Does Jupiter orbit the Jupiter-Sun barycenter or not?

[Robittybob1]

Of all the planets Jupiter being the largest has the best chance of orbiting its barycenter with the Sun, well so I thought.

 

What does the maths tell us?

How to calculate the Barycenter?

How to calculate the gravitational attraction between the Sun and Jupiter.

Comparing that gravitational force to the centrifugal force of a planet orbiting a point.

 

What do the references tell us at the moment.

[Strange]

 

Does Jupiter orbit the Jupiter-Sun barycenter or not?

 

Of course. That is the definition of barycentre.

 

 

How to calculate the Barycenter?

How to calculate the gravitational attraction between the Sun and Jupiter.

 

You have had these explained to you several times in (several) other threads. I am baffled that you keep starting new threads just to ask the same questions.

[Robittybob1]

 

Of course. That is the definition of barycentre.

 

 

You have had these explained to you several times in (several) other threads. I am baffled that you keep starting new threads just to ask the same questions.

You will note they are not the same question every time, they are specific questions that I want answered. In this thread I want to discover the barycenter between Jupiter and the Sun, and then to get Jupiter and the Sun to orbit that point perfectly, so that the gravitational attraction between the masses balances both of their centrifugal forces as they orbit the barycenter. Up to now none of the figures were giving consistent answers.

If there was just the Sun and Jupiter in the Solar System and they had perfectly circular orbits there are values whereby the Sun and Jupiter with their current calculated mass could orbit their barycenter at the exact time period.

Once you throw in all the other planets and their eccentric orbits I have a feeling it is not that easy to do.

 

These are the figures for a perfect orbit of Jupiter around the Sun both orbiting their common barycenter.

Distance apart () than best measurement)

barycenter from primary (Sun) () than best measurement).

 

If the actual figures exceed these precise values it shows something else is happening. (still being edited)

Edited February 18, 2015 by Robittybob1

[imatfaal]

You will note they are not the same question every time, they are specific questions that I want answered. In this thread I want to discover the barycenter between Jupiter and the Sun, and then to get Jupiter and the Sun to orbit that point perfectly, so that the gravitational attraction between the masses balances both of their centrifugal forces as they orbit the barycenter. Up to now none of the figures were giving consistent answers.

If there was just the Sun and Jupiter in the Solar System and they had perfectly circular orbits there are values whereby the Sun and Jupiter with their current calculated mass could orbit their barycenter at the exact time period.

Once you throw in all the other planets and their eccentric orbits I have a feeling it is not that easy to do.

 

These are the figures for a perfect orbit of Jupiter around the Sun both orbiting their common barycenter.

Distance apart 778329968793.5 m (217,231,207 m less than best measurement)

barycenter from primary (Sun) 741974768.8 m (819231 m less than best measurement).

 

Because the actual figures exceed these precise values it shows something else is happening.

Janus, SwansonT and Others have already explained the idea of precision - and more importantly spurious precision. You are seriously still giving astronomical distances to an accuracy of 10cm?

 

To show that "something else is happening" you need to calculate error ranges and show that your figures are outside these error ranges. Basically every figure will have an error (the masses, the orbits, the constants - all of them) which have been agreed - this standard uncertainty will have been quantified (this in my opinion is a wonder of science - the understanding of the nature of error, imprecision, and uncertainty). Simplisitically you need to do all the calcs biased towards a small answer (ie smallest possible on the top and largest possible on the bottom) - and then redo towards a large answer (ie big on top and small below); you will then have a range rather than a single figure, you will probably be disappointed how large this range is.

There will be problems with this as many of the figures will have been calculated in the same way and you may end up compounding error ranges un-necessarily. But you have to account for errors. You could calculate from only empirical data - but that would be a lot of legwork

[Robittybob1]

Janus, SwansonT and Others have already explained the idea of precision - and more importantly spurious precision. You are seriously still giving astronomical distances to an accuracy of 10cm?

 

To show that "something else is happening" you need to calculate error ranges and show that your figures are outside these error ranges. Basically every figure will have an error (the masses, the orbits, the constants - all of them) which have been agreed - this standard uncertainty will have been quantified (this in my opinion is a wonder of science - the understanding of the nature of error, imprecision, and uncertainty). Simplisitically you need to do all the calcs biased towards a small answer (ie smallest possible on the top and largest possible on the bottom) - and then redo towards a large answer (ie big on top and small below); you will then have a range rather than a single figure, you will probably be disappointed how large this range is.

There will be problems with this as many of the figures will have been calculated in the same way and you may end up compounding error ranges un-necessarily. But you have to account for errors. You could calculate from only empirical data - but that would be a lot of legwork

Note: I have withdrawn those figures for the moment.

[Robittybob1]

As I posted in the other Sun barycenter thread but is very pertinent here; this point in important. So this is not a repeated post.

 

Why am I using centrifugal force when I have not seen centrifugal force being used? Centrifugal (or is it centripetal force?) for I don't assign a sign to it but I just calculate "mV^2/r" but the r has to be the radius that the body is taking around the barycenter not the "r" of the gravitational attraction force., which is the distance between the bodies (that was the source of my errors yesterday).
With this simultaneous equation situation I have been able to find exact solutions to bodies going around a barycenter. (That was where I ended up last night at 1:30 in the morning! I love science to the detriment of my body.)
Like if we were to say the masses and times are exact (which they are not, but for the time let's say they are), we then can tell exactly where the masses will orbit (if they are in a true binary orbiting situation).
Using these two equations and redefining the distances between the bodies I am able to readjust the two radii of the barycenter continually refining the distances until I got the exact solution. (OK there maybe better math out there but that is how I have been able to do it using a formulated Excel worksheet which automatically recalculates the barycenter distances and both lots of forces every time I adjust the distances between the bodies looking for the situation where each body's gravitational attraction is balanced by the centrifugal force it experiences going around the barycenter with that exact period and mass.

I had my doubts about the Sun being a true binary, the reason I started that thread, but to get a lead on the situation we looked first looked at Pluto Charon as an example of a true binary, and later in this thread we are looking specifically at the Jupiter Sun situation. Trying our best to keep on topic.

[Mordred]

Excellent work excellent is a handy tool +1

Edited February 18, 2015 by Mordred

[Robittybob1]

Excellent work excellent is a handy tool +1

Thanks for the encouragement. But now I have these exact figures and I have to to see if they could be physically correct, for the exact figures I have found don't represent the real situation but the ideal situation, and now I am trying to find an explanation why the real measurements are different from the ideal. It is hard work trying to comprehend the situation. So I'm going to have to do some chores and come back to it.

 

I had better see if the ideal figures work for the Sun part of the barycenter as well. I have only calculated for the Jupiter half at this moment, but since I have a way of adjusting the ideal barycenter radii I have every confidence it will, but better to be safe than sorry.

Edited February 19, 2015 by Robittybob1

[Robittybob1]

Thanks for the encouragement. But now I have these exact figures and I have to to see if they could be physically correct, for the exact figures I have found don't represent the real situation but the ideal situation, and now I am trying to find an explanation why the real measurements are different from the ideal. It is hard work trying to comprehend the situation. So I'm going to have to do some chores and come back to it.

 

I had better see if the ideal figures work for the Sun part of the barycenter as well. I have only calculated for the Jupiter half at this moment, but since I have a way of adjusting the ideal barycenter radii I have every confidence it will, but better to be safe than sorry.

The formula is working well for both the primary and the secondary sides of the barycenter

 

So what are the parameters treating the period and mass of the Sun and Jupiter as exact.

 

These are the figures for a perfect orbit of Jupiter around the Sun both orbiting their common barycenter.

Distance apart 778329968793.6 meters, 217,231,206 meters less than current best measurement.

barycenter from primary (Sun) 741974768.8 meters, 819231.2 meters less than best measurement.

Barycenter from secondary (Jupiter) 777587994024.7 meters, 216,411,975 meters less than current best measurement.

 

My next step will be to check the orbital period of Jupiter for it seems to be orbiting a larger distance than it would as a true "ideal" binary.

Could it be just the mass of the inner planets enabling Jupiter to orbit the barycenter at the higher rate?

Edited February 19, 2015 by Robittybob1

[Fuzzwood]

Do you also take into account the effect of all other orbital bodies? What you are calculating is definitely not a 2-body system.

[Robittybob1]

Do you also take into account the effect of all other orbital bodies? What you are calculating is definitely not a 2-body system.

Not at this stage, but I did edit my previous post.

 

 

Could it be just the mass of the inner planets enabling Jupiter to orbit the barycenter at the higher rate?

Edited February 19, 2015 by Robittybob1

[Strange]

Has anyone pointed you at this, yet: https://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29

[Robittybob1]

Has anyone pointed you at this, yet: https://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29

Yes - it is partly what prompted this study.

 

Could it be just the mass of the inner planets enabling Jupiter to orbit the barycenter at the higher rate?

Adding in the extra mass of the inner planets (just added to the mass of the Sun improved the orbital distances but not by enough.

 

Distance apart 778331518127.8 meters, 215,681,872 meters less than current best measurement.

 

1.99077E+30 if the Mass of the Sun plus inner planets was 1.99077E+30 then the jupiter-sun barycenter works properly. How accurate has the mass of the Sun been calculated?

Inner planets added to 1.19E+25 kg (MVEmMA)

 

1.98855±0.00025)×10^30 kg Best answer from Wikipedia and agrees with NASA site. so it seems it will never get to 1.99077

Edited February 19, 2015 by Robittybob1

[Strange]

 

Distance apart 778331518127.8 meters, 215,681,872 meters less than current best measurement.

 

"778331518127.8 meters"? Really? I mean, really??

[imatfaal]

Solar Mass is known to about 1 part in 10000 . The solar mass parameter GM_sum is known to about 1 part in 10 billion

[swansont]

Once again, I will point out that mv²/r is not the correct equation, since the orbit is not circular. The eccentricity is around 0.05, so it's a good approximation, but there is no legitimate reason to expect agreement of calculations with the observations to better than a percent, give or take (go ahead and run the numbers). So I would not be surprised if discrepancies less than a million km are all due to the faulty assumption of a circular orbit. That's the first place to look, before questioning the physics.

 

IOW, when you knowingly use the wrong equation, you should not be surprised that you get the wrong answer.

[imatfaal]

 

"778331518127.8 meters"? Really? I mean, really??

 

And he has been told about half a dozen times already

[Robittybob1]

 

"778331518127.8 meters"? Really? I mean, really??

That was from the "idealistic Jupiter -Sun binary orbiting the common barycenter", based on the parameters of period and masses being correct.

Once again, I will point out that mv²/r is not the correct equation, since the orbit is not circular. The eccentricity is around 0.05, so it's a good approximation, but there is no legitimate reason to expect agreement of calculations with the observations to better than a percent, give or take (go ahead and run the numbers). So I would not be surprised if discrepancies less than a million km are all due to the faulty assumption of a circular orbit. That's the first place to look, before questioning the physics.

 

IOW, when you knowingly use the wrong equation, you should not be surprised that you get the wrong answer.

OK I will next explore how the eccentricity will affect orbital periods.

Solar Mass is known to about 1 part in 10000 . The solar mass parameter GM_sum is known to about 1 part in 10 billion

Thanks. The error in the Sun's mass would need to be 9 parts in 10,000 to get it up to 1.99077E+30 kg so we can count that out.

Edited February 19, 2015 by Robittybob1

[swansont]

OK I will next explore how the eccentricity will affect orbital periods.

 

I suggest you compare the circumference of the circle you are using with that of the ellipse from Jupiter's orbital parameters.

[Robittybob1]

 

I suggest you compare the circumference of the circle you are using with that of the ellipse from Jupiter's orbital parameters.

Could you explain that a bit more fully please? Do you want the difference in circumference length?

 

And he has been told about half a dozen times already

I have yet to master how I could work with Excel and having the error bars working alongside the best values.

When the mass of the Sun was adjusted upward it was surprising that the radius of Jupiter's orbit increased. This seems counter intuitive until you realise that the higher mass allows Jupiter to orbit at a greater distance in the same orbital period.

The current situation is as if Jupiter is orbiting a Sun of a slightly larger mass, but it is not the Sun's mass, so what is it?

Edited February 19, 2015 by Robittybob1

[swansont]

Could you explain that a bit more fully please? Do you want the difference in circumference length?

 

If they are different, then this explains orbit speed differences. The actual distance of travel is not what you calculate.

[imatfaal]

I was curious if I could get more details of the calculations involved (as even using the correct equations I am still a distance off from the published data); I wanted to know which data are observations and which were calculations. It seems that the explanatory guide is 734 pages long and written by scientists from some strange clandestine organisation called the US Naval Observatory!

[Robittybob1]

 

If they are different, then this explains orbit speed differences. The actual distance of travel is not what you calculate.

So far I have been calculating the distance around a circular orbit and then dividing that with the known period for Jupiter (OK I take the period as 100% accurate) "Jupiter period around Sun (Jupiter orbits the Sun every 11.86 Earth years (or 4,332 days) * seconds in a day (86400) = 374,284,800 seconds." there might be a degree of error here for they just worked it out to the nearest day.

Has anyone got a more precise period of Jupiter's orbit? It might vary that much it is only accurate to the nearest day.

So I use 374,284,800.000000 seconds heaps of zeros (15 digits per number) are added by Excel.

So with elliptical orbits there will be sections of faster travel balanced with slower periods and the orbital velocity will not always be orthogonal to the radius to the barycenter (for the centripetal force calculations), in fact the barycenter is continually shifting too (as the distance between the Sun-Jupiter is always changing), so one hopes there is a simplification for this.

[swansont]

I was curious if I could get more details of the calculations involved (as even using the correct equations I am still a distance off from the published data); I wanted to know which data are observations and which were calculations. It seems that the explanatory guide is 734 pages long and written by scientists from some strange clandestine organisation called the US Naval Observatory!

 

Gotta be those dang astronomers.

So far I have been calculating the distance around a circular orbit and then dividing that with the known period for Jupiter (OK I take the period as 100% accurate) "Jupiter period around Sun (Jupiter orbits the Sun every 11.86 Earth years (or 4,332 days) * seconds in a day (86400) = 374,284,800 seconds." there might be a degree of error here for they just worked it out to the nearest day.

Has anyone got a more precise period of Jupiter's orbit? It might vary that much it is only accurate to the nearest day.

So I use 374,284,800.000000 seconds heaps of zeros (15 digits per number) are added by Excel.

So with elliptical orbits there will be sections of faster travel balanced with slower periods and the orbital velocity will not always be orthogonal to the radius to the barycenter (for the centripetal force calculations), in fact the barycenter is continually shifting too (as the distance between the Sun-Jupiter is always changing), so one hopes there is a simplification for this.

 

That timing accuracy isn't the problem. ±1 day out of 4332 is 0.02%

 

Using the wikipedia data, you can get the semi-major axis (797.5 million km), and using the eccentricity (0.05), you get the semi-minor axis (777.3 million km)

 

That gives a circumference of 4.95 billion km.

 

Using r = 778.3 million km for a circle, you get 4.89 billion km. That's a difference of 1.2%. All of your calculations of orbital speeds will be off on the low side — Jupiter travels further in a year than you would calculate assuming a circle. Any calculations using a circle can't be expected to agree with observation to better than ~1%. Any difference smaller than this and the numbers are statistically equal. IOW, you can't argue that they are unequal.

 

You really, really, really need to learn about significant digits and measurement precision.

[Robittybob1]

 

Gotta be those dang astronomers.

 

That timing accuracy isn't the problem. ±1 day out of 4332 is 0.02%

 

Using the wikipedia data, you can get the semi-major axis (797.5 million km), and using the eccentricity (0.05), you get the semi-minor axis (777.3 million km)

 

That gives a circumference of 4.95 billion km.

 

Using r = 778.3 million km for a circle, you get 4.89 billion km. That's a difference of 1.2%. All of your calculations of orbital speeds will be off on the low side — Jupiter travels further in a year than you would calculate assuming a circle. Any calculations using a circle can't be expected to agree with observation to better than ~1%. Any difference smaller than this and the numbers are statistically equal. IOW, you can't argue that they are unequal.

 

You really, really, really need to learn about significant digits and measurement precision.

Looking it up in Wikipedia there is a statement "Note that for all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity." so that to me implies that an orbit is equivalent whether circular or elliptical provided the semi-major axis is the same (semi-major axis = radius for a circular orbit).

Now in the same article it did give the reasons orbital speeds are calculated as if the planets are orbiting the Sun rather than the barycenter (earlier issue).

http://en.wikipedia.org/wiki/Semi-major_axis

 

The orbiting body's path around the barycentre and its path relative to its primary are both ellipses. The semi-major axis is sometimes used in astronomy as the primary-to-secondary distance when the mass ratio of the primary to the secondary is significantly large (M»m); thus, the orbital parameters of the planets are given in heliocentric terms. The difference between the primocentric and "absolute" orbits may best be illustrated by looking at the Earth–Moon system. The mass ratio in this case is 81.30059. The Earth–Moon characteristic distance, the semi-major axis of the geocentric lunar orbit, is 384,400 km. The barycentric lunar orbit, on the other hand, has a semi-major axis of 379,700 km, the Earth's counter-orbit taking up the difference, 4,700 km. The Moon's average barycentric orbital speed is 1.010 km/s, whilst the Earth's is 0.012 km/s. The total of these speeds gives a geocentric lunar average orbital speed of 1.022 km/s; the same value may be obtained by considering just the geocentric semi-major axis value.

The speeds are measured in heliocentric terms quote: "the orbital parameters of the planets are given in heliocentric terms", or as in the example in geocentric terms.

 

But when I'm applying the centripetal force equation I am looking at the individual speeds separately, i.e. orbital speed of the Sun and the orbital speed of Jupiter, even though both of them added together gives the classic heliocentric orbital speed of Jupiter around the Sun, even though we assume Jupiter goes around the barycenter and not the Sun. - confused?

 

There is a new word - "primocentric".

Edited February 19, 2015 by Robittybob1