Robittybob1

They were the result of my own calculations after comparing the gravitational force with the reactive centrifugal force (they had to be equal) but the whole problem was choosing the right period from the start. The "current best measurements" were calculations of the barycenter distances (based on the barycenter formula) using the mass of the Sun and Jupiter in Wikipedia or NASA sites, (there could have been some confusion over the distance value, like was it the SMA or the average distance apart, that I should have been using). So my original period for Jupiter was the sidereal period from Wikipedia (I thought) but it was list in days only but I have just seen a NASA value "Sidereal orbit period (days) 4,332.589" the original number I was using was just based on 4332 days. but using Janus' formula and the SMA the answer for the period was 4,333.8 days, nearly 2 days longer. So maybe my original period was out by a bit. I see Wikipedia lists it as "4332.59 d" So what sort of period is being measured that gives an answer of 4333.8 days? Using this formula T = 2*Pi()*SQRT(a^3/(G(m1+m2))) I don't think you have ever told me that before, well not so precisely. Well honestly I am trying to find the most precise values and to use them in the correct formulas to see if we can show that Jupiter is orbiting the Sun Jupiter barycenter. We must be getting closer to the right answer surely. So what sort of period is being measured that gives an answer of 4333.8 days? Using this formula T = 2*Pi()*SQRT(a^3/(G(m1+m2))) That is the mystery for today for there is nothing close to that figure in Google search. For when I use this period in my calculation workbook there is no longer differences in the SMA but the position of the barycenter still does not match the statement below. This is an often quoted statement "Jupiter's mass is 2.5 times that of all the other planets in the Solar System combined—this is so massive that its barycenter with the Sun lies above the Sun's surface at 1.068 solar radii from the Sun's center." Can we reverse engineer that and find out what distances or masses they were using in the barycenter equation? I tried before and got "778,500,000,000 - 1.068*6.955E+8 = 777,757,206,000" So at that stage I was using "778,500,000,000" as the distance apart. Should I have been using 778,547,200,000 (our now agreed SMA)? No that makes the calculations worse.

I got the period of Jupiter out of Wikipedia to start with and later (late last night) I used the equation that Janus shared with us for the period the one with the a^3 in it, and it gives a totally different period for Jupiter. That is the bit I'm struggling with now for how can the same planet have two different periods? There might be an explanation but I've just got up.

I had lost concentration on the Jupiter problem, but I have just used the SMA 778,547,200,000 meters but if I use the SMA to calculate the period it is -155586.73 seconds different (nearly 2 days difference) so the reason for the error was the orbital period, some how there was another orbital period that is about two days longer 374440386.73 seconds compared to the 374284800.00 seconds which I had been using before. I'll look into this again tomorrow, as it is 1:00 AM again.

Yes I think you are right the force is equal and opposite but the Earth being so massive doesn't move so there is no work done to the Earth (Force * distance = zero for there was no distance.) By analogy are you saying in the Moon Earth system most of the energy will be going to the Moon. You could be right. I wish I could use the data I have to prove that. Have you ever used any the formulas on this page? http://en.wikipedia.org/wiki/Specific_orbital_energy I have the SMA, and, the two barycenter radii, the period and the velocity of the Moon for the next billion years (1000 * million year increments). Maybe I should have calculated the the gravitational strength, and the Earth's velocity too, but if I knew what equation I could use I'd know what values I needed.

These are the 12 things most likely to destroy the world http://www.msn.com/en-nz/news/techandscience/these-are-the-12-things-most-likely-to-destroy-the-world/ar-BBhLrv6?ocid=UP97DHP#page=12 Everyone of them would be the worst nightmare to live through.

I'm not sure, but the Earth is so much more massive any change in distance from the barycenter represents a lot of energy. Imagine trying to lift the Earth! but you are only lifting it WRT to the Moon. I am in process of converting my worksheet across to the Earth-Moon system and it is really interesting for the barycenter is based on the distance between the centers of the masses, but the distance between the Moon and the Earth varies so much what distance do you plug into the barycenter formula? I ran the sheet and it gave me the answer as being very close to the figure for the semi-major axis (SMA), 384743158 meters was my answer and the best SMA data 384748000 m. http://en.wikipedia.org/wiki/Orbit_of_the_Moon Which is only just over 4 km out. but there are no error measurements listed. The position of the barycenter which is commonly listed as 4661000 from the center of the Earth, was not confirmed, with my best result at 4676022 meters a full 15 km out. There are a variety of masses of the Earth listed. Is the period around the barycenter the same as the sidereal period of the Moon orbiting the Earth? I should be using the formula to calculate the period T = 2*Pi()*SQRT(a^3/(G*(m1+m2))) But even then I need the best SMA. For as the Moon is accelerated the period will change and this formula will calculate that so I'll take the increases in 1000 year blocks of 38 m. I increased that to million year periods to see the effect kick in a bit faster. The barycenter orbits keep getting longer as the SMA elongates, how one works out the specific orbital energy, beats me at the moment. But it is clear that whatever happens to the moon the exact amount of energy goes into the Earth as well and all this is at the expense of the rotational energy of the planet Earth, so more energy is lost that converted.

What are you measuring? If it was GPE or orbital energy or momentum I wouldn't be surprised if they were equal but distance is inversely proportional to mass, for the same force is able to move larger mass by less than a smaller one. Yes the barycenter will move away from the moon greater than it will move away from the Earth. Except the bodies are orbiting from that point ... so that is why you can look at the reactive centrifugal force from that point. It may not have been used before but I think it has been now shown to be of real value.

So you were brushing up on the physics of the situation. The bit that is never covered is if the Moon is gaining potential energy WRT the Earth the Earth is gaining the same Potential Energy too. It must for they are both orbiting a common barycenter. I've got the Excel sheet set up to analyse Jupiter and the Sun so I could swap it over to the Earth - Moon situation and increase the distance between the Earth - Moon by +38.247±0.004 mm/y and see what effect that has on the Earth. Part of that distance will be a movement of the Earth away from the Moon as much as the Moon is going away from the Earth (since both barycenter radii will be increasing). OK the change in the orbital radius of the Earth might be as little as a 0.5 mm/y but will we be able to calculate the change? Someone else must have figured this before now! When you take the Earth and Moon orbiting the barycenter the same force that accelerates the moon in its orbit around the barycenter also promotes the Earth in its orbit around the barycenter, in an equal and opposite fashion, but because the masses are not equal the effect on the Earth is minimal, and since our view is geocentric so we don't notice it, and just observe all the changes as occurring on the Moon. The situation of Reactive Centrifugal forces apply http://en.wikipedia.org/wiki/Reactive_centrifugal_force The planets and Sun orbit the barycenters as if there is a physical constraint.

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 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".

As I crashed into bed at 1:00 AM again last night, I kept on thinking about the situation I proposed way back in the OP of this thread. OK the SS is just 4.6 billion years old and has it taken Jupiter all this time to nearly build up the momentum in the Sun to orbit the barycenter perfectly, could that be the solution? If the Sun's orbital momentum has been transferred back from the planets how much momentum has Jupiter lost? As I'm writing this, I can recall that Jupiter (and the other planets) would be gaining momentum from gravitational tidal acceleration (at the expense of the Sun's rotational momentum and to counter this Jupiter and the other planets will be transferring that orbital energy back to the Sun making the Sun move in a true binary fashion around their barycenters. So instead of drifting further and further away the planets have been balancing this outward drift (tidal acceleration) against an inward drift. But what say this process has not yet finished? This is really significant and I think it is provable when one considers the tidal acceleration of the Moon. As the Moon moves further from the Earth and the Earth-Moon barycenter moves where does the Earth get its extra orbital energy from? Same question/same answer as where does the Sun get its orbital energy from?

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.

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

OK that would be like having the barycenter at the very center of the main mass so the two distances are virtually the same. Thanks for that insight.

That was from the "idealistic Jupiter -Sun binary orbiting the common barycenter", based on the parameters of period and masses being correct. OK I will next explore how the eccentricity will affect orbital periods. 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.

Yes - it is partly what prompted this study. 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

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

No, I don't think the centrifugal force is the opposite of the gravitational force but they must be equally opposite. The centripetal force is provided by the gravitational force but countering this force is the fictitious centrifugal force, and the radius of these forces is the barycenter not the distance between the masses.

The curvature? The Sun orbits a much smaller circle so in my way of thinking its curvature is greater. I think we are better sticking with orbital speed or orbital radius rather than curvature.

OK on point 3 you are correct, points 1 and 2, I still dispute. Instead of symbols use words. In the gravitational force equation it is the distances between the centers of the two masses. Whereas in the centripetal force formula it is the distance from the barycenter, and you already say these distances aren't the same but are "Proportional". But they are "reversely" proportional in fact as the barycenter is closer to higher mass. r1 = a*M2/(M1+M2) is that "reversely proportional"? M2/(M1+M2) so the distance of the barycenter to the primary is a fraction of the distance between them and is proportional to the fraction of total mass contributed by the secondary.

The gravitational force between the Sun and the Earth is equal and opposite for both. But since the masses are enormously different the acceleration from that force is vastly different.

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?

How can you give me a reference to read without the word barycenter in it? What particular part do you want me to read? Could you please expand your argument. You make a statement and your reference has limited use. F=mv^2/r centripetal force formula, it hasn't got the word centripetal or centrifugal in it either.

Stupid mirror!

That is the point I was making to Strange you can't multiply both sides by "r" because both of those "r"s were different measurements. Which ever way you look at it the barycentric r is shorter than r the difference in distance between the masses. I'm beginning to think Strange was just joking and doesn't like admitting she/he is wrong.

That's even worse! Escape velocity = v_e =sqrt(2GM/r) where r in this case is the radius of the planet. So that is another use of the variable "r". "this from wikipedia on Escape velocity: What does that mean? A zero mass test particle"?