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.. ad
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 It would have a different surface gravity - because you would be a different distance to the centre of the mass 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
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 earths 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. thats as far as the extent of density has on gravity that I know of Sincerely, 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?
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.
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)
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.
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
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.
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?
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 occuring 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.