Does Jupiter orbit the Jupiter-Sun barycenter or not?

[whiskers]

http://en.wikipedia.org/wiki/Tidal_acceleration#Angular_momentum_and_energy

 

"The Moon moves farther away from Earth (+38.247±0.004 mm/y), so its potential energy (in Earth's gravity well) increases. It stays in orbit, and from Kepler's 3rd law it follows that its angular velocity actually decreases, so the tidal action on the Moon actually causes an angular deceleration, i.e. a negative acceleration (-25.858±0.003 "/century²) of its rotation around Earth. The actual speed of the Moon also decreases. Although its kinetic energy decreases, its potential energy increases by a larger amount."

There is no centripetal force in the standard model. You use gravity to determine the motions of the bodies. If you are running into trouble using the more general terms, disregard them. http://en.wikipedia.org/wiki/Centripetal_force

[Robittybob1]

http://en.wikipedia.org/wiki/Tidal_acceleration#Angular_momentum_and_energy

 

"The Moon moves farther away from Earth (+38.247±0.004 mm/y), so its potential energy (in Earth's gravity well) increases. It stays in orbit, and from Kepler's 3rd law it follows that its angular velocity actually decreases, so the tidal action on the Moon actually causes an angular deceleration, i.e. a negative acceleration (-25.858±0.003 "/century²) of its rotation around Earth. The actual speed of the Moon also decreases. Although its kinetic energy decreases, its potential energy increases by a larger amount."

 

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.

 

There is no centripetal force in the standard model. You use gravity to determine the motions of the bodies. If you are running into trouble using the more general terms, disregard them. http://en.wikipedia.org/wiki/Centripetal_force

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.

Edited February 20, 2015 by Robittybob1

[whiskers]

Centrifugal forces

 

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.

"Even though the reactive centrifugal is rarely used in analyses in the physics literature, the concept is applied within some mechanical engineering concepts." You need to understand that different models are used to solve different problems. Motion is relative. Bodies attract one another, there is nothing going on at any barycenter.

[J.C.MacSwell]

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.

 

WRT the barycenter the gain for the moon is much greater than that for the Earth.

Edited February 20, 2015 by J.C.MacSwell

[Robittybob1]

WRT the barycenter the gain for the moon is much greater than that for the Earth.

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.

Centrifugal forces

 

"Even though the reactive centrifugal is rarely used in analyses in the physics literature, the concept is applied within some mechanical engineering concepts." You need to understand that different models are used to solve different problems. Motion is relative. Bodies attract one another, there is nothing going on at any barycenter.

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.

Edited February 20, 2015 by Robittybob1

[J.C.MacSwell]

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.

 

The increase in GPE wrt distance to the barycenter. It is less for the Earth than the Moon.

[Robittybob1]

The increase in GPE wrt distance to the barycenter. It is less for the Earth than the Moon.

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.

Edited February 20, 2015 by Robittybob1

[J.C.MacSwell]

I'm not sure, but the Earth is so much more massive any change in distance from the barycenter represents a lot of energy.

For the Earth/Moon gravity, the force is the same, but opposite, on the Earth as it is on the Moon. The Barycenter is not a massive point effecting the GPEs

 

For the Earth/Moon the changes in GPE are pretty close to proportional to displacement (marginal change in force over the distances discussed)

 

If you jump in the air, the COM between you and the Earth stays in place (or on it's path), but of course (almost) all the energy of displacement goes to you. It is not split equally even though the mass displacement is equivalent (opposite)

 

...all because the Earth is so much more massive

[Robittybob1]

For the Earth/Moon gravity, the force is the same, but opposite, on the Earth as it is on the Moon. The Barycenter is not a massive point effecting the GPEs

 

For the Earth/Moon the changes in GPE are pretty close to proportional to displacement (marginal change in force over the distances discussed)

 

If you jump in the air, the COM between you and the Earth stays in place (or on it's path), but of course (almost) all the energy of displacement goes to you. It is not split equally even though the mass displacement is equivalent (opposite)

 

...all because the Earth is so much more massive

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.

[swansont]

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).

 

Is Jupiter's semi-major axis equal to the radius you used in your calculations?

 

 

797.5 vs 778.3 million km. No, it is not.

 

 

I mean, good grief. I did the frikkin' calculation for you, since you were unwilling (or unable) to do it yourself, and yet you are clinging to the idea that you are right. That's backwards.

 

Edit: Math error. Sorry.

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.

 

In this view there is no PE of the moon vs PE of the earth. There is the PE of the earth-moon system. You don't get to double-count the potential energy.

 

PE = -GMm/r

[Robittybob1]

 

Is Jupiter's semi-major axis equal to the radius you used in your calculations?

 

 

797.5 vs 778.3 million km. No, it is not.

 

 

I mean, good grief. I did the frikkin' calculation for you, since you were unwilling (or unable) to do it yourself, and yet you are clinging to the idea that you are right. That's backwards.

 

In this view there is no PE of the moon vs PE of the earth. There is the PE of the earth-moon system. You don't get to double-count the potential energy.

 

PE = -GMm/r

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.

[swansont]

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.

 

778.5 million km is not the SMA that I had found. I'll have to check the numbers again.

 

2 days is still a small difference in calculations.

 

Edit: it looks like I made a math error. Ignore that particular objection.

[Strange]

(nearly 2 days difference)

 

Nearly 0.05% difference. Wow.

[Greg H.]

 

Nearly 0.05% difference. Wow.

 

You get larger error bars on resistors used in high tech electronics. I think we're close enough to call it.

[swansont]

Math error aside, one source of my confusion was the labeling of the distance as the average distance (in the other thread) rather than the semi-major axis. These are not necessarily identical. (it depends on how you take the average)

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.

 

 

I don't see where you've given a link to these "current best measurements" nor explained where the other numbers came from, in addition to clarifying why differences of order 0.03% are causing so much consternation.

[Robittybob1]

 

Nearly 0.05% difference. Wow.

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.

[imatfaal]

Rob - how many times have you been told in this topic, and numerous other threads, that all these calculations are based on imperfect models and have inputs which are imprecise?

[Robittybob1]

Math error aside, one source of my confusion was the labeling of the distance as the average distance (in the other thread) rather than the semi-major axis. These are not necessarily identical. (it depends on how you take the average)

 

I don't see where you've given a link to these "current best measurements" nor explained where the other numbers came from, in addition to clarifying why differences of order 0.03% are causing so much consternation.

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)))

Rob - how many times have you been told in this topic, and numerous other threads, that all these calculations are based on imperfect models and have inputs which are imprecise?

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.

Edited February 20, 2015 by Robittybob1

[whiskers]

The average of two numbers is defined to be halfway between them. So it is with the Sun/Jupiter barycenter - it is *defined* to be the center of mass of both bodies, and the center of their mutual orbit. There is no way to verify or disprove it - you can only verify vs disprove the way that you have calculated its position.

Edited February 20, 2015 by whiskers

[Robittybob1]

I searched for the origin of the "1.068 solar radii" expression and came across this paper http://www.researchgate.net/publication/200710983_On_the_estimated_precession_of_Mercury%27s_orbitthat really question the precision of these facts.

For if we can't tell how the Sun orbits its barycenter with Jupiter, and Mercury is orbiting the Sun how precisely did they calculate Mercury's precession?

That paper just looks at the maths but didn't give values of the S-J distances.

The average of two numbers is defined to be halfway between them. So it is with the Sun/Jupiter barycenter - it is *defined* to be the center of mass of both bodies, and the center of their mutual orbit. There is no way to verify or disprove it - you can only verify vs disprove the way that you have calculated its position.

I agree the barycenter is the center of mass of the two bodies but how did you define it as "the center of their mutual orbit"? If they were binary maybe.

Let's look at a contrived situation where a 2 kg mass is gravitationally orbiting a 1 kg mass in free space. What is their barycenter point and how fast do they orbit? Make the orbits perfectly circular.

In my way of thinking they will only be able to do this where the reactive centrifugal forces balanced their gravitational force.

 

There are obviously multiple answers to this and it would depend on their relative speed, so can it happen with a distance between them of 1000 m?

So the barycenter point is 1/3 and 2/3 position long the distance between them. The heavier object will orbit the barycenter at a distance 1/3*1000 meters and the 1 kg mass will be 2/3*1000 m from the barycenter but they will be 1000 m apart, so what will their G force be?

Newtonian G force = G* m1*M2/r^2 = G * 1*2/10^6 = 2G*10^-6 = 6.67384*10^-11*2.0*10^-6 = 13.34768 *10^-17 N

 

So what speed do the masses have to travel to stop them just from being gravitationally brought together?

Edited February 20, 2015 by Robittybob1

[Strange]

 

In my way of thinking they will only be able to do this where the reactive centrifugal forces balanced their gravitational force.

 

Which is always the case in a stable orbit.

[Robittybob1]

 

Which is always the case in a stable orbit.

OK

[whiskers]

I would recommend "Asimov on Astronomy".

[Robittybob1]

... So what speed do the masses have to travel to stop them just from being gravitationally brought together?

Orbital period is 444.9665645 years

Mass 1 kg velocity = 0.00000029830275 m/sec

Mass 2 kg velocity = 0.00000014915138 m/sec

In opposing directions and orthogonal to the gravity between them.

Could that have been worked out if I hadn't used reactive centrifugal forces around the barycenter point?

 

So what would happen if mass 1 (1 kg) approached the mass 2 (2 kg) at 1000 meters when mass 2 has no motion, would mass 1 go into an orbit?

Would there be a way for mass 1 getting mass 2 to start orbiting the common barycenter?

Edited February 21, 2015 by Robittybob1

[Robittybob1]

It is not as if the two masses have to approach each other at the right speed to result in a perfect circular orbit, but it is obvious it wouldn't take too much extra velocity to get to escape velocity.

But my thoughts are that as a mass passes by at a higher than required speed ("speed" is hardly the right word since they are only moving in the order of 30 microns/sec!) but less than the escape velocity the mass being overtaken will be moved along with that mass for the gravitational force will keep on acting over enormous distances. and it will curve the path of the passing mass.

So when the overtaking mass finally runs out of kinetic energy it will fall back down to the now moving primary mass but the primary mass will keep gaining kinetic energy and the relative speed as they pass for the second time will be lower.

Now it is possible this dampening down could take billions of years to circularise the orbits, the lighter mass will appear to be orbiting the primary but they have not yet established their true binary orbit. Gradually more and more energy is being transferred to the primary as it gains orbital energy at the expense of the secondary's kinetic energy.

In these situations the orbital energy formulas will not work accurately for they are not true orbiting bodies until the kinetic energy has been "equalised".

 

Could that account for the difficulty of making Jupiter's orbital parameters fit the orbital formulas? That the Sun is not in a true barycentric orbit with Jupiter as yet.

Edited February 21, 2015 by Robittybob1