Where is most "gravity", inside or out?

[nebel]

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.

Yes, and I repeat in retrospect. all my comments were regarding the g force never the potential.

Potential cannot be measured by a spring or other device that measures force. In fact, I can think of no device which measures it,

That is why I emphasize, that even in my mistakenly inverted negative Y diagrams ( springs pulled down rather than compressed ), I intended to express the g force.

OK, suppose I've measured a system at 100 m-2sec-2

You can do better than that.
In post #579 I assumed to measure the g forces sum over shell surfaces at different radii, pretending we are talking of radiation-like gravitons (hopefully mass-less). There might not be a unit for that measurement, but as an example, the total energy emitted by the sun measured at any distance, on any shell it encloses, remain constant (absorption ignored). some with the g force., field.
just because the force or radiation acts over a greater area, and measures proportionally weaker, does not mean the total has diminished with distance. It that is a trivial truth,
then there is more g force outside the surface then the interior, simply because the exterior goes on to infinity. ( the area below the red line).

I think he's just too old to take it in. He's in his nineties, I understand.

If off topic reply is allowed? born 1930, and had a great time skateboarding today, respectfully maintaining the balance of the g force and velocity--- teaming up with my 8 year old daughter. There is no fool like an old fool.

 

[Halc]

A number of us have had a go at teaching nebel the difference between gravitational force (or field) and potential. I think he's just too old to take it in.

OK, I must agree with this now. He's managed to confuse the two yet again.

Yes, and I repeat in retrospect. all my comments were regarding the g force never the potential.
...
I intended to express the g force.

You say that, but then go on and demonstrate otherwise:

In post #579 I assumed to measure the g forces sum over shell surfaces at different radii

The g force is the g-force at different radii, not the sum of it. You asserted (in post 579 that that sum is infinite, which it isn't, and that sum happens to be a measure of (wait for it ....) gravitational potential, not acceleration or g-force. So you've done it again, confusing the two. The g-force is the curve (illustrated by origin in first reply), not the area under the curve, which is potential, and quite finite.

pretending we are talking of radiation-like gravitons (hopefully mass-less).

Gravitons are irrelevant. Earth in isolation emits no gravitons, and gravitational force is not conveyed by them.

there is more g force outside the surface then the interior

Both fall to zero as you approach the middle or a large radius. Both are the same just above and just below the surface. Earth has far more g-force 500 km below the surface than it does 500 km above the surface. It also has more g-force at halfway to the center than it exerts halfway to Mars. So I don't buy your poorly thought out statement.

 

[nebel]

You asserted (in post 579 that that sum is infinite

no, the g force has infinite reach, and that it is the summation of all the all the gravity force measurements stay the same no matter the outside distance, now,
If that is true of the elusive, unmeasurable G potential, then there is also more gravitational potential out there then inside.

Gravitons are irrelevant. Earth in isolation emits no gravitons,

I was not talking about Earth, but the Sun, which emits photons, that follow the square over distance laws as do gravitons, (if they exist)

Earth has far more g-force 500 km below the surface than it does 500 km above the surface. It also has more g-force at halfway to the center than it exerts halfway to Mars. So I don't buy your poorly thought out statement.

Please refer to graph # 255, and the fat green g gradient line, for constant density, which is the relevant one when talking about theories. At 2000 km below or above the surface, the single points g forces measured there are equal. but
the inside force of ~ 4 ms^2 acts over a shell with radius 2300 kms. The same g force on the outside acts equally on all points on a surface of shell with a 10 000 km radius . using the surface of sphere formula, we get a 19 times greater value for the sum of g force on the out side.
the sum of ~ 10 m sec^2 max at the Earth surface x any Radius above the surface will not change at Mars distance, not even to infinity. so:
let it be resolved that there is, in summation, way more g force outside than inside. by far. now
I if your idea, that the red line in #579 correctly describe Gravitational potential, then that is proven for G too.
thank you. buy it please

 

[DaveC426913]

no, the g force has infinite reach, and that it is the summation of all the all the gravity force measurements stay the same no matter the outside distance, now...
let it be resolved that there is, in summation, way more g force outside than inside. by far. n

Since gravity-as-a-force has been shown to be an inaccurate model for over a century now, your hypotheses that are dependent on it are flawed as well.

If you use the correct curved spacetime model of gravity, you will see that such discontinuities do not happen.

 

[nebel]

If you use the correct curved spacetime model of gravity, you will see that such discontinuities do not happen.

what discontinuities please? and
Is there more Gravitational potential inside than outside? or perhaps equal?
Perhaps newton still works in certain cases as kepler does for asteroids?

 

[DaveC426913]

what discontinuities please? and

The cusp, in your graph. Can't happen.

Is there more Gravitational potential inside than outside? or perhaps equal?

Considering "outside" encompasses "the rest of the universe" that's a pretty ambiguous question.

Perhaps newton still works in certain cases as kepler does for asteroids?[/QUOTE]
You trying to apply it to a place where it does not work.

 

[nebel]

The cusp, in your graph. Can't happen.

what graph? posts#2, ? #255? # 575?
those were not my original graphs but by origin, others, from the public domain. so are you denying that the strength of the gravitational force g peaks at the surface R?

Considering "outside" encompasses "the rest of the universe" that's a pretty ambiguous question.

The question was not wether the universe is finite or not, but a comparison of what happens with 'g' above and below the surface.
#2, and #255 show it is at parity at about . 4 R and 1.4 R above the surface that is more on the outside before even considering the surface over which it acts or the reach.

Is there more Gravitational potential inside than outside?

It is Gravitational potential energy bsw. so from infinity to the surface is a gain of ~10 m/s^2. what is the gain from the surface in falling to the center? or if an object could keep falling, how much would it have gained speed after passing the surface, or lost as it exits the other side? that would be your inside-outside G comparison.
the real ambiguity you mentioned is when there is no defined surface. imho. bring on the slide-rules.

 

[nebel]

A number of us have had a go at teaching nebel the difference between gravitational force (or field) and potential.

on the positive side, some of the other people behind the 26 000 views, perhaps posters too, nave been taught.
In hindsight , I should have asked the sub question:

"Is there more energy required to lift a kilo from the center the earth to the surface, or to accelerate it to leave the earth to travel to infinity? " and: " how about from the inside of an empty shell of that mass?"
I had no idea asking a "g" gravity question would turn into an "G" energy query.

 

[DaveC426913]

I had no idea asking a "g" gravity question would turn into an "G" energy query.

Indeed. You have no idea.

There is only one g.
G, on the other hand is a constant - equal to 6.67430(15)×10−11 m^3⋅kg–1⋅s–2

You've already been told that, but you're just trolling for responses.

If I thought you were interested in knowing about gravity, I might stick around - but you're not. Not at all.

You have one less member to troll. Ignoring and Unwatching.

 

[nebel]

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

Yes, I was wrong to use the aeronautical meaning.

In the Gravitational energy potential question, terminal velocity could be useful in a fall into the Earth's center, and the climb back to the surface. The central terminal velocity said to be 7.9 km/sec. to get from a standstill at the surface into free space the escape velocity is 11.19 km/sec. does that mean there is still a greater Gravitational potential to overcome outside than inside?

 

[exchemist]

on the positive side, some of the other people behind the 26 000 views, perhaps posters too, nave been taught.
In hindsight , I should have asked the sub question:

"Is there more energy required to lift a kilo from the center the earth to the surface, or to accelerate it to leave the earth to travel to infinity? " and: " how about from the inside of an empty shell of that mass?"
I had no idea asking a "g" gravity question would turn into an "G" energy query.

The confusion throughout this thread has been that you persist is speaking, vaguely, of "gravity" without specifying whether you mean (i) the gravitational field, i.e. the force exerted on a unit mass at a point in space, or (ii) the gravitational potential, i.e. the work done (=energy input) to lift a unit mass to that point in space, from the point of minimum GPE in the system.

If you fail to make this distinction, you will end up taking ballocks. Which is exactly what has happened to you.

"g" is the acceleration due to gravity. Because of F=ma (a being g in this case), this is proportional to the force, i.e. the strength of the gravitational field at a point.

"G", as Halc has already told you, is NOT the symbol for gravitational potential energy. G, often called in speech "Big G", is Newton's Constant of Gravitation, which appears in his expression for the gravitational force between two masses: F = GmM/r². G is not energy. The symbol for gravitational potential is generally V. Don't try to use G for potential or you will create another source of confusion.

To find gravitational potential energy, V, you need to work out force x distance the force is applied through. So it is going to be V= ∫Fdr, which, applied to the formula I have just quoted for force gives you V= - GmM/r, but since potential is the energy per unit mass, we get V = -GM/r.

So force varies as the inverse square of distance whereas potential varies as the simple inverse of distance.

 

[nebel]

The symbol for gravitational potential is generally V. Don't try to use G for potential or you will create another source of confusion.---
So force varies as the inverse square of distance whereas potential varies as the simple inverse of distance

Did you mean to write "U" instead of V ? and, would not "simple inverse" be linear? as compared to the curving graph lines? any way,

do you have an answer to the inside/ outside escape velocities as a marker for
"more gravitational potential energy inside or out?" post# 710 ?

 

[exchemist]

Did you mean to write "U" instead of V ? and, would not "simple inverse" be linear? as compared to the curving graph lines? any way,

do you have an answer to the inside/ outside escape velocities as a marker for
"more gravitational potential energy inside or out?" post# 710 ?

No I meant V.

V is also what Halc said. But use U if you insist, it does not matter that much. Just DON'T use G.

Inverse is not the same as linear. Linear would mean proportional to r. Inverse means proportional to 1/r.

This is the trouble. If you can get confused over things like this, it gets too hard trying to explain more complicated things to you. Re escape velocities I'm not getting enmeshed in that, because I'm sure you will get in a muddle. But I will refer you to the Mexican hat drawing of gravitational potential I showed you months ago and which appears in the Wiki article on the subject: https://en.wikipedia.org/wiki/Gravitational_potential



The slope of this surface, at any point, tells you the strength of the force of gravity at that point. At the bottom of the well- i.e. at the centre of the body exerting the gravitational force - the curve is flat. So there is no force of gravity there. But you can't escape from there without climbing up the sides and as soon as you start to climb, you experience a force pulling you towards the centre.

This is why you need a clear understanding of the difference between force and potential in order not to talk ballocks.

 

[Halc]

Did you mean to write "U" instead of V ?

V is gravitational potential (a scalar), as per my post 698. Units can be expressed as meters or joules/kg
U is potential energy, (another scalar) units in say Joules. Energy and potential are different units and thus different things, and are not interchangeable, despite some websites using them incorrectly.

Earth by itself does not have potential energy. If there was, one could express what the potential energy is at the surface of Earth, but there is no such figure. Gravitational PE (U) is a relation, not a property. Gravitational potential of Earth is a scalar property of locations at various distances from the center, and is (at the surface, normalized to not-on-Earth) under -6400 km or V=-6.3e7 joules/kg. I, personally, while at home, have gravitational potential energy of U = -5.6e9 joules. I have that energy, not Earth.

Similarly, the gravitational field is expressed as a property of a location in space. Symbol: g (acceleration, units m/sec2, a vector).
Then there is gravitational force F (newtons, vector), which is a relation, not a property, so again, there is no figure for the gravitational force at Earth's surface.
The term g-force is deceptive because it refers to acceleration, not force, but it a relation with your immediate environment, not with say Earth. So the guys on the ISS experience zero g-force despite gravitational acceleration being about .95g downward there.

do you have an answer to the inside/ outside escape velocities as a marker for
"more gravitational potential energy inside or out?" post# 710 ?

There is less gravitational potential in the middle of an object than there is at the surface, as it must be since it requires work to raise an object from a hole in the ground.

 

[nebel]

At the bottom of the well- i.e. at the centre of the body exerting the gravitational force - the curve is flat. So there is no force of gravity there. But you can't escape from there without climbing up the sides and as soon as you start to climb, you experience a force pulling you towards the centre.

Thank you, and I understood that well, not just showing the force, (through slope) but the potential through depth.
In an shell with empty cavity, the flat bottom would become a larger horizontal area/line.
The surface, will stands out bette, clearer in the ' g force/acceleration' ' graphs like #2 and #574.though, and that what i was talking about all along. --- , and the OP asks an above and below surface' question.

But you can't escape from there without climbing up the sides and as soon as you start to climb, you experience a force pulling you towards the centre.

True, if you are part of the graph. but
if in reality you climb along the sides of the inside of a shell, you will not experience a force pulling you to the center, because in the graph, you would be on the floor. Only when you penetrate the wall or rise toward the surface does that start to happen.
]

 

[river]

No I meant V.

V is also what Halc said. But use U if you insist, it does not matter that much. Just DON'T use G.

Inverse is not the same as linear. Linear would mean proportional to r. Inverse means proportional to 1/r.

This is the trouble. If you can get confused over things like this, it gets too hard trying to explain more complicated things to you. Re escape velocities I'm not getting enmeshed in that, because I'm sure you will get in a muddle. But I will refer you to the Mexican hat drawing of gravitational potential I showed you months ago and which appears in the Wiki article on the subject: https://en.wikipedia.org/wiki/Gravitational_potential

View attachment 3473

The slope of this surface, at any point, tells you the strength of the force of gravity at that point. At the bottom of the well- i.e. at the centre of the body exerting the gravitational force - the curve is flat. So there is no force of gravity there. But you can't escape from there without climbing up the sides and as soon as you start to climb, you experience a force pulling you towards the centre.

This is why you need a clear understanding of the difference between force and potential in order not to talk ballocks.

But you could escape by going towards the center of the body exerting the gravitational force .

 

[nebel]

There is less gravitational potential in the middle of an object than there is at the surface, as it must be since it requires work to raise an object from a hole in the ground.

right, that work must be related to the gravitational potential, the depth of the hole. For a deep hole in the ground, to the center of the Earth, the escape velocity is ~8 km/sec. quick work in one shove. . getting the object to arrive with zero velocity at the surface to be caught, before it falls back.
Then, to lob it into far space for good, another acceleration to ~11.2 km/sec has t be imparted. work again, but more. so,
do these 2 different values not show that more U-g has to be overcome outside than inside?

 

[Halc]

For a deep hole in the ground, to the center of the Earth, the escape velocity is ~8 km/sec. quick work in one shove. . getting the object to arrive with zero velocity at the surface to be caught, before it falls back.
Then, to lob it into far space for good, another acceleration to ~11.2 km/sec has t be imparted. work again, but more. so,
do these 2 different values not show that more U-g has to be overcome outside than inside?

U-g makes no sense since you're subtracting things of different units. But yes, there is less difference in potential between the center and the surface than there is from the surface to <infinity>, as is evident by the fact that it takes a smaller shove to do one deed compared to the other.

Out of curiosity, where did you get 7.9 km/sec figure? I had a helluva time trying to google that.
If Earth had a different density curve, the interior potential difference may actually be greater than the surface potential difference. It may be true of some of the outer planets.

 

[river]

U-g makes no sense since you're subtracting things of different units. But yes, there is less difference in potential between the center and the surface than there is from the surface to <infinity>, as is evident by the fact that it takes a smaller shove to do one deed compared to the other.

Out of curiosity, where did you get 7.9 km/sec figure? I had a helluva time trying to google that.
If Earth had a different density curve, the interior potential difference may actually be greater than the surface potential difference. It may be true of some of the outer planets.

Earth has a different density , at certain depths , Geologically , Does it not ?

 

[nebel]

But you could escape by going towards the center of the body exerting the gravitational force .

Yes, if you are an introvert, or need to isolate, , escape from the world that is advancing into an unlimited future. Most people like more space ten less. unless your are inside a hollow shell, it can really get crowded with the force of the gravitationalacceleration pressing matter in from all sides.