“JOURNEY TO THE CENTER OF THE EARTH”
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).
IDENTIFY and SET UP:
From the discussion immediately above, the value of F_g at a distance r from the earth’s center is determined by only the mass M within a spherical region of radius r (Fig. 13.25). Hence Fg is the same as if all the mass within radius r were concentrated at the center of the earth. The mass of a uniform sphere is proportional to the volume of the sphere, which is \frac{4}{3} \pi r^{3} for a sphere of arbitrary radius r and \frac{4}{3} \pi R_{\mathrm{E}}^{3} for the entire earth.
EXECUTE:
The ratio of the mass M of the sphere of radius r to the mass m_E of the earth is
\frac{M}{m_{\mathrm{E}}}=\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R_{\mathrm{E}}^{3}}=\frac{r^{3}}{R_{\mathrm{E}}^{3}} \quad \text { so } \quad M=m_{\mathrm{E}} \frac{r^{3}}{R_{\mathrm{E}}^{3}}The magnitude of the gravitational force on m is then
F_{\mathrm{g}}=\frac{G M m}{r^{2}}=\frac{G m}{r^{2}}\left(m_{\mathrm{E}} \frac{r^{3}}{R_{\mathrm{E}}^{3}}\right)=\frac{G m_{\mathrm{E}} m}{R_{\mathrm{E}}^{3}} r
EVALUATE: Inside this uniform-density sphere, F_g is directly proportional to the distance r from the center, rather than to 1/r^2 as it is outside the sphere. At the surface r = R_E, we have F_{\mathrm{g}}=G m_{\mathrm{E}} m / R_{\mathrm{E}}^{2}, as we should. In the next chapter we’ll learn how to compute the time it would take for the mail pouch to emerge on the other side of the earth.
