Cosmic rain's gravity

[cosmic rain]

We can't say gravity is produced from matter. We tend to view gravity as something local, when in fact it could be something non-local. For all we know, gravity may be a separate arrow of time opposing matter. Since we don't know the nature of gravity or how gravity works, the equations we use should be treated with a fair amount of objectivity in the analysis. To understand the action of gravity doesn't require any special mathematics (unless you're talking about specific densities of a planet or a star's interior), however a basic understanding of the Law of Inverse Square is required. It also helps to realize that specific gravity and specific density are synonymous. Therein lies the confusion where we may mistake gravity for density, which is only associated with matter and energy. But it is the Law of Inverse Square at work here; the higher the density, the stronger the gravity.

 

The action of gravity moves from concentric space, or surrounding space, to the centrosymmetric position of the physical geometry within that space, i.e. planets and stars. This action holds our atmosphere in place and holds everyone on the planet simultaneously. The specific density of a planet or a star is usually stronger at the center because of pressure and compactification. Exotic stars such as neutron stars are an exception to the rule.

 

One of the most interesting things about gravity is that you will only find it where matter is present. This is another reason we so closely associate the two main forces of nature. But gravity could be holographic, dividing itself every time matter divides. It depends on whether or not there were precision based mechanics at the beginning of time. We know for example that gravity is fine-tuned and that any variance whatsoever would have produced an entirely different universe. This is why many scientists believe there are unknown physics yet to be discovered.

 

The Earth is very wide indeed. A freefall from the surface of the Earth to the core would take over three days. But the calculation is based on samples from above the Earth's surface. This is a region of space which is fairly empty. The rate may increase slightly as you approach the Earth's core, or it may increase exponentially. It all depends on the specific density of the surrounding matter. Splatttt.

 

 

Edited January 25, 2012 by cosmic rain

[ajb]

We can't say gravity is produced from matter.

 

General relativity says almost exactly that. "Stuff" acts as a source of the gravitational field.

 

We tend to view gravity as something local, when in fact it could be something non-local.

 

General relativity is a local theory.

 

Since we don't know the nature of gravity or how gravity works, the equations we use should be treated with a fair amount of objectivity in the analysis. To understand the action of gravity doesn't require any special mathematics (unless you're talking about specific densities of a planet or a star's interior), however a basic understanding of the Law of Inverse Square is required. It also helps to realize that specific gravity and specific density are synonymous. Therein lies the confusion where we may mistake gravity for density, which is only associated with matter and energy. But it is the Law of Inverse Square at work here; the higher the density, the stronger the gravity.

 

General relativity is our best theory of classical gravity. It has been tested to some huge degree of accuracy. There is no reason to think that general relativity is not a good model. Within general relativity there is the Newtonian limit. This limit basically says that as long as the masses are not too large, gravity can be well approximated by Newtonian gravity, hence the inverse square law.

 

One of the most interesting things about gravity is that you will only find it where matter is present.

 

What you mean is that far away from an isolated source the space-time is flat.

 

Anyway, learning a little general relativity would help you lots.

[D H]

The Earth is very wide indeed. A freefall from the surface of the Earth to the core would take over three days.

This (along with your entire post) is utter nonsense. Why do you do that?

 

The correct answer is about 19 minutes, not three days. You'll get about 21 minutes if you assume a constant density. A better assumption is that gravitational acceleration is constant down to about 2890 km below the surface, and then falls linearly to zero. Use the Preliminary Earth Reference Model and you'll get an even better estimate.

Edited January 25, 2012 by D H

[Widdekind]

This (along with your entire post) is utter nonsense. Why do you do that?

 

The correct answer is about 19 minutes, not three days. You'll get about 21 minutes if you assume a constant density. A better assumption is that gravitational acceleration is constant down to about 2890 km below the surface, and then falls linearly to zero. Use the Preliminary Earth Reference Model and you'll get an even better estimate.

 

Following your equations, accurately-if-imprecisely, gravity is approximately constant down to the core (>90% accuracy). Such would imply a local density, decreasing as r^-1. Might other terrestrial-like planets possess similar density profiles ? Could "g = constant above the core" be a generally-if-vaguely applicable principle ?

[Klaynos]

!

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[Tres Juicy]

 

The Earth is very wide indeed. A freefall from the surface of the Earth to the core would take over three days.

 

 

HA! This made me laugh!

[D H]

Following your equations, accurately-if-imprecisely, gravity is approximately constant down to the core (>90% accuracy). Such would imply a local density, decreasing as r^-1.

No, it doesn't. A constant acceleration means the local density is 2/3 the average density, where average density is the total mass inside the sphere out to the radial distance in question divided by the volume of that sphere. You'll end up with a nasty Volterra integral of the second kind.

Never mind. You're correct. A constant acceleration from the core to the surface would require [math]\rho® = \frac 2 3 \bar{\rho}(r_0) \frac{r_0}{r}[/math], where [math]r_0[/math] is the radius of the core and [math]\bar{\rho}(r_0)[/math] is the average density of the core.

 

Whether this is something that is at all generalizable seems rather dubious. Never extrapolate from a sample size of one.

Edited January 25, 2012 by D H

[cosmic rain]

This (along with your entire post) is utter nonsense. Why do you do that?

 

The correct answer is about 19 minutes, not three days. You'll get about 21 minutes if you assume a constant density. A better assumption is that gravitational acceleration is constant down to about 2890 km below the surface, and then falls linearly to zero. Use the Preliminary Earth Reference Model and you'll get an even better estimate.

 

 

Perhaps I should rephrase Earth's core to mean the center of the Earth. Your nineteen minutes sounds like too short of a window to cover such a distance at the rate of freefall. I would rather believe what I saw on T.V.

Edited January 25, 2012 by cosmic rain

[Tres Juicy]

Perhaps I should rephrase Earth's core to mean the center of the Earth. Your nineteen minutes sounds like too short of a window to cover such a distance at the rate of freefall. I would rather believe what I saw on T.V.

 

Surely the best way to learn...

[cosmic rain]

General relativity says almost exactly that. "Stuff" acts as a source of the gravitational field.

 

 

 

General relativity is a local theory.

 

 

 

General relativity is our best theory of classical gravity. It has been tested to some huge degree of accuracy. There is no reason to think that general relativity is not a good model. Within general relativity there is the Newtonian limit. This limit basically says that as long as the masses are not too large, gravity can be well approximated by Newtonian gravity, hence the inverse square law.

 

 

 

What you mean is that far away from an isolated source the space-time is flat.

 

Anyway, learning a little general relativity would help you lots.

 

 

You are the one who brought general relativity into the conversation. I didn't mention it once in my post. If you know so much about gravity then why can't you explain the nature of gravity and how gravity works?

[D H]

Perhaps I should rephrase Earth's core to mean the center of the Earth. Your nineteen minutes sounds like too short of a window to cover such a distance at the rate of freefall. I would rather believe what I saw on T.V.

T.V. ? You mean that silly box where you see crap about Nostradamus and ancient aliens?

 

Let's start with a non-rotating, uniformly dense spherical Earth. Drill a negligibly small tunnel that passes through the Earth. It's a fairly standard freshman physics problem to show that a test mass falling through this tunnel is a simple harmonic oscillator that has exactly the same period as the orbital period of a test mass orbiting just above the surface of the Earth, or 84.3 minutes. Falling from the surface to the center of the Earth takes 1/4 of this period, or just over 21 minutes.

 

The Earth is not a uniformly dense sphere. Density increases markedly with depth, with the most marked change at the core/mantle boundary. The simple uniform density model overestimates the required time. The correct answer is less than 21 minutes (it's about 19 minutes). I haven't the foggiest why someone would have said three hours. That is utter nonsense.

[ajb]

You are the one who brought general relativity into the conversation. I didn't mention it once in my post. If you know so much about gravity then why can't you explain the nature of gravity and how gravity works?

 

 

We can model classical gravity extremely well using general relativity. This is well established.

[zapatos]

Terminal velocity of a man in free fall, head down, is something like 200 mph near the surface of the earth.

 

Radius of the earth is something less than 4000 miles.

 

Why wouldn't it take a man about 20 hours to fall to the center of the earth?

[John Cuthber]

If he has a terminal velocity then he's not in free fall. He's being slowed down by air resistance.

[zapatos]

If he has a terminal velocity then he's not in free fall. He's being slowed down by air resistance.

Ah, thank you. Non-technical use of technical terms seems to get me every time.

[Widdekind]

No, it doesn't. A constant acceleration means the local density is 2/3 the average density, where average density is the total mass inside the sphere out to the radial distance in question divided by the volume of that sphere. You'll end up with a nasty Volterra integral of the second kind.

Never mind. You're correct. A constant acceleration from the core to the surface would require [math]\rho® = \frac 2 3 \bar{\rho}(r_0) \frac{r_0}{r}[/math], where [math]r_0[/math] is the radius of the core and [math]\bar{\rho}(r_0)[/math] is the average density of the core.

 

Whether this is something that is at all generalizable seems rather dubious. Never extrapolate from a sample size of one.

 

Please permit me to follow along:

 

[math]g = -\frac{G M_{<r}}{r^2} \approx -\frac{G M}{R^2} \equiv g_0[/math]

 

[math]G \int 4 \pi r^2 \rho® dr \approx r^2 \frac{G M}{R^2}[/math]

 

[math]G \left( 4 \pi r^2 \rho® \right) \approx 2 r \frac{G M}{R^2}[/math]

 

[math]\rho® \approx \frac{1}{2 \pi r} \frac{4 \pi \bar{\rho} R^3}{3 R^2} = \frac{2}{3} \bar{\rho} \frac{R}{r}[/math]

 

[math]\rho(x) \approx \frac{2}{3} \frac{\bar{\rho}}{x}[/math]

If so, then, from HSE, w.ht.:

 

[math]\frac{dP®}{dr} = - g \rho®[/math]

 

[math]\frac{1}{R}\frac{dP}{dx} \approx -g_0 \frac{2}{3} \frac{\bar{\rho}}{x}[/math]

 

[math]\int dP \approx -\frac{2}{3} \left( \frac{G M \bar{\rho}}{R} \right) \int \frac{dx}{x}[/math]

 

[math]P - P_0 \approx -\frac{2}{3} \left( \frac{G M \bar{\rho}}{R} \right) \ln(x)[/math]

where we have included the surface pressure [math]P_0[/math] in case the hypothetical world has an appreciable atmosphere. Let

 

[math]B \equiv \frac{2}{3} \left( \frac{G M \bar{\rho}}{R} \right)[/math]

then w.h.t.:

 

[math]\frac{\Delta P}{B} \approx -\ln\left(\frac{2}{3} \frac{\bar{\rho}}{\rho}\right)[/math]

 

[math]e^{-\frac{\Delta P}{B}} \approx \frac{2}{3} \frac{\bar{\rho}}{\rho}[/math]

 

[math]\rho \approx \left( \frac{2}{3} \bar{\rho} \right) e^{\frac{\Delta P}{B}} \approx \left( \frac{2}{3} \bar{\rho} \right) \left( 1 + \frac{\Delta P}{B} + \frac{1}{2} \left( \frac{\Delta P}{B} \right)^2 + ...\right)[/math]

Thus, to first order in applied pressure, [math]B[/math] is essentially a "Bulk Modulus", of the planet. For our earth, [math]B \approx 230 GPa = 2.3 Mbar[/math], closely comparable to rock, steel, & diamond, i.e. the constituent materials from which our rocky world is composed.

 

Therefore, this simple model explains, and accurately predicts, the BM, of common rocky materials, i.e. human-measured BM represent the lowest order "linear regime", of density increase, due to applied pressures. To date, human civil engineers have not applied multi-mega-bar pressures, to terrestrial structures, and so have yet to perceive higher-order "corrections". Such higher-order effects would plausibly be prominent, for rocky materials, residing deep within the core of Jupiter, wherein the pressures are predicted to be [math]\ge 3 TPa = 30 Mbar[/math] -- and wherein this simple model presumably is inapplicable, predicting, as it does, exponentially large densities. Still, such suggests, that the matter in the cores of large planets (& BDs ?) is dramatically degenerate.