Gravitational Forces

                As a young child, I often wondered what it would be like to dig a hole from one side of the Earth to the other.  While this task is clearly impossible for many reasons, the concept introduces an interesting modeling problem investigating the effects and interactions of gravitational force.  Newton’s law of universal gravitation states that each mass particle attracts every other particle in the universe.  Furthermore, Newton says that this force varies directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.  Mathematically, we write this as

where G is the gravitational constant, m is the mass of one particle, M is the mass of a second particle, r is the distance between the particles, and r hat tells us this force occurs along the radial axis.  While this equation works for a system of point particles, it needs some improvement for applications involving bodies of continuous mass distribution.  Thus, we can use an integral to sum up the forces due to each infinitesimally small distribution of mass.  This provides

In this variation, ρ represents the mass density of the distribution as a function of its radius.  Moreover, this volume integral allows for us to sum the entire mass of the object of interest given any unique mass distribution.

                Applying this to our problem of dropping an object down a hole spanning the diameter of the Earth, we can determine the behavior of that force over time and thus model the path of the object.  To begin, we will simplify our model by assuming the Earth is perfectly spherical and has a uniform mass distribution ρ.  Furthermore, we will designate the mass of our object the variable m.  Now, due to the symmetry of our simplified Earth, we can effectively apply a version of Gauss’s Law in determining our gravitational force.  In short, this means we only need to account for the volume of Earth’s mass within a radius r, or distance of the dropped object, to the center of our Earth.  Therefore, as our object gets closer to the center of the Earth the volume of mass integrated over will decrease and vice versa.  Upon evaluation of our gravitational force equation, we see

Interestingly, this force vector looks very similar to, and acts the same as, that of a simple harmonic oscillator.  The force for simple harmonic motion is

where k is the spring constant and x is the distance from equilibrium, or in our case the radius from the center of the Earth.  The motion of the object is modeled in the animation below.

        

Given that the object undergoes simple harmonic motion, we can determine the period of the objects flight.  The period of a simple harmonic oscillator is given by 

Substituting in our gravitational equivalence to k,

Finally, plugging in realistic values, we find that the period for the object to reach the other side of the world and return, neglecting air resistance, is approximately 84 minutes.