How does the Sun get orbital energy to start orbiting a barycenter?

[Mordred]

Another key aspect is "every measurement we make is according to a standard point." GR likes rest mass, CMB likes the CMB, they are the more consistent points of reference. Orbitals tend towards largest mass, the Sun or star.

It is tricky even in the situation of Jupiter and the Sun. You would think they would both orbit their common barycenter end of story, maybe I have messed up somewhere.

Welcome to physics, the trick is to set your baseline, then describe the change.

 

This is the basis of differential geometry.

Oft times what is more accurate depends on the reference frame, or system described.

Scalar quantities are easy, vectors get complex. Tensor is a combination of the two.

When you get down into the nitty gritty details, relations of a causes b etc , favors the mathematics that show those relations in the easier to model metrics. Occams razor. A good example, take a container of a known gas. Can we truly describe every interaction of every particle,? Of course not. We can at best model an average, and use statistics, to average the frequency.

[Robittybob1]

Another key aspect is "every measurement we make is according to a standard point." GR likes rest mass, CMB likes the CMB, they are the more consistent points of reference. Orbitals tend towards largest mass, the Sun or star.

 

Welcome to physics, the trick is to set your baseline, then describe the change.

 

This is the basis of differential geometry.

Oft times what is more accurate depends on the reference frame, or system described.

Scalar quantities are easy, vectors get complex. Tensor is a combination of the two.

When you get down into the nitty gritty details, relations of a causes b etc , favors the mathematics that show those relations in the easier to model metrics. Occams razor. A good example, take a container of a known gas. Can we truly describe every interaction of every particle,? Of course not. We can at best model an average, and use statistics, to average the frequency.

Strange thing is math is not my strongest point, yet when I listen to a lecture on math it is really interesting so I'm determined to improve. At the moment I'm just using a formulated Excel sheet with my data in column A and I use each data entry as a referenced cell, so the maths is not hard I just have to write the equations perfectly and not make the sheet too spread out that I'd get lost.

[swansont]

Yes those two figures (13068.82 m/sec = 13070 m/sec) were considered the same. I was comparing 13056.35 with 13070 (as given by Wiki or 13068.82 as calculated by me using Jupiter-Sun distance), for when I use the average Sun-Jupiter distance their orbital velocity comes out as their calculated value. Well let's see what Janus has to say. I don't know how they got their figure at this stage.

 

That's still only 0.1%. How much error do you introduce by assuming a circular orbit?

[Robittybob1]

 

That's still only 0.1%. How much error do you introduce by assuming a circular orbit?

I'm thinking we have to look at the simplest situation first. If we can find a solution to circular orbits (with just two bodies Jupiter and the Sun) then we will introduce eccentricity and complexity after that.

[swansont]

I'm thinking we have to look at the simplest situation first. If we can find a solution to circular orbits (with just two bodies Jupiter and the Sun) then we will introduce eccentricity and complexity after that.

Right, but if there's a discrepancy between a calculation and data, and you know the calculation is an approximation, that's the first place you look for the discrepancy. The average orbital speed around an ellipse is not necessarily the orbital speed around a circle. What's the circumference of the circle you used, and of the ellipse used for the average speed? Are they the same?

[whiskers]

Surely, centrifugal forces don't come into play in a discussion of orbital dynamics. When modeling the solar system, the primary forces are gravitational interactions - the rest is some solar wind and some relativistic corrections.

[Robittybob1]

Surely, centrifugal forces don't come into play in a discussion of orbital dynamics. When modeling the solar system, the primary forces are gravitational interactions - the rest is some solar wind and some relativistic corrections.

This is one of the weirdest things to me too, is that I have not seen centrifugal force being used. Centrifugal (or is it centripetal force?) for I don't assign a sign to it but 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.

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 this 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 another thread we are looking specifically at the Jupiter Sun situation.

Edited February 18, 2015 by Robittybob1

[Strange]

This is one of the weirdest things to me too, is that I have not seen centrifugal force being used.

 

Huh? Isn't that the obvious way to calculate orbital velocity?

 

For a circular orbit, we have [math]f = \frac{m v^2}{r}[/math]

But, from Newton we have: [math]f = G \frac{m M}{r^2}[/math]

So [math]\frac{m v^2}{r} = G \frac{m M}{r^2}[/math] and therefore [math]v = \sqrt{ \frac{G M}r}[/math]

[Robittybob1]

 

Huh? Isn't that the obvious way to calculate orbital velocity?

 

For a circular orbit, we have [math]f = \frac{m v^2}{r}[/math]

But, from Newton we have: [math]f = G \frac{m M}{r^2}[/math]

So [math]\frac{m v^2}{r} = G \frac{m M}{r^2}[/math] and therefore [math]v = \sqrt{ \frac{G M}r}[/math]

Envious of you your la Tex skills, but in this case how were you going to define "r"? What is "r"? The two "r" s you are are using are not the same quantity when it comes to barycenter calculations.

Edited February 18, 2015 by Robittybob1

[Strange]

Envious of you your la Tex skills, but in this case how were you going to define "r"? What is "r"? The two "r" s you are are using are not the same quantity when it comes to barycenter calculations.

 

r is the distance between the (centre of) the two bodies. The position of the barycentre, on the line joining the two centres, is proportional to the two masses.

[Robittybob1]

 

r is the distance between the (centre of) the two bodies. The position of the barycentre, on the line joining the two centres, is proportional to the two masses.

So one "r" is only a proportion of the other "r". The "r" of the gravitation force is the whole distance and the "r" of the barycenter is a proportion of that first "r" but they are not equal. Do you agree with that? so combining the equations as you've done is wrong. Agree?

Edited February 19, 2015 by Robittybob1

[Strange]

 

Do you agree with that? so combining the equations as you've done is wrong. Agree?

 

No. No. And no.

[Robittybob1]

 

No. No. And no.

I was hoping it was a yes, yes and yes. So who is going to sort this impasse out?

 

Have a look at the types of orbits around a barycenter http://en.wikipedia.org/wiki/Barycentric_coordinates_%28astronomy%29

In all cases the distance between the two masses is further than distance from any one of them to the barycenter.

So you must agree that "r" the distance between them is greater than the "r" to the barycenter point.

Edited February 19, 2015 by Robittybob1

[Mordred]

There is no impasse Strange derived the escape velocity from the common center of mass between two objects.

 

In this case it is the barycentric escape velocity

r is the distance from the center of gravity

Edited February 19, 2015 by Mordred

[Robittybob1]

There is no impasse Strange derived the escape velocity from the common center of mass between two objects.

 

In this case it is the barycentric escape velocity

r is the distance from the center of gravity

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:

 

A barycentric velocity is a velocity of one body relative to the center of mass of a system of bodies. A relative velocity is the velocity of one body with respect to another. Relative escape velocity is defined only in systems with two bodies. For systems of two bodies the term "escape velocity" is ambiguous, but it is usually intended to mean the barycentric escape velocity of the less massive body. In gravitational fields "escape velocity" refers to the escape velocity of zero mass test particles relative to the barycenter of the masses generating the field.

What does that mean? A zero mass test particle"?

Edited February 19, 2015 by Robittybob1

[Mordred]

Lol variables depend on the coordinate system being modelled. r can and does mean different measurements depending on the coordinate system it's applied in.

 

A zero mass test particle is a hypothetical particle that introduces no influence to the system being examined, in this case it is the escape velocity of a mass without adding another barycenter influence by introducing further mass. Aka zero mass

By the way Whiskers has been guiding you into this direction. All measurement models are coordinate dependant, and relative to a model dependant baseline. One can describe orbital motion setting from the barycenter or from the Sun as zero.

 

Neither is more correct, they can both give accurate modelling with the same degree of accuracy.

Or in the case of the Earth being the center of the universe. The key is what is the baseline or relative to what?

This statement applies in all aspects of physics. Any model can choose a baseline or reference point, then describe the change from that point

Edited February 19, 2015 by Mordred

[Robittybob1]

Lol variables depend on the coordinate system being modelled. r can and does mean different measurements depending on the coordinate system it's applied in.

 

A zero mass test particle is a hypothetical particle that introduces no influence to the system being examined, in this case it is the escape velocity of a mass without adding another barycenter influence by introducing further mass. Aka zero mass

By the way Whiskers has been guiding you into this direction. All measurement models are coordinate dependant, and relative to a model dependant baseline. One can describe orbital motion setting from the barycenter or from the Sun as zero.

 

Neither is more correct, they can both give accurate modelling with the same degree of accuracy.

Or in the case of the Earth being the center of the universe. The key is what is the baseline or relative to what?

This statement applies in all aspects of physics. Any model can choose a baseline or reference point, then describe the change from that point

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.

[Mordred]

No he isn't joking, you need to look carefully at the first equation.

http://www.physicsclassroom.com/class/circles/Lesson-3/Newton-s-Law-of-Universal-Gravitation

[Robittybob1]

No he isn't joking, you need to look carefully at the first equation.

http://www.physicsclassroom.com/class/circles/Lesson-3/Newton-s-Law-of-Universal-Gravitation

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.

Edited February 19, 2015 by Robittybob1

[Mordred]

The term barycenter should provide the clues.

http://en.m.wikipedia.org/wiki/Center_of_mass

 

In Newtons three laws the force of of gravity from the sun and the force of gravity from the Earth influence each other. They cannot be seperated. The barycenter is the sum of vectoral influence the two have on each other, in solar coordinates every planet has its influence on the barycenter. This is the point where the sum of vectoral forces balances out.

Look close at Newtons third law in regards to the universe gravitational equation

[Strange]

I was hoping it was a yes, yes and yes. So who is going to sort this impasse out?

 

The 'r' in both equations is the distance between the objects. (Maybe I should have used 'd'.)

 

In all cases the distance between the two masses is further than distance from any one of them to the barycenter.

So you must agree that "r" the distance between them is greater than the "r" to the barycenter point.

 

As you have been told at least once, the distance to the barycentre is proportional to the masses of the objects.

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".

 

It is not always the radius of the planet: it is the distance between the two objects. For an object on the surface, then it is the radius of the planet. Not otherwise.

[Robittybob1]

The term barycenter should provide the clues.

http://en.m.wikipedia.org/wiki/Center_of_mass

 

In Newtons three laws the force of of gravity from the sun and the force of gravity from the Earth influence each other. They cannot be seperated. The barycenter is the sum of vectoral influence the two have on each other, in solar coordinates every planet has its influence on the barycenter. This is the point where the sum of vectoral forces balances out.

Look close at Newtons third law in regards to the universe gravitational equation

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.

[Mordred]

Correct

[Strange]

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.

 

And because the acceleration of the Sun is far smaller than the acceleration of the Earth, the curvature of its path is much smaller so it moves in a small circular path. The Earth, on the other hand, moves in a larger circle. The common centre of these two circles is called the barycentre.

 

(Actually, strictly speaking they are ellipses rather than circles. But the principle is the same.)

 

The acceleration in each case is inversely proportional to the mass, which is why the distance from the barycentre is inversely proportional to the mass.

[Robittybob1]

 

The 'r' in both equations is the distance between the objects. (Maybe I should have used 'd'.)

 

 

As you have been told at least once, the distance to the barycentre is proportional to the masses of the objects.

 

It is not always the radius of the planet: it is the distance between the two objects. For an object on the surface, then it is the radius of the planet. Not otherwise.

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.