# Where is most "gravity", inside or out?

Discussion in 'Pseudoscience' started by nebel, Feb 29, 2016.

1. ### nebel

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right. From the outset, Post# 2 of origin set the tone for that. that the strengths of the field is discussed, and is shown as positive on the Y axis. not negative.
Interesting that David introduced the idea of terminal velocity, acceleration to it, into the graph. zero gravity floating not only inside the empty shell but also in the ship forever. (a line parallel to the x axis).

3. ### DaveC426913Valued Senior Member

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Just because the net gravitational pull in a given direction is zero doesn't mean potential energy is zero.

You don't get to jump out of a hollow sphere all the way to flat space for free.

This is the correct graph.

It shows both
the amount of energy required to get from one point to another,
and
the amount of gravitational time dilation experienced at any given point.

Both are maximum inside the shell.

5. ### nebel

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2,469
how so? inside the empty shell, only a small initial push will send B all the wau across to the other side, and back again, provided the inside wall is a perfect trampoline. Free travel along the x axis direction. Since it is a sphere, in any direction.
Try that on the outside.
That is why I say, there is more "gravity" force opposing your move outside then in.
time dilation ? more of it inside than out? perhaps not. with the strongest gravity effect always on the outer surface of an entity, and not limited in reach by the size of the body. or?

7. ### DaveC426913Valued Senior Member

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No. Once it reaches the wall, in your diagram, there is a precipitous dropoff. The object will literally fall down that near vertical wall - very rapidly trading potential energy for kinetic energy - causing it to go firing off away from sphere.

Let's zoom in and exaggerate the green hilit portion of this graph.

Here it is. I've changed nothing except made the inner slope more visible.

Tell me something. How is this diagram not essentially symmetrical about a vertical axis through point C?
You've got
- a slope on the outside of the shell wall, sloping down from A to C, and
- a slope on the inside (albeit much steeper), sloping down from B to C.
They would behave the same, in opposite directions.

A rock, dropped from A will fall to C i.e. it will fall and land on the ground with a thud, just as we'd expect.
If there were a hole in the ground, the rock would keep going until it hit the bottom, just as we'd expect.

But look at the other side of the curve now. According to your idea - A rock, dropped from B will "fall" from very near the inside of the shell toward C - and land on the inside of the shell with a thud.

If there were a hole in the shell, the rock would just keep going.
i.e it would shoot out into space. You have just invented a rocket launcher - and free energy to-boot!

Last edited: Jun 11, 2020
8. ### DaveC426913Valued Senior Member

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If you're referring to me, I most certainly did not.

Do you know what terminal velocity is? Can you explain any relevance to this topic?

9. ### nebel

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Correction please I meant escape velocity, a very good idea to use here, because you need to go not faster to reach far. I used the aerodynamic term inadvertently, because in my flying days, and dreams now, was always determined not to reach " v terminal" before contacting terra firma, making a perfect landing.

10. ### nebel

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David you are getting carried away from a graph lines becoming tral walls with gravity pulling to the ground below them.
A rock released inside the empty shell will just keep floating, until directed by a push, and the pusher will go off even gentler in the other direction.

Do not forget, the graphs deal with a totally isolated hypothetical sphere, not a vacuum chamber or Gauss cage somewhere on earth, or even orbit.

I had the idea written up, but erased it, because it belongs to the science fiction forum. but:
Yea, a rocket like Musk's great effort, has to accelerate the thing, its inertia mass, but in addition lift the weight against gravity.
The Rocket, inside the zero-gravity cavity would have to accelerate only the inertia mass (no gravity on that horizontal line). If the walls of the shell are made of dark matter, invisible, ephemeral, The vehicle, not a rocket bsw, could, given enough push, exit the shell at better than escape/terminal velocity, and be gone for good, surface gravity or not. and:
since the propellant burn would pollute the pristine vacuum of the cavity, why not have 2 rockets accelerate away from each other, using mechanical energy storage like a spring? 2 launches for the price of one, reuse the spring, musk must love the idea. but:
free energy? dont bet on it!

Last edited: Jun 12, 2020
11. ### DaveC426913Valued Senior Member

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OK, you are clearly way off topic.

I'm going to assume that you have no rebuttal to the explanation of the errors in your diagram.

Thus, I'd say this thread has reached its resolution.

12. ### nebel

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There is no error in origin's diagram in post #2, , or it's application in #664, describing the strength of the gravity fields in solid or empty spheres, following the shell theorem. however,
The limit of credibility for you, on this thread is being reached on this page though, , when you have rocks rolling down diagrams lines , and calling for free energy.
do we not have a humour & jokes forum?

Last edited: Jun 12, 2020
13. ### DaveC426913Valued Senior Member

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You cannot have a vertical - or even near-vertical - slope in a gravitational diagram. Full stop.

The implication of a near-vertical slope in a diagram of gravity is a near-infinite gradient - that's what slope means. The tides would rip any physical substance apart.

For this reason - among many others - your diagram - and your understanding of gravity - is wrong.

And that, by the way, is why this thread is in pseudo-science, where it belongs, rather than any actual science forum.

Last edited: Jun 12, 2020
14. ### HalcRegistered Senior Member

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350
This is a major part of the disconnect. You are using the term 'strength' which has meaning in physics, but not in any way that you are using it. Strength has do with the ability of a material to take stress without failure. Gravity doesn't have strength.
Origin points out that you have to be careful what you're discussing, but in your choice of the term 'strength', you have been completely ambiguous.

You seem to be talking about gravitational force field (Newtons) or better, gravitational acceleration (g-force) field, but using the wrong term for it. Dave is talking about the gravitational potential/energy field, which is a different animal and has a different graph.

This kind of language isn't helping. Dave is being clear what he is talking about. You are not. Use the correct terms and the disconnect goes away.

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15. ### nebel

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right! post #2:

page 13 post # 255

page29#579

16. ### nebel

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My mistake was to invert the small g image, not realizing that imagery is used for the G potential.page 16 # 304;

and then moveing the sloping gradient line to near perpendicular . near "R" to show a greater empty space in the interior of the shell. but
I will adress the OP question considering my point about "more" of a Gravity potential field on the outside.

17. ### HalcRegistered Senior Member

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This post (which I found since you referenced post 664) has many mistakes. Take figure C for instance (the hollow shell). If it is meant to graph gravitational energy/potential, the interior portion is level, but should be down at the lowest level, not up at zero. The discontinuity drawn there doesn't exist, and is what Dave has been trying to point out since he's usually talking about potential.
If it is meant to depict gravitational force or acceleration, then it is inverted. Acceleration is highest at the surface of such an object, but you've got it graphed as a minimum there.

You don't actually say what you're graphing, calling it simply 'gravity' in your text, which is ambiguous.

The picture you posted in post 692 fixes that: It graphs gravitational acceleration (not strength, not potential, and not gradient), and has it at a maximum at or near the surface, not a minimum.
Based on post 693, you seem to be aware of the inversion issue.

More issues:
At one point you say #4 'identifies a g gradient', but it actually (seems to) identify a negative g force, whereas the 'g gradient' would be the first derivative of that curve (the change in g force over distance), which would be a level line for a ball with uniform density.
Can't figure out what fig d is trying to depict. There seems to be a radius to it since the curve (whatever it is) starts going down when it gets to R, but then there's this black hole spike in the middle, which shows a 'deep well' that is somehow approached by a decreasing well, suggesting that gravity force decreases while approaching a black hole, at least until you get very close to it. Ouch... Point 6 is a Nan (not a number), not zero like you have it depicted.

18. ### nebel

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Yes, as I said, my mistake was to invert these graphs. Looking at them upside down will make that clear. They were also never meant to depict gravitational "potential". Post 692 gives examples of what is meant.
" g Gradient" is meant as shown by the green line in post # 255. That gradient of increasing g becomes steeper as the central cavity expands.
The gravitational acceleration gradient is from zero at the center, gradually rising to max at the surface. In an empty shell, it rises from the inside of the wall to the outside surface. Steepness depending on the shell wall thickness.

The inability of mine to put that down comprehensibly for layman, but not to everyone's satisfaction, exacting standards, should not detract from the answers to the posed question and proposition that

There is more gravitational acceleration effecting greater volumes above the surface of entities than their interiors. let that be resolved.

In 2 cases, the singularity of Black Holes, and the inside of empty shells, - there is no interior g at all, all g is in the vast exterior volume reaching to infinity. or?

Last edited: Jun 13, 2020
19. ### DaveC426913Valued Senior Member

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And again, you're back to ambiguous terms.

Sure. Whatever. You're not really saying anything, so it's not even wrong.

20. ### nebel

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"g" is the symbol used for gravitational acceleration, contrasting the the G for gravitational potential. "g" that set the tone since Post #2. expressed as a position on the Y axis. as measured locally by an accelerometer, in it's simplest form a spring scales. with which
A litre of water weighs a kilo on the surface of the Earth, but is weightless at the centre of our globe, weights 250 grams at 2R and zero again at an infinity distance. All that volume above 6400 km altitude filled with a field with only 1/4 yield.
That mass of 1 litre water would be weightless right up to the interior wall in any cavity of any shell, even Earth' size and mass.

My naive mistake was thinking rather only of compressed spring in a weight scale that gives the positive value for g on the y axis as in posts #2, #255 and #579, but that using a pulled spring instrument would give the same values but downward. and showing below the x axis. I did not realize i was entering gravity potential territory. # 575

Novel ideas might come expressed in unconventional terms. like total measurable gravitational acceleration per volume.

Last edited: Jun 14, 2020
21. ### HalcRegistered Senior Member

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350
It is, but it is called g-force, not g-gradient. Gradient describes a totally different thing, and I've seen no graph of it.

G is the symbol for the gravitational constant, which is 6.67E-11 N m2 kg-2, not a unit of gravitational potential (symbol V or V_g) which would be in units of something like meters or joules/kg. The latter is more correct, but translates 1-1 with meters, and the best diagram of potential I've ever seen (the xkcd one) was expressed in meters.
Example: V on Earth has greater magnitude (is more negative) than V on Saturn, despite the fact that I weigh more on Saturn (by only ~8%) and the escape velocity of Saturn is more than thrice that of Earth. So if I somehow managed to set up a siphon hose between the surfaces of those two planets (held stationary to eliminate centripetal effects), the fluid would run from Saturn to Earth, not the other way.

A pulled spring measures the exact same thing (force, or positive weight) as does a compressed spring. Potential cannot be measured by a spring or other device that measures force. In fact, I can think of no device which measures it, or else there would be a published number somewhere as to what the typical gravitational potential is here on actual Earth, not some hypothetical Earth that exists in an otherwise empty universe.

OK, suppose I've measured a system at 100 m-2sec-2
What can you then tell me about that system, and how exactly would I go about getting that measurement? Translation, how is what you just said not just word salad?

Last edited: Jun 14, 2020
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22. ### DaveC426913Valued Senior Member

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OK, so yeah, you don't know what you're talking about.