You're working on the script of a movie whose plot involves a hole drilled straight through Earth's center and out the other side. You're asked to determine what will happen if a person falls into the hole. You find that the gravitational acceleration inside Earth points toward Earth's center, with magnitude given approximately by \(g(r)=g_{0}\left(r / R_{\mathrm{E}}\right),\) where \(g_{0}\) is the surface value, \(r\) is the distance from Earth's center, and \(R_{\mathrm{E}}\) is Earth's radius. What do you report for the person's motion, including equations and values for any relevant parameters?

Based on Newton's second law, we know that the force \( F \) acting on the body is equal to the mass \( m \) of the body multiplied by its acceleration \( a \). In this case, the force is the gravitational force, which is equal to the mass of the body multiplied by the gravitational acceleration \( g \). So, we have the equation \( F = m \cdot a = m \cdot g \). Substituting \( g \) with \( g(r) \) given in the problem, we get \( F = m \cdot g_0 \left( r / R_E \right) \).

The equation from step 1 is a second-order differential equation and represents a simple harmonic motion. Such an equation can be solved using standard methods for linear ordinary differential equations. In this case specifically, we can use the standard solution for simple harmonic motion, which gives us: \( x(t) = A \cos(\omega t + \phi) \) where \( A \) is the amplitude, \( \omega \) is the frequency and \( \phi \) is the phase difference. The period \( T \) of the motion is given by the equation \( T = 2 \pi / \omega \). However, we still need to find \( \omega \). To do so, we will have to look at the relationship between the period \( T \) and gravitational acceleration \( g \). After some manipulation, we can derive the formula \( T = 2 \pi \sqrt{R_E/g_0} \). Then, the angular frequency \( \omega \) (which is the reciprocal of the period) is \( \omega = \sqrt{g_0/R_E} \). And so we obtain \( x(t) = A \cos(\sqrt{g_0/R_E} \cdot t + \phi) \).

The problem states that the body falls into the hole, which suggests the initial condition is at the surface of the Earth, where \( r = R_E \) and thus \( x(0) = R_E = A \cos(\phi) \). Because of the symmetry of the Earth around the center, \( R_E \) will also be the amplitude of the motion (the maximum distance the person will counter from the center of the Earth). Therefore, \( A = R_E \). Furthermore, as the motion starts at maximum displacement \( x(0) = R_E \), this gives \( \cos(\phi) = 1 \) from which \( \phi = 0 \). Therefore, the final equation of motion will be \( x(t) = R_E \cos(\sqrt{g_0/R_E} \cdot t) \).