If you could travel to the center of the Earth (or any planet), would you be weightless there?
[Physics] Would you be weightless at the center of the Earth
earthgeophysicsgravitynewtonian-gravityplanets
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Forget about force. Force is a bit much irrelevant here. The answer to this question lies in energy, thermodynamics, pressure, temperature, chemistry, and stellar physics.
Potential energy and force go hand in hand. The gravitational force at some point inside the Earth is the rate at which gravitational potential energy changes with respect to distance. Force is the gradient of energy. Gravitational potential energy is at it's lowest at the center of the Earth.
This is where thermodynamics comes into play. The principle of minimum total potential energy is a consequence of the second law of thermodynamics. If a system is not in its minimum potential energy state and there's a pathway to that state, the system will try to follow that pathway. A planet with iron and nickel (and other dense elements) equally mixed with lighter elements is not the minimum potential energy condition. To minimize total potential energy, the iron, nickel, and other dense elements should be at the center of a planet, with lighter elements outside the core.
A pathway has to exist to that minimum potential energy state, and this is where pressure, temperature, and chemistry come into play. These are what create the conditions that allow the second law of thermodynamics to differentiate a planet. As a counterexample, uranium is rather dense, but yet uranium is depleted in the Earth's core, slightly depleted in the Earth's mantle, and strongly enhanced in the Earth's crust. Chemistry is important!
Uranium is fairly reactive chemically. It has a strong affinity to combine with other elements. Uranium is a lithophile ("rock-loving") element per the Goldschmidt classification of elements. In fact, uranium is an "incompatible element", which explains the relative abundance of uranium in the Earth's crust.
Nickel, cobalt, manganese, and molybdenum, along with the most extremely rare and precious metals such as gold, iridium, osmium, palladium, platinum, rhenium, rhodium and ruthenium, are rather inert chemically, but they do dissolve readily in molten iron. These (along with iron itself) are the siderophile (iron-loving) elements. In fact, iron is not near as siderophilic as the precious metals. It rusts (making iron is a bit lithophilic) and it readily combines with sulfur (making iron a bit chalcophilic).
This is where pressure and temperature come into play. Pressure and temperature are extremely high inside the Earth. High pressure and high temperature force iron to relinquish its bonds with other compounds. So now we have pure iron and nickel, plus trace amounts of precious metals, and thermodynamics wants very much to have those dense elements settle towards the center. The conditions are now right for that to happen, and that's exactly what happened shortly after the Earth formed.
Finally, there's stellar physics. The Earth would have a tiny little core of rare but dense elements if iron and nickel were as rare as gold and platinum. That's not the case. Iron and nickel are surprisingly abundant elements in the universe. There's a general tendency for heavier elements to be less abundant. Iron (and to a lesser extent, nickel) are two exceptions to this rule; see the graph below. Iron and nickel are where the alpha process in stellar physics stops. Everything heavier than iron requires exotic processes such as the s-process or those that occur in a supernova to create them. Moreover, supernova, particularly type Ia supernovae, are prolific producers of iron. Despite their relatively heavy masses, iron and nickel are quite abundant elements in our aging universe.
(source: virginia.edu)
The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. When you go inside the body the geometry is described by the (less well known) Schwarzschild interior metric.
The exact form of the interior metric depends on how the density of the spherical body changes with depth, so for real objects is has a rather complicated form. However for a body with constant density it simplifies a bit and looks like:
$$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} + r^2 d\Omega^2 \tag{1} $$
In this equation $r$ is the distance from the centre of the object with mass $M$, and $R$ is the radius of the object. So this metric applies for $r \le R$.
If you ignore relativity for a moment and consider just Newtonian gravity, then the radial acceleration of an object near a spherical body is:
$$ a = \frac{d^2r}{dt^2} = -\frac{GM}{r^2} $$
In GR the situation is rather more complicated (which I'm sure comes as no surprise) but we can define an analogous quantity called the four-acceleration. In particular we want the radial component $d^2r/d\tau^2$. Although this looks like the Newtonian acceleration above, it's the second derivative with respect to proper time $\tau$ not the coordinate time $t$.
We get the four acceleration using the geodesic equation:
$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} $$
where the $\Gamma^\mu_{\alpha\beta}$ are called Christoffel symbols (of the second kind) and they depend on the spacetime curvature. This all looks rather complicated, but for spherically symmetric objects all but one of the sixteen terms on the right hand side are zero and the radial four acceleration of a stationary object is simply:
$$ \frac{d^2r}{d\tau^2} = - \Gamma^r_{tt} u^t u^t \tag{2} $$
This is the equation that tells us how the (four) acceleration varies with depth, and we can use it to show that you are weightless at the centre of the object. Equation (2) is going to go to zero if either $u^t$ becomes zero or $\Gamma^r_{tt}$ becomes zero. The quantity $u^t$ is the time component of the four-velocity, which you can think of as the time dilation factor. This is simply $dt/d\tau$ for constant $r$, $\theta$ and $\phi$, and we get it from the metric (1) simply by setting $dr = d\theta = d\phi = 0$ to give:
$$ ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 $$
Since $ds^2 = -d\tau^2$ a simple rearrangement gives:
$$ u^t = \frac{dt}{d\tau} = \frac{1}{\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}} $$
Let's just note that this does not go to zero as $r \rightarrow 0$, and move swiftly on to look at the other term $\Gamma^r_{tt}$. Calculating this involves some painful algebra, but Mathematica is good at this sort of thing and Danu helpfully used Mathematica to do the calculation for me. The result is:
$$\Gamma_{tt}^r= \frac{r}{2R^6}\left[2M^2r^2+MR^3\left(3\sqrt{1-\frac{2Mr^2}{R^3}}\sqrt{1-\frac{2M}{R}}-1\right)\right] $$
Yet another fiendishly complicated expression, but note that the whole thing is multiplied by $r/2R^6$ and that means if $r = 0$ the whole complicated expression is just zero.
And there's our result!
When $r = 0$ the Christoffel symbol $\Gamma_{tt}^r$ is zero and that means the radial four-acceleration is zero and that means you're weightless.
Best Answer
Correct. If you split the earth up into spherical shells, then the gravity from the shells "above" you cancels out, and you only feel the shells "below" you. When you are in the middle there is nothing "below" you.
Refrence from Wikipedia Gauss & Shell Theorem.
{I am using some simplistic terms, but I don't want to break out surface integrals and radial flux equations}
Edit: Although the inside of the shell will have zero gravity classically, it will also have non zero gravity relativistically. At the perfect center the forces may balance out, yielding an unstable solution, meaning that a small perturbation in position will result in forces that exaggerate this perturbation.