Please Help - Question About Gravitational Pressure

jinydu · 2004-03-21, 04:19

According to Newton's Law of Gravitation, gravitational force is inversely proportional to distance.

But Pressure = Force/Area, and Area is proportional to the square of linear distance.

Therefore, gravitational pressure is inversely proportional to the fourth power of distance.

Did I make a mistake somewhere, or is this right?

jinydu · 2004-03-21, 04:43

Sorry, I don't think I made that question very clear.

I meant, the gravitational pressure at the surface of an imploding star. By distance, I meant the instantaneous radius of the star.

Uncwilly · 2004-03-21, 05:29

Gravity is not a pressure force per se. It is an attractive force. gravity can create pressure, like the 1 atmosphere that we live in, but that is because the mass of the air.

Where gravity should be put in an equation of this sort is:
F=mA
Force = mass * acceleration

Gravity is an acceleration force. The local gravity at the surface of the star will tell you how many N per Kg that you would get.

As the star shrinks the radius to the center of the star shrinks. This reduction in radius is the change in distance that you want to focus on. The area of the star is of little or no import. The local gravitational force does not depend on the surface area (the area is a by product of the radius, not the other way around).

jinydu · 2004-03-21, 05:50

The reason I'm looking for this is to compare the gravitational effects with electron degeneracy effects.

http://scienceworld.wolfram.com/phys...yPressure.html

The link above gives the pressure exerted by electron degeneracy. But how can I compare force with pressure (since they aren't even measured with the same units)?

Thanks

Fusion_power · 2004-03-21, 13:55

How to compare?

Sit down and do the math. You are trying to make a quick and dirty formula that unfortunately is incorrect.

Think of a star as a series of shells like an onion. As you get closer to the center of the star, the gravitational field becomes weaker (presumes no planets to induce a barycenter) but the pressure exerted by the layers above becomes greater. This is the opposite of what occurs above the star's surface. One of your mistakes is in presuming that gravity becomes stronger as you go deeper into the star. In other words, think 3 dimensionally.

Lets define degenerate matter as mass that has lost the nucleus/electron shell structure and has turned into a polyglot of free electrons, protons, and neutrons. How much "pressure" would have to be exerted to achieve this state and how long would it last? Calculate the point at which the gravitational pressure exceeds the electrical force to approximate this state. Remember that gravity is many times weaker than the electro/weak and that I said "approximate"!

Also, Washington State has a decent physics site. Unfortunately, you have to search around to find the interesting stuff.

Fusion

cheesehead · 2004-03-21, 17:55

Quote:

Originally Posted by jinydu

Did I make a mistake somewhere

Yes.

Quote:

According to Newton's Law of Gravitation, gravitational force is inversely proportional to distance.

Nope. It's inversely proportional to the square of the distance. g = G*M1*M2/R^2

Start over. :)

(BTW, in his Principia, Newton also examined other possibilities for gravitational force proportionality to distance. He proved mathematically what shapes the orbits of planets (well, of mutually-attracting bodies, that is) would have to be under these alternative laws. E.g., if gravity were inversely proportional to the first power of distance, the orbits would still be ellipses, but with the Sun at the center of the ellipse, not at one of the foci.)

jinydu · 2004-03-22, 09:12

Quote:

Originally Posted by jinydu

According to Newton's Law of Gravitation, gravitational force is inversely proportional to distance.

Whoa, I really was typing with my brain turned off!

jinydu · 2004-03-22, 09:40

In any case, what I was thinking was that according to the link, electron degeneracy pressure, P, is proportional to density, p, raised to the 5/3 power.

Since p = m/V, and V = r^3, P should be proportional to r^(5/3 * -3). That is, r^(-5).

Gravitational force is proportional to r^(-2). Since Pressure is F/A, I thought that gravitational pressure might be proportional to r^(-2-2), which is r^(-4). Obviously, r^(-5) is stronger than r^(-4) as r approaches 0.

Maybeso · 2004-03-22, 19:21

jinydu,

As strange as it sounds, you were almost right. Although the force of gravity is not linear, gravitational pressure is (an inverse square force applied to a square area). Both atmospheric pressure (compressible) and water pressure (incompressible) are proportional to pGh.

For a star of radius R, P(r) = pG(R-r).

This equation implies that the pressure at the 'surface' is 0. I suspect that the 'surface' that you're considering is a sort of 'sea level', where the pressure P of the gas above causes the solar material to act more like a liquid than a gas.

What is the minimum pressure required to keep solar plasma in a 'liquid' state? Let's call that Po. Then our equation becomes:
P(r) = Po + pG(R - r).

From the way you stated your question, it sounds like you're more interested in the changing behavior of the density p and surface pressure Po as the star collapses. Then we would need an equation for p and Po in terms of temperature, etc.

FeLiNe · 2004-04-05, 20:11

It might be easier to point you to the right direction if you spelled out what exactly you're trying to compute. A stability criterion for white dwarves? An upper mass limit? A rule for the formation of neutron stars? Are you looking for an overview or an in-depth analysis or a detailed example or what?

The usual good starting point for all kinds of astronomical things is Kenneth R. Lang, "Astrophysical Formulae", Springer 1980. In this case, you might be interested in section 3.3.2 and in particular 3.3.2.3. Any astronomy library or general university library should have that. However, Lang only compiles results and references, so if you want to know how any one equation was actually obtained, you'll have to follow the references to the original work.

Working your way through these sections should give you an appreciation of the many things you left unspoken in your question: are we talking about relativistically degenrate gas, for example, and if so are we talking about specially relativistic gas? In a general relativistic solution, the effects of gravity itself would be intricately woven into the fabric of the equation of state of the gas, for example, and any attempt at mixing classical (nonrelativistic) Newtonian gravity in with it is guaranteed to produce meaningless results.

Sorry if this may seem unhelpful, but if you're trying to figure out complex things, you really have to ask more complex questions...

2004-03-21, 04:19	#1
jinydu Dec 2003 Hopefully Near M48 2×3×293 Posts	Please Help - Question About Gravitational Pressure According to Newton's Law of Gravitation, gravitational force is inversely proportional to distance. But Pressure = Force/Area, and Area is proportional to the square of linear distance. Therefore, gravitational pressure is inversely proportional to the fourth power of distance. Did I make a mistake somewhere, or is this right?

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2004-03-21, 04:43	#2
jinydu Dec 2003 Hopefully Near M48 3336₈ Posts	Sorry, I don't think I made that question very clear. I meant, the gravitational pressure at the surface of an imploding star. By distance, I meant the instantaneous radius of the star.

2004-03-21, 05:29	#3
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 23072₈ Posts	Gravity is not a pressure force per se. It is an attractive force. gravity can create pressure, like the 1 atmosphere that we live in, but that is because the mass of the air. Where gravity should be put in an equation of this sort is: F=mA Force = mass * acceleration Gravity is an acceleration force. The local gravity at the surface of the star will tell you how many N per Kg that you would get. As the star shrinks the radius to the center of the star shrinks. This reduction in radius is the change in distance that you want to focus on. The area of the star is of little or no import. The local gravitational force does not depend on the surface area (the area is a by product of the radius, not the other way around).

2004-03-21, 05:50	#4
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	The reason I'm looking for this is to compare the gravitational effects with electron degeneracy effects. http://scienceworld.wolfram.com/phys...yPressure.html The link above gives the pressure exerted by electron degeneracy. But how can I compare force with pressure (since they aren't even measured with the same units)? Thanks

2004-03-21, 13:55	#5
Fusion_power Aug 2003 Snicker, AL 1677₈ Posts	How to compare? Sit down and do the math. You are trying to make a quick and dirty formula that unfortunately is incorrect. Think of a star as a series of shells like an onion. As you get closer to the center of the star, the gravitational field becomes weaker (presumes no planets to induce a barycenter) but the pressure exerted by the layers above becomes greater. This is the opposite of what occurs above the star's surface. One of your mistakes is in presuming that gravity becomes stronger as you go deeper into the star. In other words, think 3 dimensionally. Lets define degenerate matter as mass that has lost the nucleus/electron shell structure and has turned into a polyglot of free electrons, protons, and neutrons. How much "pressure" would have to be exerted to achieve this state and how long would it last? Calculate the point at which the gravitational pressure exceeds the electrical force to approximate this state. Remember that gravity is many times weaker than the electro/weak and that I said "approximate"! Also, Washington State has a decent physics site. Unfortunately, you have to search around to find the interesting stuff. Fusion

2004-03-22, 09:40	#8
jinydu Dec 2003 Hopefully Near M48 2×3×293 Posts	In any case, what I was thinking was that according to the link, electron degeneracy pressure, P, is proportional to density, p, raised to the 5/3 power. Since p = m/V, and V = r^3, P should be proportional to r^(5/3 * -3). That is, r^(-5). Gravitational force is proportional to r^(-2). Since Pressure is F/A, I thought that gravitational pressure might be proportional to r^(-2-2), which is r^(-4). Obviously, r^(-5) is stronger than r^(-4) as r approaches 0.

2004-03-22, 19:21	#9
Maybeso Aug 2002 Portland, OR USA 2×137 Posts	jinydu, As strange as it sounds, you were almost right. Although the force of gravity is not linear, gravitational pressure is (an inverse square force applied to a square area). Both atmospheric pressure (compressible) and water pressure (incompressible) are proportional to pGh. For a star of radius R, P(r) = pG(R-r). This equation implies that the pressure at the 'surface' is 0. I suspect that the 'surface' that you're considering is a sort of 'sea level', where the pressure P of the gas above causes the solar material to act more like a liquid than a gas. What is the minimum pressure required to keep solar plasma in a 'liquid' state? Let's call that Po. Then our equation becomes: P(r) = Po + pG(R - r). From the way you stated your question, it sounds like you're more interested in the changing behavior of the density p and surface pressure Po as the star collapses. Then we would need an equation for p and Po in terms of temperature, etc.

2004-04-05, 20:11	#10
FeLiNe Dec 2003 23 Posts	It might be easier to point you to the right direction if you spelled out what exactly you're trying to compute. A stability criterion for white dwarves? An upper mass limit? A rule for the formation of neutron stars? Are you looking for an overview or an in-depth analysis or a detailed example or what? The usual good starting point for all kinds of astronomical things is Kenneth R. Lang, "Astrophysical Formulae", Springer 1980. In this case, you might be interested in section 3.3.2 and in particular 3.3.2.3. Any astronomy library or general university library should have that. However, Lang only compiles results and references, so if you want to know how any one equation was actually obtained, you'll have to follow the references to the original work. Working your way through these sections should give you an appreciation of the many things you left unspoken in your question: are we talking about relativistically degenrate gas, for example, and if so are we talking about specially relativistic gas? In a general relativistic solution, the effects of gravity itself would be intricately woven into the fabric of the equation of state of the gas, for example, and any attempt at mixing classical (nonrelativistic) Newtonian gravity in with it is guaranteed to produce meaningless results. Sorry if this may seem unhelpful, but if you're trying to figure out complex things, you really have to ask more complex questions...