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Hi.

What is the pressure at the center of a planet? For example, the pressure at 3 foot depth measured from the earth's surface under rock would be approximately 2.5 pounds per square inch.

It is commonly accepted that the gravity at the planet center is zero, equal mass pulling on either side cancels out in the middle. If an object at the center has no weight, then it cannot push on an adjacent object, so there should be no pressure exerted.

I believe the pressure distribution curve looks like half a cycle of a sine wave, the part above the horizontal line on the graph, with the depth being the horizontal axis and the left side being the surface, peaking somewhere around 1/2 or 2/3 the radius, dropping to zero at the center.

Most accounts claim high pressures and temperatures at the planet center, with the pressure causing the high temperature. I theorize the temperatures are high not because of pressure, but because the planets still have not cooled off after being created.

I am not looking for a numerical answer, just an explanation of how the pressure can be anything but zero in a zero gravity field. An assumption I have made is the planet is basically a pile of gravel held together by it's own gravity, not a prestressed structure similar in construction to a prestressed concrete beam, where there would be stress in the middle regardless of gravity. Most accounts claim planets form from the aggregation of material so assume no pre-stressing elements in the construction of a planet.

thanks,
Answer 1:

The reason why it is said there is "pressure" in the center of the earth is due primarily to force balance. There is no assumption of anything about planetary formation: you get pressure through regular force-balance. Whether or not the core is liquid or solid or whatnot is a problem involving the equation of state and some dynamics. I will give you the punch line now since I am sometimes bad at explaining (I am not an educator like yourself!): it is a balance between weight of the rock sitting on top of you and the contact forces with rock beneath you. These two forces provide a stress. That is where the pressure comes from.

Newtons second law says that if an object is stationary or moving at constant velocity (not accelerating) in an inertial frame, then the net force upon it must be zero. Therefore all the contact forces must balance.

So imagine you're a piece of rock sitting near the center of the earth. The gravity you experience must be pretty close to zero. However, what does provide a force on you is all the rock sitting on top of you. That stuff actually provides a great deal of force. Although gravity drops off as you go deeper in the earth, you still have to support all the rock on top of you that experiences more gravity than you do. Its like in the ocean: even though gravity is dropping off (just a tiny bit) as you go lower, you experience a lot more pressure as you go deeper. The reason why is because you have more and more water sitting on top of you. And even though each piece as you goes deeper experiences less and less gravity, you have to sum up all the pieces of mass on top and how much gravity each of those pieces experiences. Likewise, as you go deeper in the earth, although you experience less gravity, you have more and more rock sitting on top of you that you must add up.

A sort of ridiculous analogy is to imagine that someone is triangle shaped (massive torso and really skinny legs). His torso weighs 180 lbs and his legs weigh 70 lbs. Now imagine that he is standing on top of you for a circus act. Even though as you see that as you move down, each piece of his body weighs less, the fact of the matter is that you still have to carry 250 lbs, which is more than either his legs or his torso alone. This is because (I am assuming since you are a good circus performer) that you guys are not accelerating and you guys are standing still (and not falling down). Otherwise if you were falling, then you don't have to support any of his weight.

Now, in our situation, nothing is falling/accelerating. And in fact there is a simple integral we can do if we want to be qualitative. Here, we imagine that the earth is a sphere, and we parameterize our location by our depth/radius r, and latitude and longitude. Imagine I have rho(r) is my density as a function of radius and we assume that the mass is isotropic in the angular coordinates. then the amount of pressure due to gravity at some radius R is then the following spherical integral with the boundaries being the surface of the earth down to the point R that you are interested in calculating:

integral_(R_surface --> R) rho(r) g(r) r2 dr = P * R21/R2 * [integral_(R_surface --> R) rho(r) g(r) r2 dr] = P

I know this is hard to read, and I'm not sure if you remember a lot from calculus, but in case you do, this is relatively enlightening.

Very roughly, rho(r) is the density of earth as a function of the depth r. g(r) is the amount of gravitational force experienced at that radius (depth) by each piece of rock. And now we integrate over the spherical surface of interest. The left hand side gives you the amount of force due to the entire rock sitting on top of you. (If you are diligent, you will notice I left out the factor of 4pi on both sides. If we assume the problem doesn't depend on the angular location on the earth (the latitude and longitude you sit at) and only depends on the depth you're at (the radius), then the angular portions of the integral don't change anything overall).

The right hand side is basically something that must balance this weight. This is the pressure (as we call it) due to the rock sitting below radius R. Remember! We, as a piece of rock minding our own business at radius R. We see lots of force coming from above. Since we aren't moving, something must be balancing us from below! Essentially what happens is the stress of gravity against the contact force below causes rock to deform, and that deformation pushes back (think of how hard a rock pushes back when you squeeze it! You're applying a lot of force, and it applies a lot of force right back!). This is the pressure, and it is due to the material properties of rock. And again, this is just a statement of force balance! Balancing the weight on top of you with the contact forces from the rock beneath you that keeps you from accelerating in the dire


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