I didn't say that centrifugal force is the cause of volcanic activity. I said that it might be responsible for lower density magma in the area where the rotational axis of the planet is. Do you read before criticizing or just scan?
If you even knew what the "centrifugal force" was, you would know that you are using it erroneously. The centrifugal force is fictitious. What you are thinking of is the Centripetal Force. I did skim your post, partly because your posts contain little to no real science and consist mostly of opinions and misconceptions.
QUOTE
Further, wouldn't the magma be pulled not just by centrifugal force but also by gravity since the largest portion of the Earth's mass would be in the direction of the surface after a certain depth was reached?
Where do you get your facts? Everything you say is completely made up on the spot! Where are your sources? The centripetal force would not be strong enough to counter gravity, since the Earth isn't spinning fast enough to have a significant centripetal force. All of the atoms in the Earth are most strongly attracted to the center. As you descend into a sphere, there is no gravitational attraction to layers above you, since the whole sphere balances out. At the very center, there is no net pull of gravity.
QUOTE (->
| QUOTE |
| Further, wouldn't the magma be pulled not just by centrifugal force but also by gravity since the largest portion of the Earth's mass would be in the direction of the surface after a certain depth was reached? |
Where do you get your facts? Everything you say is completely made up on the spot! Where are your sources? The centripetal force would not be strong enough to counter gravity, since the Earth isn't spinning fast enough to have a significant centripetal force. All of the atoms in the Earth are most strongly attracted to the center. As you descend into a sphere, there is no gravitational attraction to layers above you, since the whole sphere balances out. At the very center, there is no net pull of gravity.
Mining a tunnel through the axis would be increasingly like mining up through a mountain as more of the Earth's mass was below you instead of above. Plus, since explosives would have the effect of breaking a hole in the wall where magma would leak out into the tunnel, the mining process would require little digging and mostly well-planned explosions. It could be done with deep-sea robots.
AAAAAAND no. As you descend, the combined pull of the rock above you would be balanced out by the pull of the rock below you.
That is your answer, if you wish for clarification, ask. The final answer is that all atoms in the Earth are attracted towards the center.
Maks
8th September 2009 - 07:16 PM
This formula from wikipedia for gravity is not correct formula. You may see Gauss’s law for Electricity.
Y = S *K = q/eps
S = 4piR^2
Because this Gauss’s law is valid for all interaction and has same form , the formula for gravity flux also must have same form.
Y = S * g = m * G - according analogy 1/eps is same as G and K is same as g.
So there is no mistake at all.
According present formula for Earth mass flux is throw Earth surface S = R*R. Which part of Earth is that? About Paris or London, maybe?
v^2 *R = v^2 *R *R/R = R*R *v^2/R = S *g = GM = Y ?
Is this correct?
AlexG
8th September 2009 - 08:03 PM
You are aware that electricity and gravity are two different things, obeying two different sets of laws?
Maks
8th September 2009 - 10:37 PM
Hi AlexG,
If you read my post on 06.09.2009 carefully you will find answer there.
There is no differences between fields in electricity and gravity. For example do you know what is the charge.Of course no. No one knows.
But Gauss' law, as I said before, as mother of all formulas is base for calculations for all interactions , fields, forces, energy, orbital parameters for planets and for particles and so on. This formula is valid also and for nuclear forces too. There is no exception. Only the time is limit to find the an answered questions.
AlexG
9th September 2009 - 02:09 AM
The answer to my question is 'no, I don't know the difference."
Maks
9th September 2009 - 09:37 PM
AlexG
This writing of Gauss’s formulas with S is not as naive as is look like.
For example until present days there is still dilemma about rationalization of the SI units. The problem is 4 pi. At first seems that 4 pi is unnecessary value in formulas. There were suggestions for change physical unit or physical value.
But that is not necessary. There is no need for rationalizations of the units. The value 4pi is not part of m, q, G or eps. This value 4 pi is part of S for particle or planets with spherical shape.
In special cases for example in electricity, field strenght K between two charged plates has form:
Y = S * K = S * V / d = q / eps
S – Plates surface [m2]
d – Distance between them [m]
As you can see in electricity using S is no problem.
But in mechanics of planets it’s big problem. So 4pi must be added in those formulas.
So the real Earth mass, you like or not, is still
M = 7.511333506 exp25 [kg]
AlphaNumeric
10th September 2009 - 07:08 AM
Have you ever actually done the calculus proof of Gauss's theorem? Obviously not because if you had you'd know you're wrong. Gauss's theorem isn't just stated blindly by current physicists, it's a result you learn the proof to when you're an undergrad.
light in the tunnel
11th September 2009 - 01:51 AM
QUOTE (flyingbuttressman+Sep 7 2009, 12:33 AM)
all atoms in the Earth are attracted towards the center.
So if there was a hollow cavity in the center of the Earth, would you float in the center of it or would you stand on the perimeter? Would you weigh the same as you do on the surface?
AlphaNumeric
11th September 2009 - 09:00 AM
http://en.wikipedia.org/wiki/Shell_theoremA shell of material of uniform density and total mass M will have a gravitational field such that outside the shell the field is exactly the same as if the mass is concentrated at a point at the center of the shell. Inside the shell, in the hollow cavity, the gravitational field is such that there is
zero resultant force on a person, they are weightless
anywhere in the shell, not just the center.
O_o
11th September 2009 - 09:34 AM
This is not conducive to large scale structure formation.
light in the tunnel
11th September 2009 - 04:27 PM
QUOTE (AlphaNumeric+Sep 11 2009, 09:00 AM)
http://en.wikipedia.org/wiki/Shell_theoremA shell of material of uniform density and total mass M will have a gravitational field such that outside the shell the field is exactly the same as if the mass is concentrated at a point at the center of the shell. Inside the shell, in the hollow cavity, the gravitational field is such that there is
zero resultant force on a person, they are weightless
anywhere in the shell, not just the center.
Excellent. So how would the gravitational field vary as the cavity was expanded in the direction of the surface? Would it make a difference if the expansion of the cavity was along the rotational axis?
light in the tunnel
11th September 2009 - 04:31 PM
QUOTE (AlphaNumeric+Sep 11 2009, 09:00 AM)
http://en.wikipedia.org/wiki/Shell_theoremA shell of material of uniform density and total mass M will have a gravitational field such that outside the shell the field is exactly the same as if the mass is concentrated at a point at the center of the shell. Inside the shell, in the hollow cavity, the gravitational field is such that there is
zero resultant force on a person, they are weightless
anywhere in the shell, not just the center.
Excellent. So how would the gravitational field vary as the cavity was expanded in the direction of the surface? Would it make a difference if the expansion of the cavity was along the rotational axis?
Also, what would the effect on air-pressure be? Assume there was a ventilation duct from the Earth's surface into a large cavity in the center of the Earth. Would atmospheric pressure force air into the cavity at the center? Or would the air become thinner as gravitation decreases? Maybe it would become more pressurized, the way water is as depth increases.
flyingbuttressman
11th September 2009 - 04:46 PM
QUOTE (light in the tunnel+Sep 11 2009, 12:31 PM)
Excellent. So how would the gravitational field vary as the cavity was expanded in the direction of the surface? Would it make a difference if the expansion of the cavity was along the rotational axis?
The weightless effect works best when we are talking about a sphere hollowed out of the exact center of another sphere. If you make it non-spherical, you would be gravitationally attracted to the thicker walls.
QUOTE
Also, what would the effect on air-pressure be? Assume there was a ventilation duct from the Earth's surface into a large cavity in the center of the Earth. Would atmospheric pressure force air into the cavity at the center? Or would the air become thinner as gravitation decreases? Maybe it would become more pressurized, the way water is as depth increases.
I'm not sure that you're understanding. The "cavity" has to be at the exact center of the sphere, and be spherical in shape for weightlessness to occur. If you're talking about digging a hole to the center of the Earth, then I think you need to write a few equations to figure out the air pressure. It's probably going to be higher than surface-level air pressure.
Air has a gravitational pull. Given that the sphere is sealed, the air would group up somewhere within the sphere and have a decreasing air pressure as you move out from the center of the air "cloud."
Maks
11th September 2009 - 06:53 PM
Continue for post on 09.09.2009
For AlexF and people with open mind
Fiction story:
Image that you live in Newton’s time and you know no nothing about electricity (charge). [
Moderator: Redundant, since Ben Franklin (c. 1752) is responsible largely for the unification of atmospheric and laboratory electrical discharges as a common effect and for the convention of charge as plus and minus, was to a large extent motivated by Newton's principles of physics (c. 1687).]
Same how, you know the proton mass and the first electron traectory velocity and radius, in Hydrogen atom. [
Moderator: In what sense would I know this compound parcel that requires knowledge of hydrogen as an element (1766), the existence of the electron (1897) with a certain charge/mass ratio, the existence of atoms (1905) and their masses (1905-1913), the structure of atoms (1909), the existence of a proton (1919) and the correct relativitic quantum field theory to be able to express that trajectory (1940s). -- Oh, I see, you want to use the semi-classical toy version of the Bohr model (1913), and not something resembling the truth. So you have m_p, r_Bohr = 5.29×10^-11m and a velocity v = αc ≈ c/137]
So you are curios what is the proton gravitational field constant, same as Newton’s gravitational constant for the Earth. [
Moderator: Except noone in Newton's time knew the mass of the Earth, either. It wasn't until 1798 when Cavendish provided the first measurement of the proportionality constant, G. And electricity is not the same as gravity. And the numbers you are using can't be measured because they come from a theory that doesn't work.]
So you take same formula and you will get:
Gp = v^2*R /mp = 1.514172943 exp29 [
Moderator: Missing units, so it's wrong. http://www.google.com/search?q=what+is++5....proton+mass+%3F It doesn't explain positronium or magnetism, so it's wrong. It doesn't explain Deuterium or any of the other elements, so it's wrong. You neglect to explain why one mass is used here.]
So what are the conclusions from this value?
1. That this value is huge comparing Earth gravitational constant G [
Moderator: Yes, but since you are doing this in Newton's time you are taking made-up numbers and comparing them to made up numbers, which makes us think you have no point, just blather. Another fault is not even Newton said gravity was the only force in the universe, so it seems you are chasing a straw-man and couldn't even defeat it.]
2. That the relation between them is:
Gp/G = 2.269 exp39
3. That the larger body has smaller constant. [
Moderator: And yet you assume Newton's third law. FAIL.]
4. That body with larger density has larger constant [
Moderator: You have failed to demonstrate, or even document as "knowledge" that the electron or proton has a radius. FAIL.]
5. That maybe, every body or particle, depending on his volume or density, has his own gravitational field constant [
Moderator: FAIL. Electricity is not gravity as the first person to experiment with like charges will see.]
6. And so on
So what is this?
Is this only poor mathematics calculation or its way for understanding the real concept of the charge, based on Gauss’s formula or law? There is no tree Gauss’s formula, for gravity, electricity and magnetic. There is only one formula, with his own basic postulates. The problem is how to apply them properly.
Maks,
[
Moderator: Suspended 15 days for a pointless hypothetical excursion.]
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