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.

What I've done so far is separate the equation to

p(r) = p(r+dr) + ?drg(r)

but I don't really know what to do with p(r+dr) and dr.

Explanation / Answer

Here is an equation which will work for almost all celestial bodies:

Pressure = (Mass^2 * G) / (Volume * Radius)

Earth Mass is: 5.98E10^24 [kg]

Earth Radius is: 6.37E10^6 [m]

Newton's Constant G is: 6.67E10^-11 [m^3/kg*s^2]

V = (4 * pi * R^3) / 3 = 1.08E10^21 [m^3]

M^2 * G = (5.98E10^24 [kg])^2 * 6.67E10^-11 [m^3/kg*s^2) = 2.39E10^39 [kg*m^3/s^2]

V * R = 1.08E10^21 [m^3] * 6.37E10^6 [m] = 6.90E10^27 [m^4]

Earth Pressure = 2.39E10^39 [kg*m^3/s^2] / 6.90E10^27 [m^4] = 3.47E10^11 [kg/m*s^2]

which is: 347 giga Pascals of pressure at the center of the earth!

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