Do gravity and density have a direct relationship?

In summary: Read moreIn summary, the conversation discusses the relationship between density and gravity, specifically at different distances from the center of an object. It is noted that for surface gravity, there is a direct relationship between density and strength of the gravity field. However, when considering gravity at any given distance from the center, the relationship is not as straightforward and is affected by factors such as the mass and volume of the object. The conversation also explores the idea of a mathematical model to describe this relationship, but it is acknowledged that density is only one factor and other variables, such as distance, also play a role.
  • #1
mintparasol
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Can we say that there is a direct relationship between the density of an object and the strength of it's gravity field? Can we come up with a mathematical model to describe the relationship between gravity and density?

Say we have an object that contains the same amount of matter as the black hole at the centre of the Milky Way. Say that this object takes up a lot of space, 2x the size of the Milky Way for example. This object, tho identical in mass, would have a very different gravity field to the black hole. For starters, this object would exert less gravitational force on another object than the black hole would at the same distance.

Can we say that there may be a logarithmic relationship between the density of a body and the amount of spacetime curvature the body will cause?

I hope these questions don't sound stupid!

Thanks in advance for any replies..
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  • #2
mintparasol said:
Can we say that there is a direct relationship between the density of an object and the strength of it's gravity field? Can we come up with a mathematical model to describe the relationship between gravity and density?
For the surface gravity yes it's easy.
g = GM/r^2 and M = density * 4/3pi * r^3
so g = 4/3 pi G density * r

Say we have an object that contains the same amount of matter as the black hole at the centre of the Milky Way. Say that this object takes up a lot of space, 2x the size of the Milky Way for example. This object, tho identical in mass, would have a very different gravity field to the black hole.
It would have a different surface gravity - because you would be a different distance to the centre of the mass

For starters, this object would exert less gravitational force on another object than the black hole would at the same distance.
No - at the same distance the same mass exerts the same force. It's just that for denser objects you can get closer so the distances can get less and the force larger
 
  • #3
mgb_phys said:
For the surface gravity yes it's easy.
g = GM/r^2 and M = density * 4/3pi * r^3
so g = 4/3 pi G density * rIt would have a different surface gravity - because you would be a different distance to the centre of the mass

Say we go beyond surface gravity. Can we describe a relationship between density and gravity at any given distance from the centres of the objects (within or without the object) outlined in my OP?

mgb_phys said:
No - at the same distance the same mass exerts the same force. It's just that for denser objects you can get closer so the distances can get less and the force larger

Can we come up with any math that describes this increase in force at smaller distances? It seems to me that a mathematical law of this nature should hold equally true for objects/distances at the quantum level as it would for objects/distances at the cosmological level. It also seems to me that the density of the object in question would be crucial to this math because at the quantum level, density and distance are totally interdependent..

I'm just a lay nut tho so I'm probably way off here!
 
  • #4
mintparasol said:
Say we go beyond surface gravity. Can we describe a relationship between density and gravity at any given distance from the centres of the objects (within or without the object) outlined in my OP?



Can we come up with any math that describes this increase in force at smaller distances? It seems to me that a mathematical law of this nature should hold equally true for objects/distances at the quantum level as it would for objects/distances at the cosmological level. It also seems to me that the density of the object in question would be crucial to this math because at the quantum level, density and distance are totally interdependent..

I'm just a lay nut tho so I'm probably way off here!

Um no this would not hold true. Say this object was the earth. And then the same amount of mass but 2x the volume. Now at a distance say the radius of the larger one, surface gravity of the larger one, the force would be the same for either one. But once you go past the surface of the larger one, the force of the larger one would become less and less. But you have not reached Earth's surface yet so the force due to the Earth-object would increase. The reason that it decreases the farther down you go is because the mass above you is pulling you up and therefore cancelling out some of the mass that is pulling you down. There is a point in the center of any symetrically round object, wether it be a sun or moon or planet, the the gravitational force acting on you is zero because everything pulling you one way is counteracted by the mass pulling you in the opposite direction. So density only has to do with how strong the surface gravity is. A black hole is very very VERy high density so you can get very very VERy close to the center making its surface gravity gigantic. that's as far as the extent of density has on gravity that I know of


FoxCommander
 
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  • #5
FoxCommander said:
Um no this would not hold true. Say this object was the earth. And then the same amount of mass but 2x the volume. Now at a distance say the radius of the larger one, surface gravity of the larger one, the force would be the same for either one. But once you go past the surface of the larger one, the force of the larger one would become less and less. But you have not reached Earth's surface yet so the force due to the Earth-object would increase. The reason that it decreases the farther down you go is because the mass above you is pulling you up and therefore cancelling out some of the mass that is pulling you down. There is a point in the center of any symetrically round object, wether it be a sun or moon or planet, the the gravitational force acting on you is zero because everything pulling you one way is counteracted by the mass pulling you in the opposite direction. So density only has to do with how strong the surface gravity is. A black hole is very very VERy high density so you can get very very VERy close to the center making its surface gravity gigantic. that's as far as the extent of density has on gravity that I know of


FoxCommander

Thanks for that!

What I'm getting here then is that gravity depends on mass, density and some sort of 'field' that permeates all of everything. Matter curves spacetime, or does spacetime curve matter?
 
  • #6
I have a question, probably me being stupid in my tiredness, but what is the relation to and object's density to its gravity if mass increases and volume stays the same?
 
  • #7
RandyLowe said:
I have a question, probably me being stupid in my tiredness, but what is the relation to and object's density to its gravity if mass increases and volume stays the same?

If mass increases, but the Volume remains the same, then the intensity of gravity increases in direct proportion to the increase in the Density within the given volume. After all, Density multiplied by Volume yields the quantity of mass:

m = DV

Therefore, if Density doubles, mass also doubles, if Density triples, mass also triples hence, the increase in mass yielded by the increase in Density yields a proportionately stronger gravitational field per the same distance from the center of mass.
 
  • #8
If you are far enough away from any object then you can consider all the gravitational forces as acting through its centre of mass.

But for objects other than spheres made up of shells with spherical symmetry, you may need to Integrate the gravitational affect of all the elemental parts, which is quite a complicated calculation. (There may be some short cuts for particular shapes)

If an irregular object subtends more than a few degrees from the observer then its gravitational effect could probably be distinguished from that of an equivalent point mass. For example, the observer's orbit would probably not be a perfect ellipse any more.
An orbit round a nebula could be pretty different from that around the star that eventually forms from it. (Not that you'd live long enough to get round all the way)
 
  • #9
mintparasol said:
Can we say that there is a direct relationship between the density of an object and the strength of it's gravity field? Can we come up with a mathematical model to describe the relationship between gravity and density?

Say we have an object that contains the same amount of matter as the black hole at the centre of the Milky Way. Say that this object takes up a lot of space, 2x the size of the Milky Way for example. This object, tho identical in mass, would have a very different gravity field to the black hole. For starters, this object would exert less gravitational force on another object than the black hole would at the same distance.

Can we say that there may be a logarithmic relationship between the density of a body and the amount of spacetime curvature the body will cause?

I hope these questions don't sound stupid!

Thanks in advance for any replies..
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this is not true gravity is highest at the center and becomes less as you get further from the center the surface has nothing to do with the gravity if your model was true then the Earth would be hollow.
 
  • #10
FoxCommander said:
Um no this would not hold true. Say this object was the earth. And then the same amount of mass but 2x the volume. Now at a distance say the radius of the larger one, surface gravity of the larger one, the force would be the same for either one. But once you go past the surface of the larger one, the force of the larger one would become less and less. But you have not reached Earth's surface yet so the force due to the Earth-object would increase. The reason that it decreases the farther down you go is because the mass above you is pulling you up and therefore cancelling out some of the mass that is pulling you down. There is a point in the center of any symetrically round object, wether it be a sun or moon or planet, the the gravitational force acting on you is zero because everything pulling you one way is counteracted by the mass pulling you in the opposite direction. So density only has to do with how strong the surface gravity is. A black hole is very very VERy high density so you can get very very VERy close to the center making its surface gravity gigantic. that's as far as the extent of density has on gravity that I know of


FoxCommander

this is not true gravity is highest at the center and becomes less as you get further from the center the surface has nothing to do with the gravity if your model was true then the Earth would be hollow.
 
  • #11
chuxxx said:
this is not true gravity is highest at the center and becomes less as you get further from the center the surface has nothing to do with the gravity if your model was true then the Earth would be hollow.
Nonsense. Refer to the shell theorem to see why.
Here's a link for your convenience:
http://en.wikipedia.org/wiki/Shell_theorem
 
  • #12
chuxxx said:
this is not true gravity is highest at the center and becomes less as you get further from the center the surface has nothing to do with the gravity if your model was true then the Earth would be hollow.
What you said is of course nonsense on both accounts. In units where G = 1, [itex]\Phi = -\int_{\mathbb{R}^{3}}\frac{\rho d^{3}x'}{\left \| x - x' \right \|}[/itex] and for a thin spherical shell of mass, after choosing spherical coordinates and orienting the axes so that the field point is along the z - axis, we get [itex]\Phi = -\frac{M}{2 R^{2}}\int\frac{\delta (R - r')r'^{2}\sin\theta 'dr'd\theta '}{\sqrt{r'^{2} + r^{2} - 2rr'\cos\theta '}}[/itex]. Inside the shell, the result is then [itex]\Phi = -\frac{M}{R}[/itex] hence [itex]\vec{a} = -\triangledown \Phi = 0[/itex] (of course you could arrive at this much faster using the classical cone argument).

Fox was describing a solid sphere of approximately uniform mass density, in which case it is easy to show using a plethora of methods that [itex]F_g\propto r[/itex] within the sphere, where again [itex]r[/itex] is the radial coordinate from the origin centered around the sphere. This is in agreement with what Fox said.
 
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  • #13
chuxxx said:
gravity is highest at the center
Which direction does the force of gravity have at the center?
 
  • #14
He could be confusing gravity with the pressure you would find inside a planet / star centre.
 
  • #15
the equation is wrong r is not at the center of the out side of the shell it is at the center of the shell itself.if not then where is the gravity coming from?
 
  • #16
chuxxx said:
the equation is wrong r is not at the center of the out side of the shell it is at the center of the shell itself.if not then where is the gravity coming from?
Alright then. Solve Poisson's equation for a solid spherical volume of uniform mass density and prove to us that the gravitational field inside is strongest at the center. Also, you didn't answer AT's question: if the field was strongest at the center then obviously it would have some direction. Why should this direction be any more special / preferential than any other direction, given the spherical symmetry about the center?
 
  • #17
mass density is not uniform in a solid planet as mass density increases so will gravity.
 
  • #18
the force of gravity always pulls towards the center of greatest mass AT
 
  • #19
next your going to say well what happens to the matter at the core
 
  • #20
I'm going to assume here that you're actually asking why it isn't so, not telling everybody, starting with Newton, that they're wrong about basic physics.

So, you've obviously skipped over the shell theorem. Here's a quick breakdown.

Yes, you're right in that density varies in actual planets, and that the deeper parts tend to be denser than the surface.
But it's a minor factor when compared to the simple observation that the deeper you go, the less mass there is under your feet pulling you down.
The shell theorem, if you scroll down through the wiki article, states not only that spherical masses can be treated as point sources, but also that inside a sphere of matter, the gravitational field from the outside shells is exactly zero.
So if you go down to half a radius, you've got a 1/8th of the mass pulling you down. But since now you're also so much closer to the centre of the Earth, the gravitational force from that remaining mass is 4 times stronger than it was on the surface.
Overall, the relationship between gravitational force and radius for uniform spheres is linear - half the radius means half the gravity.
To have the density overcome this effect, you'd need it to grow at least linearly as well, which gets harder and harder to imagine as you get closer to the centre. For example, at less than 1/10th of the radius, you'd need the Earth interior made out of lead. At 1/100th, out of something 100 times denser than what you get on the surface, which is beyond the density of any naturally occurring material. It only gets worse the deeper you go.

Now, before you go on to repeat the same thing people are telling you is wrong, hold on for a moment an consider if maybe your understanding is flawed?
Words are nice and all, but at some point you just need to do some math to grok the physics, and studying in-depth the wikipedia article is a good start.
 
  • #21
the answer to that is if you could increase the mass 1,000,000 times thermonuclear reaction.earth would become a sun.but with the mass as it is :now get this: i don't know
 
  • #22
Shell theorem r is not at the center there is nothing at the center of a empty shell
 
  • #23
chuxxx said:
this is not true gravity is highest at the center and becomes less as you get further from the center the surface has nothing to do with the gravity if your model was true then the Earth would be hollow.

You're getting off to a bad start here. You posted something completely wrong on a long-dormant thread.

If you have questions, it is better to ask them than to say wrong things hoping someone will correct you.
 

Related to Do gravity and density have a direct relationship?

1. Does an increase in density lead to an increase in gravitational pull?

Yes, an increase in density does lead to an increase in gravitational pull. This is because the force of gravity is directly proportional to the mass of an object, and density is a measure of the mass per unit volume of a substance.

2. Is there a specific formula to calculate the relationship between gravity and density?

Yes, the formula for calculating the gravitational force between two objects is F = (G * m1 * m2) / d^2, where G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between them. Density can be calculated by dividing the mass of an object by its volume.

3. How does the density of an object affect its weight on different planets?

The density of an object does not directly affect its weight on different planets. Weight is a measure of the force of gravity acting on an object, and it depends on the mass of the object and the strength of the gravitational pull on that planet. However, an object with a higher density will have a greater mass and therefore a greater weight on any planet.

4. Is there a limit to how dense an object can be?

There is no known limit to how dense an object can be. However, for an object to exist in a stable state, there are certain limits to its density. For example, if the density of an object is too high, it may collapse under its own gravitational force.

5. How does the density of Earth affect its gravitational pull on objects?

The density of Earth does not have a significant impact on its gravitational pull on objects. The mass of Earth is the main factor that determines its gravitational pull. However, the distribution of mass within Earth does affect the gravitational pull at different points on its surface, such as the poles and equator.

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