In Example 13.10 (Section 13.6) we saw that inside a planet of uniform density (not a realistic assumption for the earth) the acceleration due to gravity increases uniformly with distance from the center of the planet. That is, g(r) = g_sr/R, where gs is the acceleration due to gravity at the surface, r is the distance from the center of the planet, and R is the radius of the planet. The interior of the planet can be treated approximately as an in-compressible fluid of density ρ.

(a) Replace the height y in Eq. (12.4) with the radial coordinate r and integrate to find the pressure inside a uniform planet as a function of r. Let the pressure at the surface be zero. (This means ignoring the pressure of the planet’s atmosphere.)

Equation 12.4:

dp/dy = – ρg

(b) Using this model, calculate the pressure at the center of the earth. (Use a value of ρ equal to the average density of the earth, calculated from the mass and radius given in Appendix F.)

(c) Geologists estimate the pressure at the center of the earth to be approximately 4 × 10¹¹ Pa. Does this agree with your calculation for the pressure at r = 0? What might account for any differences?

In Example 13.10:

Imagine that we drill a hole through the earth along a diameter and drop a mail pouch down the hole. Derive an expression for the gravitational force F_g on the pouch as a function of its distance from the earth’s center. Assume that the earth’s density is uniform (not a very realistic model; see Fig. 13.9).