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

[Strange]

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".

 

The centripetal force IS the the gravitational force, that is the point I was making.

Don't take my word for it: http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html

 

 

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.

 

Exactly.

Edited February 19, 2015 by Strange

[Robittybob1]

 

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.

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.

[Strange]

If you are interested in the centrifugal force due to rotation around the barycentre, then this might help: http://tidesandcurrents.noaa.gov/restles3.html

This one goes into much more detail of the calculations: https://squishtheory.wordpress.com/the-tides/

The curvature? The Sun orbits a much smaller circle so in my way of thinking its curvature is greater.

 

Depends how you define "curvature", I suppose. I was thinking in terms of the radius of curvature. I suppose you are thinking in terms of "how bent it is" (or something) which seems harder to define formally. But the point is, I was just showing how it all ties together in a consistent way.

Edited February 19, 2015 by Strange

[Robittybob1]

 

The centripetal force IS the the gravitational force, that is the point I was making.

Don't take my word for it: http://hyperphysics.phy-astr.gsu.edu/hbase/orbv.html

 

 

Exactly.

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.

[swansont]

 

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]

 

Rb1 is correct here — this derivation assumes M>>m, which the allows you to assume the r from Newton's law is the same as the r from the centripetal force equation. But the r in v²/r is the distance to the center of the circle (i.e. the orbit) and the r in gravity is the distance to the centers of the masses. They do not identically cancel — they only (approximately) cancel under the given mass assumption.

[Robittybob1]

 

Rb1 is correct here — this derivation assumes M>>m, which the allows you to assume the r from Newton's law is the same as the r from the centripetal force equation. But the r in v²/r is the distance to the center of the circle (i.e. the orbit) and the r in gravity is the distance to the centers of the masses. They do not identically cancel — they only (approximately) cancel under the given mass assumption.

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.

[Robittybob1]

From the OP .....

Here is my first attempt:

As the mass of material that formed Jupiter gathered together the mass of the Sun would gravitate toward it, but that mass is orbiting so it would drag the Sun behind it gradually build up momentum, so that means jupiter would have to migrate inward losing gravitational potential energy to make up for the energy transferred to the Sun.

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?

Edited February 19, 2015 by Robittybob1