Where is most "gravity", inside or out?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by nebel, Feb 29, 2016.

  1. nebel Valued Senior Member

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    In any entity, gravity falls to zero at the center, but reaches that value only at an indefinite distance. At a heights equal to the radius, there is still 1/4 of the gravitational surface pull to the center, whereas the same distance down from the "surface", it is already balanced out of existence. What does this mean for large scale structures?, their orbital velocities "in" and around them?
     
    Last edited: Feb 29, 2016
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  3. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I think this graph will clearly answer your question:

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    The origin is the center of the body and gs is the force felt from gravity or gravitational acceleration at the surface of the body.

    The only place where the force is 0 is at the exact center of the mass. You have to be careful of what you are discussing: gravitational force, gravitational potential, etc.
     
    Last edited: Feb 29, 2016
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  5. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    The graph is correct, but it may be helpful to explain why for r<R it is a linear rise from zero. First note with unifrom spherical mass distribution 7/8 of the mass is between R/2 and R and this mass has zero net pull on a point at r = R/2. Only the mass at R/2 or less radius pulls on that point. That mass is only 1/8 of the total mass and acts as if it were all at the origin center. As R/2 is closer to the center the gravity, from that 1/8 mass is 4 times more effective (inverse square law at work). So at the R/2 point the force of gravity ie (1/8)x4 = 1/2.

    This same logic, math applies at any r < R but is not as simple to do except for the r =R/3, r = R/4, etc. cases. I. e. at r = R/n the gravity will be 1/n what it is at r = R.
    Hint: At R/3 point only 1/27 of the mass acts on that point, but it is 9 times more effective. So there the gravity force is (1/27)x 9 = 1/3 and right on the linear line. etc.
     
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  7. nebel Valued Senior Member

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    Thank you , so: why are we surprised that we have comparative small orbital velocities in the inner, and greater movement in the outer reaches of entities, where most of the gravity is?
    PS: and what would a graph of orbital velocities look like in such an idealized system? with low mass orbiting bodies?
     
    Last edited: Feb 29, 2016
  8. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I am not sure what you are talking about. The orbital velocity is faster the closer the bodies are to each other not lower. That is exactly what we see for our solar system. What do you mean most the gravity is in the outer reaches?

    You understood on the graph that R is the surface of the planet, right?
     
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  9. Janus58 Valued Senior Member

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    If you are talking about the orbital speeds for stars in a galaxy then you have to realize that the graph given that shows gravity increasing linearly as you move out from the center until you reach the surface is for a spherical object of uniform density. A typical galaxy is neither of these things. To work out the expected orbital velocities for a typical spiral galaxy, you need to take its shape and density distribution into account. A spiral galaxy consists of a dense central bulge and a thinner surrounding disk. If you plot the expected orbital speeds as you move outward from the center based on the visible shape of the galaxy, you would expect the velocities to increase at first while you stay in the bulge region similar to it does inside the uniform sphere, and then fall off as you leave it and enter the disk region a lot like it does it would outside of the sphere's surface. The fact is that this does not happen. The orbital velocities do increase while inside the bulge, but they don't decrease like they should once you leave the bulge region, depending on the galaxy, they either decrease too slowly, don't decrease at all or even increase slightly. So not only do the orbital velocities behave as if there is more mass than we can see, but that that much of that mass is distributed a lot differently than the visible shape of the galaxy.
     
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  10. expletives deleted Registered Senior Member

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    Janus58:

    That is a very interesting explanation. Thanks. I am confused now, about that part of the orbital velocity profile from inner disk towards outer disk. I understand the part involving the spherical central bulge profile, but I would have thought that, since unlike what a spherical distribution would do to the orbital speeds, the disk is a disk, not a sphere, so the majority of the mass pulling on the outer stars would be along the plan of the disk, without much at all either side of the galaxy like there would be if it were a spherical, not disk, galactic outer region. Am I making sense? My reasoning is that if a galaxy is spherical all the way from inner central region to its outermost regions, then the gravitational acceleration and orbital velocity profile would be as you described for a spherical body and for the central bulge of the spiral galaxy. But settled spiral galaxies are not spherical all the way to the outermost regions, so why would the gravitational and velocity profiles be expected to drop if more and more galactic matter is along the plane, rather than on either side as in a spherical case, as we move outwards along the disk? That is my question and reasoning. I hope I haven't made a physics or logics blunder. If I haven't, can you or some other learned member explain the unusual expectation of decreasing velocity rather than increasing velocity as increasingly more mass is "below" the stars as we go from inner to outermost orbitals in the disk along the plane of the disk (not as in spherical distribution)? Thanks.
     
  11. paddoboy Valued Senior Member

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    As usual janus58, a nice concise explanation.
     
  12. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    If you are a star outside a spherical galaxy, then you are pulled towards the center more strongly than if only the stars closer to the center were pulling on you. Imagine two cases: A = you are out side a sphere & B = you are outside a disk of thickness R/10, where R is your distance from the center.

    In case A there are many stars not in the disk that are pulling you directly towards a point that is either above the region of the disk or below the region of the disk and they both have a component of their pull directly towards the center of the disk. Their components orthogonal to the disk cancel out, but the toward disk components add. This is why case A has greater force pulling towards the center than case B with only stars in the disk pulling on you.

    Or stating it reversely: the force on you outside a spiral galaxy, missing the net added pull to the center as has no stars above and below the disk, is less less than outside a spherical galaxy. I'm not sure I understood you question, but if I did that should answer it.
     
    Last edited: Feb 29, 2016
  13. Janus58 Valued Senior Member

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    The orbital speed depends on the mass(M) you are orbiting and your distance from the center of that mass. Increasing M has the effect of increasing the orbital speed and increasing the distance has the effect of decreasing it If you are in a spherical body, or a disc, the only mass contributing to M is that part of the body closer to the center than you are. With a spherical body the mass closer to the center increases by the cube of your distance from the center, so when you move away from the center, the effect of increasing M exceeds the effect of increasing the distance, so orbital speed goes up until you are outside the body. Then M remains constant even while the distance increases and orbital speed starts going down. With a disk, M does not go up as fast with r, and in fact, the two effects would cancel and you would expect the orbital speed to stay constant with distance. A spiral galaxy is neither purely spherical or disc shaped. It is a disc with a spherical bulge in the center. Inside the bulge, the effect on orbital speed in what you would expect to find in a sphere. Once you get out of the bulge, you have to consider both the effects the mass of the bulge has and the increasing mass added by the disk as you move outward. The effect of moving away from the mass of the bulge has the effect of decreasing the orbital speed, The addition of mass due to the disc will lessen this effect but will not be enough to counter it completely. Thus you should see the orbital speed fall off with distance as you increase distance from the bulge. Not as fast as it would if all the mass were concentrated in the bulge, but still falling off.

    The whole point is is that when you take into account all the mass of the visible galaxy and the way it is distributed, you should get a certain pattern to the way orbital speed changes with distance. The fact that the actual orbital speeds we measure do not follow this pattern indicates that there is mass involved that we do not see. And it is just not a matter of more mass. If it were just a matter of getting the scaling factor of the mass wrong, you would get different values for the orbital speeds, but the pattern would stay the same. It is the fact that the pattern itself differs that indicates that the extra mass does not follow the same distribution as the matter we see in the galaxy.
     
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  14. brucep Valued Senior Member

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    Very informative post. Hope somebody is paying attention
     
  15. nebel Valued Senior Member

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    paddoboy, yes the graph expresses my idea/ question, the "surface" R could be a planet's or any entity. even in a Galaxy with no visible outer boundary, there must be points that define it's size. Surely, a galaxy, or star cluster might not be like our Solar system where ~99.8% is in the very centre, and even then, in the center of the sun there is zero gravity. When I said " more gravity outside" I meant above the surface at R. I refer to the the fact, that gravity is a rapidly vanishing quantity toward the center, but the area representing the "outside gravity" tapers off to eternity. That is why I would like to see (for all viewers) a model put up where the gravity values from Zero at the center to max at the surface and off into the distanc are expressed in orbital velocities, not any matter supported by underlying compressed material. For example, is not in open star clusters very little speed of the inner stars required to keep from collapsing them into the center? and of course as Janus 58 brought out in his great post, , Galaxies have complex density gradients(oofen only estimated) that would have the orbital velocity diverge from the ideal pattern I would like to see.
     
    Last edited: Mar 1, 2016
  16. paddoboy Valued Senior Member

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    Please Register or Log in to view the hidden image!

    That's all I'll say for now.
     
  17. nebel Valued Senior Member

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    It occurred to me, that there is a curious similarity to Johann Kepler's second law's illustrations, and the graph that origin put up. In both cases there is convergence, minimum area, at the Center of Gravity, or rather the center of non-gravity, and of course Kepler's illustrations could be amended to show gravity's value on the outside of the ellipse, or sphere, and the area would be showing to converge to zero again in the very far distance. The implications of this could best be seen in open star clusters, and elliptical galaxies, where stars surely must follow individual ~ stable orbits, devoid of a common rotation, with speeds that are very "slow" near the inevitable zero-gravity center(s) and show the greatest speed at the point or "envelope" of maximum commun self gravitation, equivalent to origin's peak at gs/R. If there would be any gas left in these old outer peripheries, , its particles should show orbital velocities that mirror the outside triangles area, illustrating this outer realm of greater gravity. -- any good illustrators out there?
     
    Last edited: Mar 1, 2016
  18. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    The gravity distribution of a solar system is vastly different than a galaxy.

    This is just a rehash of your earlier posts I'm now realize.

    Here is a graph of the rotational velocity of a galaxy.

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  19. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    That difference in post 15's graph, between green and orange curves leaves us in the dark - pun intended.
     
  20. nebel Valued Senior Member

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    indeed, it leaves us in the valid dark matter theory variants. Here is my question though: can it be said there is always more gravity outside the maximum point of gravity of any system, than toward it's (bari) centre? as an example, is there more gravity outside the Pluto surface than the one that faces away on a corresponding, equidistant point on it's orbit opposite the sun? Yes, a possible rehash, but with some spice added. because how is the fact that there is more gravity outside expressed in the movements we observe? origin's curves show that there is even more gravity outside than expected. and how about those open and ellipticals?
     
  21. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    How can there be a place that has 'more gravity' than the maximum? The maximum is the maximum.
    If you mean is the force of gravity from Pluto on the surface of Pluto less when that surface is facing the sun, the answer is yes, but of course it is miniscule.
    There is much more mass in the galaxy than is observed. So the unobserved matter must be dark.
    The velocity profiles indicate there is more mass than what is observered, hence the theory of dark matter.
    What do you mean?
     
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  22. exchemist Valued Senior Member

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    Seconded. In fact this is a very high quality thread.

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  23. nebel Valued Senior Member

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    correction: there is clearly more gravity inside the pluto orbit than outside, with gravity having peaked at the sun's "surface".
     

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