earths orbital center

If we now place Jupiter at Mars we find the BC is 3 times nearer the center of the Sun and yet the force upon Earth by Jupiter is much greater

Similarly if we place an earth size orbit 4 light years from the Sun and assume it will orbit the Sun the BC is now 113,000,000km from the Sun.

So as far as i can see there is no relationship at all between the SSBC and where an object will orbit the solar system - apart from the Sun that is.

We also know that the gravitational difference across a satellite creates a torque and for larger distances the center of gravity is inside the center of mass.

As i say i could be totally muddled up......................

Last edited: Jun 9, 2015
18. Jun 9, 2015

### Andrew1955

Thanks i had sort of figured that out earlier but I was struggling with the idea even while seeing F was M times A

For some reason my head begins exploding when i think about anything other than arithmetic

I had a look at this calculator earlier so it should be simple to get this result

http://astro.unl.edu/classaction/animations/renaissance/gravcalc.html

Last edited: Jun 9, 2015
19. Jun 9, 2015

### Bandersnatch

You need to step back and say precisely what you mean when you make a statement such as 'A orbits B'.

20. Jun 9, 2015

### Andrew1955

Thanks. I did read your earlier Frame of Reference text and it was helpful to me.

In the first instance I am considering what part of the Solar system the Earth is being accelerated towards, where it seems the center of mass is not helping me to know the answer when gravity is inversely proportional to the square of the distance. Surely the earth orbits the Sun Earth BC with perturbations, just like the ISS orbits the ISS Earth BC with perturbations?

According to my calculations at the surface of the Sun, the suns gravity pulling downwards is 131 million times more powerful than jupiters gravity pulling upwards

Last edited: Jun 9, 2015