Page 4 of 6 FirstFirst ... 23456 LastLast
Results 91 to 120 of 169

Thread: Is the Solar System's Barycenter Inside or Outside the Sun?

  1. #91
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by Robert Tulip View Post
    The source for your question, quoted again below, is speaking of the stability of the SSB pattern.

    What I meant is that the SSB stability is driven by the triple conjunction pattern, not by the actual conjunction events.

    Each successive 178.9 year wave function of the solar distance is almost exactly the same. This pattern stability of the SSB is solely because for any two dates separated by 178.9 years, Jupiter, Saturn and Neptune are in almost exactly the same relative positions, advanced by thirty degrees of arc, with the exactness of the triple conjunctions drifting in and out over overlapping periods of about two thousand years.

    Twelve of these periods equal 2148 years, which by coincidence is a zodiac age, close to one twelfth of earth’s precession period of 25771 years.

    Over each zodiac age, Jupiter, Saturn and Neptune form successive triple conjunctions in each of the twelve signs within each active conjunction family.

    The triple conjunction next decade is part of a family that was tightest on 17-20 July 769 AD. Over the last 1250 years its conjunction period has increased from three days to about five years.
    I still don't get the meaning, if any, of "SSB stability is driven by the triple conjunction pattern." My hunch is that you are using familiar words in unorthodox ways that are at best meaningless and at worst potentially confusing to novices who may be lurking here. No further questions on this detail.

  2. #92
    Join Date
    Dec 2006
    Posts
    2,004
    Quote Originally Posted by Hornblower View Post
    I still don't get the meaning, if any, of "SSB stability is driven by the triple conjunction pattern." My hunch is that you are using familiar words in unorthodox ways that are at best meaningless and at worst potentially confusing to novices who may be lurking here. No further questions on this detail.
    It is really very simple, and I wonder if you looked at the attachments I have provided in any detail?

    Each of the fifteen segments of each 178.9 year depiction of the barycenter distance to the sun are graphed in the same colour. They hardly change at all in successive iterations. That is stability. Your suggestion that it may be "meaningless" is wrong.

    As per the Fourier Transform, another simple depiction of this data, the causes of this stability are directly measurable as the structure of the orbits of Jupiter, Saturn and Neptune.

  3. #93
    Join Date
    Dec 2006
    Posts
    2,004
    Here is another depiction of the stable 178.9 year structure of the distance from the sun to the barycenter.

    Each row is a successive depiction of 179 years of data from NASA Jet Propulsion Laboratory of the solar system barycenter.

    As easily seen, the change from one row to the next is small, proving that this period is stable.

  4. #94
    Join Date
    Mar 2009
    Posts
    1,535
    Sir Fred Hoyle's in his book on Copernicus, writes of course that the Earth does not revolve around the Sun, but rather the Earth and Sun both revolve around the center of mass of the Earth-Sun system, which is quite a few miles from the sun's central axis.

    Hoyle points out that one must factor in all objects, starting with the nearest stars, to recalculate the true center-of-mass. He concludes that once one has properly applied the barycentric argument to all other entities in the universe the center-of-mass can easily be at the Earth's location, making it impossible to disprove the geocentric hypothesis.

    Hoyle says the barycentric argument is only properly applied when every object in the universe has been factored into the center-of-mass calculation, a calculation that has not been done. Although Hoyle was not a geocentrist, he believed that consistent application of the barycentric argument, layer by layer, places the center-of-mass farther away from the Sun and closer to the Earth, and he concludes that the Barycentric Argument, like the Speed of Light Argument, cannot be used as a refutation of geocentrism (with a small 'g'), which is of course as it has to be if Relativity is correct.
    Last edited by wd40; 2017-Mar-23 at 11:24 AM.

  5. #95
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by wd40 View Post
    Sir Fred Hoyle's in his book on Copernicus, writes of course that the Earth does not revolve around the Sun, but rather the Earth and Sun both revolve around the center of mass of the Earth-Sun system, which is quite a few miles from the sun's central axis.

    Hoyle points out that one must factor in all objects, starting with the nearest stars, to recalculate the true center-of-mass. He concludes that once one has properly applied the barycentric argument to all other entities in the universe the center-of-mass can easily be at the Earth's location, making it impossible to disprove the geocentric hypothesis.

    Hoyle says the barycentric argument is only properly applied when every object in the universe has been factored into the center-of-mass calculation, a calculation that has not been done. Although Hoyle was not a geocentrist, he believed that consistent application of the barycentric argument, layer by layer, places the center-of-mass farther away from the Sun and closer to the Earth, and he concludes that the Barycentric Argument, like the Speed of Light Argument, cannot be used as a refutation of geocentrism (with a small 'g'), which is of course as it has to be if Relativity is correct.
    To calculate the position of what Hoyle is calling the true center of mass of the universe, we would need to know the exact amounts and positions of all matter in the universe, which we do not. The barycenter of the observable universe could be at the center of the Earth, but it could just as easily be closer to the Andromeda galaxy. For the full theoretical extent of the universe under the current mainstream theory it becomes undefined. Regardless, it would be of no practical consequence in calculation of the local orbital motions of the Sun and planets relative to one another, as the gravitational gradient across the local system is vanishingly small.

    It remains a fact that in any inertial frame of reference that is useful for solar system orbital mechanics, the Earth is moving in large loops with a period of one year, while the Sun is moving in much smaller loops as a result of the gravitational action of the planets. These loops are roughly centered on the local barycenter, which is either nearly stationary or moving at nearly constant velocity, depending on the frame of reference.

    It is my understanding that in general relativity, a frame of reference in which the Earth is stationary and non-rotating is admissible, but it remains my educated guess that calculations of orbital dynamics in the solar system, not to mention the galaxy, would be horrendously messy and of no practical value.

  6. #96
    Join Date
    Sep 2003
    Location
    The beautiful north coast (Ohio)
    Posts
    45,853
    As this thread is no longer tightly orbiting the sun of Q&A, an orbital correction to move it to Astronomy is executed.
    At night the stars put on a show for free (Carole King)

    All moderation in purple - The rules

  7. #97
    Join Date
    Sep 2003
    Posts
    11,561
    Reprieve is over; I’m back.

    Quote Originally Posted by Hornblower
    Perhaps your thought process is being cluttered up by concepts of moment arms and leverage,…
    Yep, too many FBDs from too long ago, perhaps. It’s the linear relation with distance to determine the barycenter that seems to be where I stumbled since it matches the moment arm equation (see saws).

    ow consider a pair of bodies in free fall and orbiting one another. The line connecting them corresponds geometrically to the beam, but there actually is no beam present. The only forces acting on the bodies are gravitational ones on each one directed toward the other along the line, not crossways. The bodies have velocity vectors crossways and in opposite directions, and the gravitation causes the bodies to curve toward each other instead of flying off in straight lines. With just the right velocities they will be in a pair of stable orbits.
    Using Newton and thinking only in terms of a two-body model with circular orbits about the barycenter, the orbital motion itself seems to be the answer to the radius linearity (inverse) in calculating the barycenter, as opposed to the inverse square associated with gravity. Namely, centrifugal force which is simply – mv2/r. The two forces of gravity and centrifugal force are in balance, no doubt, and this too makes this a linear (inverse) relationship when equating each other. This gets us off the see-saw, where we belong. Twirling an unbalanced baton might be the better analogy.

    Quote Originally Posted by grapes
    Quote Originally Posted by George
    ) Saturn's Epimetheus and Janus exchange orbits close to every 4 years. As they near each other, excluding Saturn for a second, their mutual barycenter is easily determined, if we know their mass and locations, and we do. In a heavy-thinking (bull-headed) leverage model, the faster inner moon should swing past the slower moon and outward enough to become the outer orbiting moon. Vice-versa for the other moon. This is exactly what doesn't happen. And I don't think the kinematics improve by adding Saturn to this dynamic. The gravity/Kepler model, however, does work.
    I've never looked into that, what does happen when they exchange orbits?
    The inner moon gets boosted into the higher orbit before it is aligned with the outer moon. A “leverage” model would be opposite of this.
    We know time flies, we just can't see its wings.

  8. #98
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Robert Tulip View Post
    Here is another depiction of the stable 178.9 year structure of the distance from the sun to the barycenter.

    Each row is a successive depiction of 179 years of data from NASA Jet Propulsion Laboratory of the solar system barycenter.

    As easily seen, the change from one row to the next is small, proving that this period is stable.
    Impressive! I
    We know time flies, we just can't see its wings.

  9. #99
    Join Date
    Dec 2004
    Posts
    14,372
    George,

    I'd say neither motion, moment, momentum, angular momentum, nor
    leverage have anything to do with determining location of a barycenter.
    The questions are: 1) The barycenter of WHAT? and 2) How are you
    using this barycenter?

    The barycenter of the Earth-Moon system is useful for some things,
    the barycenter of the Sun-Earth system is useful for some other things,
    and the barycenter of the Andromeda-Milky Way system is useful for
    yet other things. I can't imagine what the barycenter of the Sun-Deneb
    system could be useful for, or the Andromeda-Sun system. I'd say
    both of those are useless. Even though they can both be defined.

    Quote Originally Posted by George View Post
    The inner moon gets boosted into the higher orbit before it is aligned
    with the outer moon.
    I wasn't aware of that. It astonishes me! I expected that they change
    places as they pass each other. Is it based on gravitational modelling
    or on observation? Do you have a reference?

    -- Jeff, in Minneapolis
    http://www.FreeMars.org/jeff/

    "I find astronomy very interesting, but I wouldn't if I thought we
    were just going to sit here and look." -- "Van Rijn"

    "The other planets? Well, they just happen to be there, but the
    point of rockets is to explore them!" -- Kai Yeves

  10. #100
    Join Date
    Dec 2004
    Posts
    14,372
    Wait a minute.... Thinking about it, the faster-moving inner moon
    would be sped up as it approaches the slower-moving outer moon,
    causing it to rise away from the planet, and the slower-moving
    outer moon would be slowed down as the faster-moving inner
    moon approaches it, causing it to fall closer to the planet. Now
    I don't see why they don't usually crash into each other!

    -- Jeff, in Minneapolis
    http://www.FreeMars.org/jeff/

    "I find astronomy very interesting, but I wouldn't if I thought we
    were just going to sit here and look." -- "Van Rijn"

    "The other planets? Well, they just happen to be there, but the
    point of rockets is to explore them!" -- Kai Yeves

  11. #101
    Join Date
    Dec 2004
    Posts
    14,372
    Basically it looks like the inner moon is rising from periapsis yet still
    gaining speed while the outer moon is falling toward periapsis yet
    still losing speed as they approach each other. The excess speed
    of the inner moon makes it rise enough and the deficient speed of
    the outer moon makes it fall enough that they change places before
    they pass. Then the new outer moon begins decelerating rapidly
    and the new inner moon begins accelerating rapidly. So I shouldn't
    have been surprised.

    -- Jeff, in Minneapolis
    http://www.FreeMars.org/jeff/

    "I find astronomy very interesting, but I wouldn't if I thought we
    were just going to sit here and look." -- "Van Rijn"

    "The other planets? Well, they just happen to be there, but the
    point of rockets is to explore them!" -- Kai Yeves

  12. #102
    Join Date
    Jul 2005
    Posts
    14,099
    You can deal with this as a pair of circular orbits around Saturn. The inner satellite gains on the outer satellite. As the two satellites get closer to each other, the gravity of the inner satellite slows the outer satellite, which therefore progressively drops into a lower orbit. Meanwhile, the gravity of the outer satellite accelerates the inner satellite, which progressively rises into a higher orbit. They are at their closest together when they've each migrated to the same orbit, but the interaction continues so that the previous outer satellite drops into an inner orbit and the previous inner satellite rises into an outer orbit. They exchange momentum, and each interconverts potential and kinetic energy as it moves.
    Although they're commonly said to "exchange orbits", they are different in mass so the exchange is not perfectly symmetrical - each has its own characteristic orbital radius when inner and when outer.

    I don't think a barycentric approach is much help in thinking about this scenario - the usual approach is to consider circular coplanar orbits and apply perturbations, as here (600KB pdf).

    Grant Hutchison

  13. #103
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by George
    Using Newton and thinking only in terms of a two-body model with circular orbits about the barycenter, the orbital motion itself seems to be the answer to the radius linearity (inverse) in calculating the barycenter, as opposed to the inverse square associated with gravity. Namely, centrifugal force which is simply – mv2/r. The two forces of gravity and centrifugal force are in balance, no doubt, and this too makes this a linear (inverse) relationship when equating each other. This gets us off the see-saw, where we belong. Twirling an unbalanced baton might be the better analogy.
    My bold. Once again, the location of the barycenter of any two bodies at a given time is purely a function of their positions at that time and of their masses. Their velocity does not figure in calculating that position. No. Nada. Nyet. Zilch.

    Velocity does come in for showing that the barycenter is indeed the focus of their orbits, and that it does not move in other than a straight line in the absence of outside forces. If you have any further questions on that point, I will be happy to show some math.

  14. #104
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Jeff Root View Post

    I'd say neither motion, moment, momentum, angular momentum, nor
    leverage have anything to do with determining location of a barycenter.
    Ok why not? That's really what I am trying to get at. Place two objects in space with no relative motion and derive the inverse linear equation for their barycenter, especially using an inverse square law
    We know time flies, we just can't see its wings.

  15. #105
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Hornblower View Post
    My bold. Once again, the location of the barycenter of any two bodies at a given time is purely a function of their positions at that time and of their masses. Their velocity does not figure in calculating that position. No. Nada.
    . The c.g. math is easy enough, but how is it derived...my real question I am struggling with, though it may be simple. See my question in my prior post to Jeff.
    We know time flies, we just can't see its wings.

  16. #106
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by George View Post
    . The c.g. math is easy enough, but how is it derived...my real question I am struggling with, though it may be simple. See my question in my prior post to Jeff.
    I, along with others, have shown the definition. I have shown how it works both in leverage systems and in orbital motion systems. I suppose we could say it was derived in the sense that the geometric construction and definition thereof were inspired by what ancient people such as Archimedes discovered in applications that did involve leverage. I don't know what else to do to get you over a conceptual hump. Perhaps someone else can find the key.

  17. #107
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Hornblower View Post
    I, along with others, have shown the definition. I have shown how it works both in leverage systems and in orbital motion systems. I suppose we could say it was derived in the sense that the geometric construction and definition thereof were inspired by what ancient people such as Archimedes discovered in applications that did involve leverage. I don't know what else to do to get you over a conceptual hump. Perhaps someone else can find the key.
    The simple barycenter math just looks like something that would be derived from known equations of phenomena and not a separate phenomena equation. Am I the only one that thinks this is likely?

    I would guess that two initially static and unequal masses that begin their free-fall to one-another could be shown by calculus to meet at the barycenter, and that the barycenter could be located using, of course, the established equation for it but derivable from the calculus of motion, which happens to match the see-saw equation.

    Or, the balance of forces equation (gravity balanced by centrifugal force) to maintain a perpetual orbit might be used to derive the barycenter.
    We know time flies, we just can't see its wings.

  18. #108
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by George View Post
    The simple barycenter math just looks like something that would be derived from known equations of phenomena and not a separate phenomena equation. Am I the only one that thinks this is likely?

    I would guess that two initially static and unequal masses that begin their free-fall to one-another could be shown by calculus to meet at the barycenter, and that the barycenter could be located using, of course, the established equation for it but derivable from the calculus of motion, which happens to match the see-saw equation.

    Or, the balance of forces equation (gravity balanced by centrifugal force) to maintain a perpetual orbit might be used to derive the barycenter.
    My bold. I am not sure what you mean by that phrase.

    In principle, the initially static masses in free fall could have been the first phenomenon to inspire the development of the mathematical definition of the barycenter, but someone like Archimedes in ancient times would have had no means of performing it, while analyzing a static seesaw would have been easy. I can imagine Newton doing it as a thought exercise, just as he did with orbital mechanics for details of planetary motion that could not be observed directly from an inertial frame of reference. He surely already knew the barycenter principle as handed down from exercises such as static loading of a seesaw in ancient times.

    It is my understanding that Archimedes discovered the principle of the lever. In principle he could have performed experiments with unequal dumbbells rotating rapidly end over end about their barycenters, but I don't know whether or not he actually did so.

  19. #109
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Hornblower View Post
    My bold. I am not sure what you mean by that phrase.
    We develop equations to define what happens with a specific phenomena. Newton's inverse square law gives us the force action between static (or dynamic) masses. If we want to know, say, how far an object falls, we simply derive the distance equation from the inverse square law calculus. My hunch is that the simple barycenter equation, even if it is a perfect match for a seesaw, is something derived in a similar fashion, or could be derived. Since we already know it works, then who cares? ... Me, though I'm really only curious and not suggesting something exceptionally interesting here. [Added: I would guess this is shown in some celestial mechanics texts, but I only recently acquired one.]

    ... He surely already knew the barycenter principle as handed down from exercises such as static loading of a seesaw in ancient times.
    The seesaw only works because of the summation of moments = 0 (at balance), which is due to the physical elemental moment transfer along the lever. But, in space, as you have correctly pointed-out, there is no actual "leverage". There is no apparent reason that the math is the same, though it is not illogical that they match.
    Last edited by George; 2017-Mar-24 at 05:22 PM.
    We know time flies, we just can't see its wings.

  20. #110
    Join Date
    Jul 2005
    Posts
    14,099
    Isn't it obvious why the maths is the same? The two bodies describe ellipses of the same eccentricity, sweeping out equal areas in equal times, and for that to happen they've got to stay constantly on either side of some shared focal point, through which the vector of their mutual gravitational attraction is always directed. Each body feels the same force (GMm/r2), but the less massive will accelerate faster under that force (F=ma). The ratios of accelerations therefore go inversely with the ratios of masses. So in a given time the masses covers orbital distance in inverse proportion to their mass (d=at2/2), so the size of their orbits must scale in inverse proportion to their mass if they are to complete their mutual orbits in the same time period. It's just the lever in another guise; and yes, if you halted these masses simultaneously in their orbit so that the fell towards each other, they'd meet at the barycentre.

    Grant Hutchison
    Last edited by grant hutchison; 2017-Mar-24 at 05:43 PM.

  21. #111
    Join Date
    Dec 2004
    Posts
    14,372
    Grant,

    To help George understand this, you described the orbits of two
    bodies around each other. You find the barycenter of that two-body
    system. What if there are other massive bodies present, but you are
    not considering them as part of the system? Such as the Sun and its
    planets, but not this huge interloper, Bronson Alpha, and its smaller
    companion, Bronson Beta, which are passing through the inner Solar
    System. Since we aren't considering them as part of the system, do
    they have any influence on the location of the barycenter? What if
    we then change our minds and decide to include them?

    -- Jeff, in Minneapolis
    http://www.FreeMars.org/jeff/

    "I find astronomy very interesting, but I wouldn't if I thought we
    were just going to sit here and look." -- "Van Rijn"

    "The other planets? Well, they just happen to be there, but the
    point of rockets is to explore them!" -- Kai Yeves

  22. #112
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by grant hutchison View Post
    Isn't it obvious why the maths is the same? The two bodies describe ellipses of the same eccentricity, sweeping out equal areas in equal times, and for that to happen they've got to stay constantly on either side of some shared focal point, through which the vector of their mutual gravitational attraction is always directed. Each body feels the same force (GMm/r2), but the less massive will accelerate faster under that force (F=ma). The ratios of accelerations therefore go inversely with the ratios of masses. So in a given time the masses covers orbital distance in inverse proportion to their mass (d=at2/2), so the size of their orbits must scale in inverse proportion to their mass if they are to complete their mutual orbits in the same time period. It's just the lever in another guise; and yes, if you halted these masses simultaneously in their orbit so that the fell towards each other, they'd meet at the barycentre.
    Yes, I knew it was obvious, but I just couldn't see it.

    I'm not sure that we even need Kepler or ellipses to help us. From your F=ma, and the force is equally applied to both masses, we get m1/m2 = a2/a1 and a is d2r/dr2. So I think there is something also obvious here (to derive the barycenter equation).... but I'm not seeing it right-a-way.
    Last edited by George; 2017-Mar-24 at 06:43 PM.
    We know time flies, we just can't see its wings.

  23. #113
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by Jeff Root View Post
    Grant,

    To help George understand this, you described the orbits of two
    bodies around each other. You find the barycenter of that two-body
    system. What if there are other massive bodies present, but you are
    not considering them as part of the system? Such as the Sun and its
    planets, but not this huge interloper, Bronson Alpha, and its smaller
    companion, Bronson Beta, which are passing through the inner Solar
    System. Since we aren't considering them as part of the system, do
    they have any influence on the location of the barycenter? What if
    we then change our minds and decide to include them?

    -- Jeff, in Minneapolis
    If these massive interlopers are massive enough to cause significant perturbations of the planets' orbits, then yes indeed, they must be included.

  24. #114
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by George View Post
    We develop equations to define what happens with a specific phenomena. Newton's inverse square law gives us the force action between static (or dynamic) masses. If we want to know, say, how far an object falls, we simply derive the distance equation from the inverse square law calculus. My hunch is that the simple barycenter equation, even if it is a perfect match for a seesaw, is something derived in a similar fashion, or could be derived. Since we already know it works, then who cares? ... Me, though I'm really only curious and not suggesting something exceptionally interesting here. [Added: I would guess this is shown in some celestial mechanics texts, but I only recently acquired one.]

    The seesaw only works because of the summation of moments = 0 (at balance), which is due to the physical elemental moment transfer along the lever. But, in space, as you have correctly pointed-out, there is no actual "leverage". There is no apparent reason that the math is the same, though it is not illogical that they match.
    I am still having a hard time figuring out just what it is you are having trouble with conceptually. The barycenter equation could in principle have been invented as a mathematical curiosity in the absence of observing a physical phenomenon, but my educated guess is that observations came first. It is also my educated guess is that something along the lines of a seesaw or a beam balance, or perhaps a crowbar, would have come first.

  25. #115
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Jeff Root View Post
    Grant,

    To help George understand this, you described the orbits of two
    bodies around each other. You find the barycenter of that two-body
    system. What if there are other massive bodies present, but you are
    not considering them as part of the system? Such as the Sun and its
    planets, but not this huge interloper, Bronson Alpha, and its smaller
    companion, Bronson Beta, which are passing through the inner Solar
    System. Since we aren't considering them as part of the system, do
    they have any influence on the location of the barycenter? What if
    we then change our minds and decide to include them?
    The barycenter is simply the point for a given set of masses. A Sun-Jupiter barycenter will not accurately give us the actual motion of the Sun around this point since there are other masses that do influence its motion significantly.
    We know time flies, we just can't see its wings.

  26. #116
    Join Date
    Sep 2003
    Posts
    11,561
    Quote Originally Posted by Hornblower View Post
    I am still having a hard time figuring out just what it is you are having trouble with conceptually.
    Earlier, I associated the seesaw (leverage) equation as something that did possibly cause leverage in space, but we're past that hiccup. My question now goes back to an early question raised, IIRC, in essence which asks how an inverse square law for distances between masses can be reduced to a linear equation to find the barycenter? I thought, perhaps, the orbital factor (centrifugal force) might be needed, but perhaps not. If not, it more explicitly supports your non-orbital view for the barycenter. [Of course, we can find the barycenter without motion as you've stated, but I'm curious if it takes orbital motion to get us to that point.]
    We know time flies, we just can't see its wings.

  27. #117
    Join Date
    Mar 2007
    Location
    Falls Church, VA (near Washington, DC)
    Posts
    7,309
    Quote Originally Posted by George View Post
    Earlier, I associated the seesaw (leverage) equation as something that did possibly cause leverage in space, but we're past that hiccup. My question now goes back to an early question raised, IIRC, in essence which asks how an inverse square law for distances between masses can be reduced to a linear equation to find the barycenter?
    The force in question need not follow an inverse square formula for its magnitude. A pair of bodies, initially stationary when released in free fall, could be connected by a thought-exercise spring for which the force diminishes as the bodies close in. What happens here is that the forces on each body are equal in magnitude and opposite in direction, as in Newton's action-reaction law. Thus their accelerations are inversely proportional to their masses, as are their starting distances from their mutual barycenter, and thus they will meet at the barycenter.
    I thought, perhaps, the orbital factor (centrifugal force) might be needed, but perhaps not. If not, it more explicitly supports your non-orbital view for the barycenter. [Of course, we can find the barycenter without motion as you've stated, but I'm curious if it takes orbital motion to get us to that point.]
    My bold. As I understand it, no. We can get the concept of the barycenter purely from static loading considerations, as in placing the masses on a trailer. It just happens that the dynamics of gravity and inertia/momentum are such that the same barycenter turns out to be the center of orbital motion if we put the bodies in free fall with some transverse relative motion rather than having them stationary on the ground.

  28. #118
    Join Date
    May 2005
    Posts
    7,581
    Quote Originally Posted by George View Post
    Earlier, I associated the seesaw (leverage) equation as something that did possibly cause leverage in space, but we're past that hiccup. My question now goes back to an early question raised, IIRC, in essence which asks how an inverse square law for distances between masses can be reduced to a linear equation to find the barycenter? I thought, perhaps, the orbital factor (centrifugal force) might be needed, but perhaps not. If not, it more explicitly supports your non-orbital view for the barycenter. [Of course, we can find the barycenter without motion as you've stated, but I'm curious if it takes orbital motion to get us to that point.]
    I think I see what you're asking for, good question.

    All sorts of calculations could show it. Maybe the easiest way to see it is through energy (how many times have I said that?)

    Force is the derivative of energy .

    There's the inverse linear relationship you've been looking for (and, it's proportional to mass).

    Conservation of energy then has that center of mass not changing, without some other ("external") force, whatever the motion of the objects.

  29. #119
    Join Date
    Sep 2003
    Posts
    11,561
    It's easier than expected. Using R=1/2at2, and T1 = T2 since they, in free-fall, will meet at the barycenter, we get m1r1 = m2r2! Summation of moments (seesaw)from inverse square law and F = ma. I didn't think we could do it.

    iphone
    Last edited by George; 2017-Mar-25 at 04:51 PM.
    We know time flies, we just can't see its wings.

  30. #120
    Join Date
    Jul 2005
    Posts
    14,099
    Maybe a direct way to see the relationship to the see-saw is to consider the sun and the Earth-moon system. The sun's gravity attracts Earth and moon in proportion to their mass, just as the Earth's gravity attracts the balanced masses at either end of a see-saw. The net effect of the sun's gravity is therefore to attract the Earth-moon system as if the force were being applied to their centre of gravity, which is necessarily in the position you'd need to put an astronomical-sized see-saw fulcrum to balance the two bodies under the sun's gravity. (In addition to this radial force on the barycentre, The net force also decomposes into a tidal component that causes the solar-tidal evolution of the moon's orbit.)
    So it's the Earth-moon barycentre that describes the smooth Keplerian ellipse around the sun every year, while Earth and moon lollop around it.

    Grant Hutchison

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •