Where is most "gravity", inside or out?

How can there be a place that has 'more gravity' than the maximum? The maximum is the maximum.
Thank you for that prompting: what I am looking at, is the total gravity as expressed by the areas in origin,s graph. There is more "total" 0utside gravity at 2R than in the whole interior., and the rest to infinity is gravy.
and how about those open and ellipticals?
What do you mean?
. -- As I understand it, these globulars, clusters and ellipticals have no common rotation,, no central bulge, perhaps no center black hole. yet the members can not just hang there, so they must have orbital velocities that reflect the gravity at their position in their particular orbits. At Inner orbit positions velocities should be low, in the periphery the speeds should be high. ergo, ; a situation opposite what Kepler's law predict for the central mass Solar system observations. so: for these old and loose conglomerations, there should be an anti Kepler law in force. (no farce).

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
The triangular areas expressing Kepler's second law are pure constructs of our human mathematics, but so are the terms in the special theory equations;-- origin's graph's areas have no counterpart in nature either, yet represent values that could have interesting implications when applied., seen in other contexts. for example ( see the "c" factor in general relativity)

This link http://www.cosmotography.com/images/galaxy_formation_and_evolution.html gives a very recent theory* about dark matter and its role in galaxy fromation (with review from the big bang). If you click on the bold text: Lambda Cold Dark Matter (or LCDM) it leads you to this interesting 13 minute youtube video:

* More on that theory here: https://www.sciencedaily.com/releases/2015/09/150924142851.htm article published September 24, 2015.

I can't evaluate it all.
Excellent video.

As I understand it, these globulars, clusters and ellipticals have no common rotation,, no central bulge, perhaps no center black hole.
A globular cluster is a long lasting generally spherical conglomeration of old stars that reside in a galaxies halo and orbit the galactic center. New information shows that many globular clusters do have rotation of the interior stars. Clusters, I will assume you mean Open Clusters, are a short lived grouping of young stars in the star forming regions of galaxies that are all associated because they all formed from the same molecular cloud. Interactions with other stars tend to break up open clusters in a relatively short time. There of course is no center bulge or black hole associated with an Open Clusters (such as Pleides or the Beehive). Globular Clusters do not have a center bulge since they are not galaxies and probably do not have a blackhole. Ellipticals are odd shapped large galaxies that are probably the result of 2 or more galxies merging resulting in a 'messy' galaxy.
ergo, ; a situation opposite what Kepler's law predict for the central mass Solar system observations. so: for these old and loose conglomerations, there should be an anti Kepler law in force. (no farce).
Unfortunately, this is taking a shift to the fringe, just like your other threads have also taken.

.

The triangular areas expressing Kepler's second law are pure constructs of our human mathematics,
And the journey to the fringy area picks up speed...
Kepler's second law is based on and explains the observations of orbit dynamics, it is not just some mathematical construct.

... At Inner orbit positions velocities should be low, in the periphery the speeds should be high. ergo, ; a situation opposite what Kepler's law predict for the central mass Solar system observations. so: for these old and loose conglomerations, there should be an anti Kepler law in force. (no farce).

The force of gravity outside the mass unifrom distribution is C/(r^2) where C is a constant. An object in circular orbit with velocity v has an acceleration a towards that central pulling force of (v^2)/r and the force making it is F=ma so
C/(r^2) = m (v^2)/r and multiplying both sides by r^2 gives C =mr(v^2)
Now for non-relativistic speed m is a constant, so C' = rv^2 where that constant has been merged with C to make new constant C'.

Thus if r is increased four fold the velocity will decrease by a factor of two . Or in general the greater the r is then the SLOWER, the orbiting object goes. That is why the more distant planets have much longer years. Not only is their path around the sun, longer, but they are going slower.

Your statement: "At Inner orbit positions velocities should be low, in the periphery the speeds should be high." is wrong for planets orbiting a central mass.

Although you can not use this math about circular orbits for planet in an elliptical orbit, the "equal areas in equal times" tells you the planet is going more slowly when more distant from the sun (central force) than when near it.

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Kepler's second law is based on and explains the observations of orbit dynamics
I was not referring to the success of the law, but the fact that the equal size triangular areas of K#2, swept out by the planets, are not a feature of the natural world. neither are the rectangular areas in origin's coordinate system of post#2..

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At Inner orbit positions velocities should be low, in the periphery the speeds should be high." is wrong for planets orbiting a central mass.
Have you not read that I was referring to elliptical, clusters, that do not have a central mass? In these bodies, any "simple single" central star could be stationary, but the outer members would have to be really moving along to combat the gravity, that would be equal to all the combined mass to be at the theoretical center.

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Thus if r is increased four fold the velocity will decrease by a factor of two .
so, can you find that represented by a graph somewhere, or superimpose that on the curve origin posted? thank you.

And the journey to the fringy area picks up speed...
Kepler's second law is based on and explains the observations of orbit dynamics, it is not just some mathematical construct.
I'm not 100% sure, but think Kepler did not know the math that explained why this OBSERVATION of his was true. It is derived in Newton's great work. The principles of mathematics.

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Have you not read that I was referring to elliptical, clusters, that do not have a central mass? In these bodies, any "simple single" central star could be stationary, but the outer members would have to be really moving along to combat the gravity, that would be equal to all the combined mass to be at the theoretical center.
There is always a center of mass. The further away from it the orbiting star is the slower its averge orbital speed will be. The part of your text I made red is wrong. Also gravity comes from mass, is not "equal to all the combined mass" I know what you were trying to say. You should try to say what you mean more correctly, not falsely.

When you spoke of / said: "At Inner orbit positions velocities should be low, in the periphery the speeds should be high." you are clearly referring to star orbiting, not necesarily in a circular or elliptical orbit, but at least going around so one can refer to "inner orbits" vs. more "peripherial ones."

One speak of "dynamic interactions" or "scattering" not "orbiting" when a star's path takes it close to another star of even a local collections of stars in a much larger stellar mass.

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Janus58:

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.

That is a very good outline of the currently held understanding of the gravitational mass and gravitational acceleration and orbital velocity profiles expected in the cases you describe. Thanks very much, Janus58.

I am still perplexed though. Here is why. I understand the spherical parent body case has greater gravitational acceleration on a test body as we go towards the surface, where it reaches a maximum because all the parent body's matter is "below" the test body; because the matter on either side as well as on opposite side is at maximum with respect to test body position at surface.

However I also know that the gravitational acceleration effects vectors from the side material adds to a lesser, resultant value, not a straight line maximal value as it would be if that same matter was in a disk configuration as in the Spiral Galaxy case: in which latter, as the test body proceed outwards to the outer edge, all the that mass's gravitational acceleration, not just a lesser resultant portion of gravitational acceleration as in the Spherical case, is acting on the test body; and so it would experience comparatively greater (from all the mass) acceleration in the Spiral distribution case than it would in the Spherical (net resultant mass effect only) distribution case.

Hence the the test body in a Spiral Galaxy (disk distribution) would have to move more speedily in its orbit if it was to remain in orbit. This is because there would be greater gravitational acceleration from the disk form at that radial altitude compared to the gravitational acceleration at the same altitude from the Spherical form of the same mass.

That is my reasoning for why I am still perplexed by the conventional expectations comparison between the spherical and disk cases which you outlined. I hope I explained it intelligibly? If not, please ask for clarification and I will do what I can to comply as soon as I can. Thanks, Janus58.

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The following may help in any confusion in relation to rotational curves and any silly outrageous anti-Kepler law pseudoscience....

http://arxiv.org/pdf/1108.5459.pdf
Constraining A String Gauge Field by Galaxy Rotation Curves and Perihelion Precession of Planets

ABSTRACT
We discuss a cosmological model in which the string gauge field coupled universally to matter gives rise to an extra centripetal force and will have observable signatures on cosmological and astronomical observations. Several tests are performed using data including galaxy rotation curves of twenty-two spiral galaxies of varied luminosities and sizes, and perihelion precessions of planets in the solar system. The rotation curves of the same group of galaxies are independently fit using a dark matter model with the generalized Navarro–Frenk–White (NFW) profile and the string model. Remarkable fit of galaxy rotation curves is achieved using the one-parameter string model as compared to the three-parameter dark matter model with the Navarro-Frenk-White profile. The average χ 2 value of the NFW fit is 9% better than that of the string model at a price of two more free parameters. Furthermore, from the string model, we can give a dynamical explanation for the phenomenological Tully-Fisher relation. We are able to derive a relation between field strength, galaxy size and luminosity, which can be verified with data from the 22 galaxies. To further test the hypothesis of the universal existence of the string gauge field, we apply our string model to the solar system. Constraint on the magnitude of the string field in the solar system is deduced from the current ranges for any anomalous perihelion precession of planets allowed by the latest observations. The field distribution resembles a dipole field originating from the Sun. The string field strength deduced from the solar system observations is of a similar magnitudes as the field strength needed to sustain the rotational speed of the sun inside the Milky Way. This hypothesis can be tested further by future observations with higher precision.

Billy T:

Thanks, Billy T, for your kind interest in my questions, discussion points and reasonings for same.

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.

Please see my post #33 to Janus58. I hope that explains why I am perplexed with the conventional expectations comparison between the spherical and disk case of same mass. Do you or any other learned member conversant with the relevant physics principles have any comments regarding the validity or otherwise of my reasonings for my questioning of why the conventional expectations comparisons are as they are? It is those conventional assumptions used for interpreting the spiral galaxy orbital speeds curve that perplexes me in light of what I have just posted to Janus58 that does not accord with what has been explained to me to date. Thanks.

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galaxy rotation curves of twenty-two spiral galaxies of varied luminosities and sizes, and perihelion precessions of planets in the solar system. The rotation curves of the same group of galaxies are independently fit using a dark matter model with the generalized Navarro–Fre
To repeat for emphasis, In the drop to central zero discussion, I am referring not to spiral galaxies or the Solar system, but to entities that come closest to the even-matter distribution model used in the graph above. Kepler's laws are predicated on a dominant central mass like the sun, and global clusters, ellipticals lack those. This would result in a reverse order of dynamics, , slow speeds in the center, maximum velocity in the outskirts, hence my term anti, or reverse kepler.

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.
And the journey to the fringy area picks up speed...
Kepler's second law is based on and explains the observations of orbit dynamics, it is not just some mathematical construct.
It's called mathematical physics. The key word is physics. Mathematics is a tool for describing the physics. It's the physics that we're talking about. Anyway I never think of gravity in terms of force. That's what Newton had to work with. Einstein was more fortunate. What I'm going to say isn't a proof but instead a loose correlation. For the metric solutions of GR the curvature component is 2M/r. The first derivative of the curvature component is g=M/r^2. So the correlation is between these two values in the local spacetime geometry with the local spacetime curvature being gravity. For inside the planet g is strongest at the planet surface and decreases as r-> 0. Outside the planet it also decreases for r > r_planet surface. Pretty much the same way you'd describe the g_field using Newton's theory. Using this correlation the local spacetime curvature, gravity, is greatest at the planet surface. So when you think of gravity as local spacetime curvature you shouldn't be confused by which side of the planet is closest to the Sun. Like you said the Sun's gravitational contribution to the local spacetime curvature at the planet surface is minuscule.

I do not consider the generic diagram in post 2 as representing a planet only, but more of a body with more even mass distribution like a open globular star cluster, where there is no concentration of denser matter in the center. In Newton's center of gravity as it turns out, is no gravity at all. and the external gravity, however thought of, as spacetime warp, exchange of gravitons, is more far reaching than what happens in the interior. Gravity pictured as a field depression would have all entities interior as a goose pump. should not even the central black holes have a gravity neutral center ?

There is always a center of mass.
yes, But I was not speaking of the center of mass but an absent central mass, like a black hole. or a bulge. In some ellipticals, clusters, there is no central mass, and in that case the gravity gradient and the orbital speeds must correspond to the linear slope in origin's illustration. without the central mass, Kepler's laws can not apply, inner orbit vs must be near 0, outer orbit' vs max?

the disk is a disk, not a sphere,
good point. In an "ideal" disk, a near 2 dimensional pancake, light, or gravity would not decrease with the square of the distance, but decline in a near linear fashion, in effect doubling the outer observed orbital velocities. There is even more gravity outside a disk than outside a sphere. or?