Calculate the pressure at the center of the earth. Assume that the interior of t

Question

Calculate the pressure at the center of the earth. Assume that the interior of the earth is an incompressible fluid. The density is constant: ? = M/V. The pressure p(r) depends on the distance r from the center of the earth. The equation for static equilibrium of a self-gravitating fluid sphere is p(r) ?A ? p(r+dr) ?A ? ? dr ?A g(r) = 0, where g(r) = G (? 4? r3/3) /r2. (If you use any other equations, explain them.) (A) Calculate the pressure within the earth p(r) as a function of distance from the center r. (B) Using appropriate values for earth's mass and radius, calculate the pressure at the center. Express the answer in tons per square inch.

Explanation / Answer

Following method will be followed the earth is mostly solid, but the equation works) is the weight of the fluid above a surface divided by the area of the surface. The surface can have any area and, through the magic of algebra, disappears from the equation so that we are left with the product of density (?), gravity (g), and height (h in swimming pools and blood vessels, r in astronomical situations like this). Now for the calculus. You can't assign a value for gravity in this situation. It varies from 9.8 m/s2 on the surface to zero at the center. We can reduce the amount of variation if we examine just a part of this total distance (?r). We can reduce it even more if we examine an even smaller part. And we can reduce the variation to nothing if we examine an infinitesimal part (dr). Now the product of density (?), gravity (g), and height (dr) works again. All we have to do is add up the contributions to the pressure made by the infinite number of infinitesimal parts from the surface of the earth down to its center. The process of adding infinitesimals is called integration. P = ? R ?g(r)dr = ? R 3m Gmr dr = 3Gm2 [ r2 ] R r r r 4pR3 R3 4pR6 2 and here's our equation … P = 3Gm2 ( R2 - r2 ) 8pR6 which reduces to … P0 = 3Gm2 8pR4 at the center where r = 0. Let's do it. For the earth … P0 = 3Gm2 = 3(6.67 × 10-11 Nm2/kg2)(5.98 × 1024 kg)2 8pR4 8p(6.34 × 106 m)4 P0 = 1.7 × 1011 Pa = 170 GPa = 1.7 million atmospheres

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