I apologize if this sounds stupid, but as someone without a lot of physics training I was wondering…Regarding gravitational forces, since it is this that brings the dust and rocks together to form planets— say you were able to place something in a void at the exact center of the Earth. Would it be crushed by the surrounding forces, or float since there is equal force on it from all directions, or both? It would appear to me that gravity is a function of mass, so my guess would be both; And how much would this force be affected by the speed of rotation of the earth?
[Physics] Pressure and gravity at the center of the earth
earthforcesnewtonian-gravityplanetspressure
Related Solutions
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...
Your reasoning demonstrates precisely why formal logic alone is insufficient to study nature. In particular, it lacks the ingredient of inductive inference that is a cornerstone of empirical science.
A cosmic ray striking the Earth is not some random act of the gods that can have any imaginable consequence whatsoever. It is a cosmic ray striking the Earth. Sure, by formal logic we cannot see 100 trillion black ravens safe cosmic rays and conclude anything about the next raven cosmic ray. But everything we know in physics indicates the collisions in the LHC should be similar to the ones in the atmosphere -- otherwise, you'd probably be forced to believe that subatomic particles care about where in the universe they are with respect to human beings and their constructs.
Note that your reasoning could be used to disprove the safety of anything whatsoever. I assure you that never before in the history of the observable universe has user number 40492 ever asked a question about particle collider safety on Stackexchange. Sure, millions of other users have asked questions, and sure billions of things have been done on the internet, and sure no electronic device has ever been known to initiate a supernova and destroy its solar system, but perhaps you asking this very question could be the trigger for such an event. And by the anthropic principle we would never know until we detonated the planet.
Best Answer
The question, and many of the answers, are confusing two forces, pressure and gravity. Pressure is a surface force. By itself it does not cause motion, but if the pressure is not constant, then there will a force in the (negative) direction of the gradient. Gravity is a body force, and will always try to produce acceleration.
Think of the Earth as composed of shells, as suggested. Inside the Earth, if we not at the center, we are inside some of the shells, and these exert no gravitational force on us. But we are outside of other shells, which pull us toward their center, which is always the center of the Earth. There is therefore always an attraction to the center, except at the center itself, and the Earth has a tendency to collapse under its own weight. But it does not collapse, because there is a pressure gradient. If pressure depends only on depth, and increases with depth, there will be an outward force, essentially due to Archimedes principle, that just balances gravity. So the poor guy down there experiences no force tending to move him in any direction, but he will experience a pressure that crushes him with all of the weight of rock above.
ADDITION By pressure, I mean normal stress, such as would be present even if the interior of the Earth were liquid. If we assume a completely rigid material for the interior such that stress produces no strain, then the astronaut would be perfectly comfortable inside his little bubble. In fact all it would take would be a little ball of completely rigid material. However, at these pressures, things are usually liquid.