Where is most "gravity", inside or out?

Of course not, but it is correct in this specific case which is what the question was about.
The question was about whether there is more [accumulated] gravity toward the interior or the exterior of a symmetrical spherical mass, ideally ignoring for simplicity sakes local (black hole) peculiarities, and: is it not true, that gravity is mainly on the outside? Even in an evaporating black hole, the interior gravity has to fall to zero at the center, theoretically. or? That mass of dark matter too is inert gravitationally toward the center. imho.
so: if you think that the higher outer orbital velocity have to be the result of dark matter in the peripheral neighbourhood, ?? re-examine that please! perhaps it is really because all the gravity is there , compliments of the old fashioned matter and the shell theorem?
PS: Even outside the Schwarzschild radius, the same mighty gravity of the Black hole mass that falls quickly to the near center, will only slowly peter out approaching infinity. Adding mass will not change the fact that all the gravity acts toward the outside. Sphagettification is reversed once you pass the event horizon, or max gravity.
 
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Take the case of the singularity, with that small interior dimension, all the mass is there, but no gravity, all the force or projected tension is to the outside, to eternity, but, the force reverses direction through the center, so there must be null gravity in the center and with diminishing gravity inward, there has to be a reverse tidal effect,or?
 
Take the case of the singularity, with that small interior dimension, all the mass is there, but no gravity, all the force or projected tension is to the outside, to eternity, but, the force reverses direction through the center, so there must be null gravity in the center and with diminishing gravity inward, there has to be a reverse tidal effect,or?

There is a concept of central pressure for a spherical object. This central pressure is non zero.
 
There is a concept of central pressure for a spherical object. This central pressure is non zero.
Pressure is the result of gravity acting from the outside --(because that is where it is)--inward. While pressure , as kinetic energy can be a contributing source of gravity, it does not cause gravity, a pulling condition/force to go to maximum at the center. or?
PS: A more pressure-compressed mass would not reduce the amount of gravity (intensity x distance) on the outside would it, but would confine its gravity to a smaller volume. so: could we say that the more dense the body the greater the outside to the inside gravity ratio?
 
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I think this graph will clearly answer your question:

2762529.png
If you compressed the mass with size R into a smaller and smaller volume, The peak of the surface gravity would rise with decreasing radius, but the original curve, from R to 2R to 3R would remain. The interior gravity pictured by the area of the 0Rgs triangle, would have a sum of the same value, but an additional outside gravity potential would be created, defined by area inclosed by the newly added steeper section of curve . so:
can it be said that there is more gravity outside than inside? more so as the central mass becomes denser?
 
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Like creatures? Why would they follow different laws
entities like galactic superclusters, the solar system, the Earth. there is no gravity at their center of gravity, most gravity is outside their :surface, the denser they are, the more gravity is at the outside, stretching to infinity, do not be surprised if a lot is happening outside. faster than we might think. whether they are directly created, creatures, is another, non-technical matter.
 
There is related material in a thread in the Alternate Theory forum:
"ALMA," lookback time to oldest galaxies. .
In there is proposed a model of the universe as seen having moved away from the BB in all directions through time, to now exist in an expanding sphere membrane.
In such a membrane, all the gravity or other force fields would be on the outside, tapering off into infinity the future, that is outside the universe, none would be left in the empty past, even the BB location , the point in timespace where it all started.
startling?
Why the big bang was not the beginning
First hints are emerging of a universe that existed before our own: an alien world of chaos where time, space and geometry were yet to form --quote from New Scientist mag issue 14 March 2018.


perhaps this condition "world of chaos" outside, and before the present universe, before this universe in timespace was susceptible to gravity?
There was even more gravity outside then inside then?
 
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There is related material in a thread in the Alternate Theory forum:
"ALMA," lookback time to oldest galaxies. .
In there is proposed a model of the universe as seen having moved away from the BB in all directions through time, to now exist in an expanding sphere membrane.
In such a membrane, all the gravity or other force fields would be on the outside, tapering off into infinity the future, that is outside the universe, none would be left in the empty past, even the BB location , the point in timespace where it all started.
startling?
Please don't drag pseudoscience from the Fringe section into the Science section.
 
If you compressed the mass with size R into a smaller and smaller volume, The peak of the surface gravity would rise with decreasing radius, but the original curve, from R to 2R to 3R would remain. The interior gravity pictured by the area of the 0Rgs triangle, would have a sum of the same value, but an additional outside gravity potential would be created, defined by area inclosed by the newly added steeper section of curve . so:
can it be said that there is more gravity outside than inside? more so as the central mass becomes denser?
I doubt that very much. Gravity itself is caused by mass, regardless of size.

What you are talking about is the size of the gravity field becomes larger with greater mass. But the gravity field itself does not add to the mass of the object, it is caused by the object. It is a result of mass.

But i'd like to ask a counter question: If mass causes a warping of spacetime, would it follow that the smaller the size of the massive object, the greater the warping of spacetime. IOW the gravitational well becomes deeper instead of more spread out.

If I visualize a very small neutron star with the same mass as say, Sirius, which is a very large star, what happens to their gravitational fields? Will the neutron star create a smaller, but deeper well , than Sirius, which would create a larger but shallower warping of spacetime?

Either way the gravitational fields might be equal in overall volume, but differ in dimensional shape?
 
I thought an item in New Scientist would be worthy enough to be considered here, even though on a topic that has so far escaped heading south.
I am talking about your own stuff in there.
 
I doubt that very much. Gravity itself is caused by mass, regardless of size.

What you are talking about is the size of the gravity field becomes larger with greater mass. But the gravity field itself does not add to the mass of the object, it is caused by the object. It is a result of mass.

But i'd like to ask a counter question: If mass causes a warping of spacetime, would it follow that the smaller the size of the massive object, the greater the warping of spacetime. IOW the gravitational well becomes deeper instead of more spread out.

If I visualize a very small neutron star with the same mass as say, Sirius, which is a very large star, what happens to their gravitational fields? Will the neutron star create a smaller, but deeper well , than Sirius, which would create a larger but shallower warping of spacetime?

Either way the gravitational fields might be equal in overall volume, but differ in dimensional shape?

Here's a the cross sections of the gravity wells of Sirius, compared to the a equal mass neutron star. It is plotted as gravitational potential along the vertical and distance from the center along the horizontal. The black vertical line marks the center for both bodies and the green lines where the surface of Sirius would be.
gravwell.png
Note that outside of the green lines, the wells are identical. It is inside the lines where they differ. The well for Sirius begins to level out again as you pass beneath its surface. The well for the neutron star gets steeper and plunges a lot deeper (once below its surface, it will also start to flatten out.)
The slope of the lines are representative of the local field strength (strength of gravity) at that point. Horizontal =0, Vertical would be infinite. For Sirius, it increases as you get closer to the star, but then decreases after you pass below the surface and becomes zero at the center. With the smaller neutron star, gravity continues to increase after you pass the distance equal to the radius of Sirius up until you reach its surface.
 
Either way the gravitational fields might be equal in overall volume, but differ in dimensional shape?
good points. The OP question was very general. as you said so much depends on the shape of the object associated with the gravitational field, the strength measured everywhere and plotted.
Take the case, discussed somewhere else, of a universe sized model having expanded into an empty outer shell. No gravity inside at all, only on the outside all the forces reaching to infinity.
Even the various graphs on paper do not give the total strength/volume picture, because even at equal value at a distance from the center, the 3 dimensional inner versus outer volume projection shows that gravity is a phenomenon of the perimeter on, and not of the interior.
 
Here's a the cross sections of the gravity wells of Sirius, compared to the a equal mass neutron star. It is plotted as gravitational potential along the vertical and distance from the center along the horizontal. The black vertical line marks the center for both bodies and the green lines where the surface of Sirius would be.

Thank you for all the graphs.
with due respect, all the curves shown should go back up to the zero level at the top of the intersection of the horizontal line and dotted and vertical black line. Is that not so?, because you said correctly:
but then decreases after you pass below the surface and becomes zero at the center.
bsw, a similar graph was shown as what timespace would look like if a dense subject stopped moving through time, and the rest of the universe kept moving out from the beginning.
 
Here's a the cross sections of the gravity wells of Sirius, compared to the a equal mass neutron star. It is plotted as gravitational potential along the vertical and distance from the center along the horizontal. The black vertical line marks the center for both bodies and the green lines where the surface of Sirius would be.
View attachment 1885
Note that outside of the green lines, the wells are identical. It is inside the lines where they differ. The well for Sirius begins to level out again as you pass beneath its surface. The well for the neutron star gets steeper and plunges a lot deeper (once below its surface, it will also start to flatten out.)
The slope of the lines are representative of the local field strength (strength of gravity) at that point. Horizontal =0, Vertical would be infinite. For Sirius, it increases as you get closer to the star, but then decreases after you pass below the surface and becomes zero at the center. With the smaller neutron star, gravity continues to increase after you pass the distance equal to the radius of Sirius up until you reach its surface.
Thanks for that clear illustration of identical masses creating the same gravitational force outside of the wells.

But now the logical next question would be what happens to the spacetime fabric inside the wells. From the illustration it would appear that spacetime becomes much more stretched in the neutron star's well, than for Sirius' well.

So the question would be if the volume of each gravitational well is also equal, due to the much larger circumference of Sirius' gravitational well even as the spacetime fabric seems to be less warped than for the neutron star?

Can this be measured? I guess this is basically the same question as *nebel* posed in #97.
 
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Thank you for all the graphs.
with due respect, all the curves shown should go back up to the zero level at the top of the intersection of the horizontal line and dotted and vertical black line.
No. Gravity has zero net force at the centre, but you are still deep in a gravitational well. If you could drop a probe to the centre of Sirius to take measurements, you would find that time dilation applies.
 
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