A tunnel is dug through the center of a perfectly spherical...

A tunnel is dug through the center of a perfectly spherical and airless planet of radius $R$. Using the expression for $g$ derived in Gravitation Near Earth's Surface for a uniform density, show that a particle of mass $m$ dropped in the tunnel will execute simple harmonic motion. Deduce the period of oscillation of $m$ and show that it has the same period as an orbit at the surface.

Key Concepts

Uniform Density Sphere

Assuming a uniform density for the sphere allows for a straightforward calculation of the gravitational field inside the sphere, as mass enclosed within a radius r is proportional to r³. This assumption simplifies the problem by enabling the use of Gauss's law to derive a linear relationship between the gravitational field and the distance from the center.

Gravitational Field Inside a Uniform Sphere

For a uniformly dense sphere, the gravitational field inside increases linearly with distance from the center (g ? r). This arises from applying Gauss's law for gravity or integrating the contributions from spherical shells, leading to a force law that acts as a restoring force if displaced from the center.

Simple Harmonic Motion

The linear variation of the gravitational force inside the sphere (F ? -r) is the hallmark of simple harmonic motion (SHM). In SHM, the restoring force is directly proportional to the displacement from equilibrium, ensuring that a particle dropped in the tunnel will oscillate sinusoidally about the center of the sphere.

Period of Oscillation and Orbital Period Equivalence

The period of oscillation for a particle executing SHM in this context is derived from the linear force constant. Remarkably, this period turns out to be identical to the orbital period of a satellite in a circular orbit just above the planet's surface—a result that connects the properties of gravitational potential inside a uniform sphere with Kepler's laws of planetary motion.

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