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so I had my first approximation to the gravity equation

\begin{equation} F=GmM/r^2 \end{equation} and some questions arose that my teacher couldnt respond:
if r approximates to 0 with mM being constant, then the force should tend to infinity, right? so that being true, then if we would travel to the center of the earth, we would be crushed by gravity? if so, how can there be anything right in the center of the earth? wouldnt anything there be compressed and destroyed by the infinite force? having that in mind, my next question is as gravity force is measured from the center of an object, any part of the same object which is not at the center would be affected by the same object's gravity? if so, how can we differentiate between two objects that are not in a vacuum? my question originated thinking in us standing on the earth, in contact with it, are we, in a gravitational sense, one object? this lead to the last question, if gravity is a property of matter, then how applying the same logic as before, how does it work between two neutrons?(chose neutrons because i assumed it would be more difficult obtaining gravitational force if there was electrostatic forces from charges inbetween) do two newtrons in contact count as 1 gravitational object? and how thats its own gravity affect the newtron from center and out?

sorry for the big text.

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$GmM$ is not constant along the path to the center of the Earth. $M$ starts to decrease once you are underground, and becomes zero right at the center of the Earth. – leongz Jun 19 '13 at 5:35
possible duplicate: – leongz Jun 19 '13 at 5:47
Please change the title to something like "Gravity at the centre of a massive body" instead because this title seems to suggest that you are talking about two point particles being infinitesimally far from each other. – centralcharge Jun 19 '13 at 6:28
Also use better punctuation (capitalisation, spaciing) and paragraph it out because many people would just runaway seeing the gigantic block of text. – centralcharge Jun 19 '13 at 6:29
up vote 6 down vote accepted

In your question, I see 3 different context, where considering gravitational forces :

a) 2 point-like objects

b) 1 point-like object and one extended spherical symmetric object (not too dense)

c) A auto-gravitating extended spherical symmetric object (not too dense)

a) If you take 2 point-like objects, and take the limit $r \rightarrow 0$, in fact, at some value of $r >0$, you create a black hole, because the ratio $\frac{Energy}{Radius}$ cannot excess a constant value $\sim \frac{1}{G}$ (in $c=1$ units). Note that mass is a kind of energy. So you do not have a problem with $r=0$, because you create a black hole before.

b) If you consider a problem of a point-like object and a extended spherical symmetric object like Earth (not too dense), a theorem states that a object at distance $r$ only feels the gravitational force of masses inside the sphere of radius $r$.

That is, for instance, if the point-like object is inside the earth at radius $r < R_{Earth}$, it feels only the gravitational force of masses inside the sphere of radius $r$.

If we suppose a constant density $$\rho = \frac{M_{Earth}}{4/3 \pi R_{Earth}^3}$$, then the force will be $$ F(r) = \frac {G m M(r)}{r^2} = \frac {G m (\rho ~4/3 \pi r^3)}{r^2}$$

So, you have a linear force : $$F(r) \sim r$$

So, when $r\rightarrow 0$, nothing bad, about gravitation, appears. (of course, temperature and pression increase very much...)

If the spherical object is very dense, it is an other story, because you have a black hole, and you may have a "singularity": it is thought that something very bad happens to objects reaching the singularity (tidal forces, roasted, etc..). But you are here in the context of general relativity.

c) The last problem is an auto-gravitating extended spherical symmetric object. I will just give this reference Of corse, as usual, if the object is too dense, you need general relativity, black holes, etc...

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Thanks! very clear and well organized, but i still have the doubt on how to differentaite between two or more objects in a gravitational sense, for example, the earth, the atmosphere and a human standing on the surface. The other doubt i have, is are ALL objects self-gravitating? because as you answer point C) it appears to me that its an special case, not the rule. – Manuel Herrera Jun 19 '13 at 18:37
Everything which has energy/momentum is a source for gravitation. So earth, atmosphere, human beings have energy/momentum, and so are a source for gravitation. For extended objects, each part of the extended object has energy/momentum, so each part is a source for gravitation, so you have a gravitationnal force between 2 different parts, and you have a global resultant gravitational energy for the whole extended object. A point-like object is in fact a idealization (because with a non-null mass means an infinite density, which is not realist) – Trimok Jun 19 '13 at 18:44
I just want to point out that (b) isn't a problem, as one might be tempted to think. Reason being, you can easily establish that a black hole won't be created unless you have a minimum of two Planck mass particles, about one Planck length away from each other. In other words, these kind of distances/energies/masses are necessarily in the domain of quantum gravity by definition. – Alan Rominger Jun 19 '13 at 19:01
Thanks for the clarification. Now your last comment rose a new question, if everything which has energy/momentum is a source of gravitation then what is the behavior of gravity in atoms and subatomic particles? I assume here we step into quantum gravity. – Manuel Herrera Jun 19 '13 at 19:25
See for instance this ref, you will see that the gravitational force is neglectible comparing to other forces at short distances. – Trimok Jun 19 '13 at 19:29

Actually, no. Once you are inside the earth, the forces of gravity are acting from different directions. A picture explains this well: enter image description here Consider the hypothetical scenario where the entire earth is a shell. Then, what would be the gravitational force inside this shell at a position $\vec x$? Clearly, it is: $${\vec F_{resultant}} = \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\nolimits_{} {G\frac{m}{{{{\left\| {\vec r - \vec x} \right\|}^2}}}{\text{d}}M} = Gm\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\nolimits_{} {\frac{{{\text{d}}M}}{{{{\left\| {\vec r - \vec x} \right\|}^2}}}} $$ It can be shown that this amounts to exactly 0. Now, this is for a spherical shell. For the earth, it is not exactly 0 (since this integral doesn't even hold) If you are exactly at the centre of the earth, then clearly, there is an equal force from each direction and thus the net force is exactly 0.

So to answer your question, no, the force is actually 0.

Another interesting fact is that if you are buried in the Earth, then you can divide the mass above you into shells, so that means that those masses actually cancel out, and the only mass that does have a gravitational influence on you is the mass below you. More clearly, the light brown part can be divided into small shells of infinitesimally small width, so that effectively, only the part below you actually makes a gravitational influence on you:

enter image description here

This is Newton's Shell theorem..

Edit: (answer to new (or old (or whatever)) questions)

So, for a system of two objects: enter image description here

For the yellow person, clearly, he feels gravity in the vertical direction only (since the horizontal components from the red and black stars cancel out, assuming they have the same mass). As for the blue person, he is clearly not feeling any gravitational force. $$\vec F_{\operatorname{Resultant For The Blue}} =\vec F_{\operatorname{On Blue From Red}}+\vec F_{\operatorname{On Blue from Black}}=0$$

$$\vec F_{\operatorname{Resultant For The Yellow}}=\vec F_{\operatorname{On Yellow From Red}}+\vec F_{\operatorname{On Yellow from Black}}=2\left(\vec F_{\operatorname{On Yellow From Red/Black}}\cdot\hat\jmath\right)\hat\jmath$$

Now, consider an orange horse/donkey inside the black/ (or red, but let us just say, black, for now, since that doesn't really make any difference). Now, what would be the net force on him? If the objects were shells, clearly the only force acting on him is from the red body. But if they are not shells, then clearly, the forces acting on him are from the mass of the black object below him and that from the red object: enter image description here

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"A picture explains this well" ... I'm just glad you didn't also use MS Paint for the equations. – RedGrittyBrick Jun 19 '13 at 9:18
@RedGrittyBrick: Maybe that would have been a better idea given that the closed loop surface integrals don't look good in LaTeX :) – centralcharge Jun 19 '13 at 9:33
thankyou, that answers the first set of questions, but not all. how do we differentiate betwee 2 objects not in a vacuum? because as you said, going inside the earth would decrease M and gravity acts in different directions, that means that gravity force is being exerted by the portions of earth we have 'passed' on our way to the center. Then is it correct to say gravity acts on the object itself(earth in this example)? – Manuel Herrera Jun 19 '13 at 16:29
@ManuelHerrera: Done. I ha've editted my answer. – centralcharge Jun 20 '13 at 4:28
It's a reference to the hippie movement, predominant in the late 1960s United States, and the psychedelic art that was part of it. :) – Michael Brown Aug 17 '13 at 13:57

I don't know what are your philosophical goals with the question , but as long as you "enter" the object (distance <= 0?), the object isn't a single one anymore. In fact, you standing on its center, each piece of earth influence would be canceled by the sum of all other pieces.

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maybe im wrong here, but as I understand distance between objects in gravity, as in electrostatics, is measured from the centers of the objects. of course if we are working with single dots in space this problem disappears. – Manuel Herrera Jun 19 '13 at 5:17

What happens to the force of gravity as the separation between two masses approaches 0, is that it increases. Although it can attain a large value, in practice, it never goes to infinity. This is due to the fact that the masses, even if very small, have a radius. Therefore, the closest they can be is r1 + r2, were r1 & r2 are the radii of the masses m1 & m2. In the case of 2 neutrons, it would be the diameter of a neutron, since the radius are the same. At this distance, the force would be at its maximum since the "centers" can't get any closer.

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Our subatomic particles don't really have a radius as you have described it. A neutron is not a solid sphere. It appears that way due to the nuclear forces taking effect at a certain radius from the neutron, giving it a repulsive force that appears as a barrier. However the gravitational force could overcome this repulsive force, and collapse the two objects into a black hole. – Nuclear_Wizard Apr 3 '14 at 4:15

There's an implicit question here, which is when does a conglomerate of particles count as a single object for the purposes of gravitation? The answer, in theory, is never.

The earth isn't a single object; it's a conglomerate of many particles. When you're pulled by the earth's gravity, you're really being pulled by each particle individually. If you were standing one meter away from the center of the earth, the entire earth wouldn't pull you towards its own center with a humongous gravitational force. Instead, each of the earth's individual particles pulls you toward it with a tiny force. It turns out that these tiny forces mostly cancel out (since they're going in all directions approximately equally), so you're left with a very small overall force.

(Of course, the temperature and pressure near the earth's center are both immense, so you'd be crushed and burned.)

This is also how the earth's gravitational attraction to itself works. Every particle in the earth is attracted to every other particle in the earth.

Now, it happens to be true that if you're far away from an object, then the gravitational force is approximately the same as if the entire object were concentrated at a single point. As far as I know, this approximation doesn't hold true when you're close to an object.

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protected by Qmechanic Aug 12 '14 at 18:23

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