AP Physics C Mechanics: Gravitational Field Inside a Sphere

Understanding gravity inside a planet isn't just a theoretical curiosity—it's the key to predicting the motion of objects in mines, designing deep-space missions, and grasping the fundamental structure of celestial bodies. For the AP Physics C student, mastering this topic solidifies your command of Newtonian gravitation, calculus, and oscillatory motion, providing a powerful tool for solving complex, non-uniform force problems.

The Foundational Tool: Newton's Shell Theorem

Any analysis of gravity inside an object begins with Newton's Shell Theorem. This theorem, which you must be able to state and apply, has two critical parts:

1. A uniform, spherical shell of mass attracts an external point mass as if all the shell's mass were concentrated at its center.
2. The gravitational field inside a uniform spherical shell is exactly zero.

The second part is the crucial insight for our problem. It means that for any point inside a solid sphere, only the mass that lies at a radius smaller than your position contributes to the net gravitational force. All the mass in the spherical shells outside your radius produces no net gravitational effect at your location. This theorem transforms an immensely complicated triple integral into a simple proportional relationship.

Deriving the Gravitational Field Inside a Uniform Solid Sphere

Consider a uniform solid sphere of total mass $M$ and radius $R$. Its uniform density $\rho$ is constant: $\rho =M/(\frac{4}{3}\pi R^3)$.

Now, place yourself at a point a distance $r$ from the center, where $r<R$. By the shell theorem, only the mass interior to your position, $M_{enc}$, pulls you inward. This enclosed mass is simply the density times the volume of a sphere of radius $r$:

$$M_{enc}=\rho \cdot V_{sphere}(r)=\rho \cdot \left(\frac{4}{3}\pi r^3\right)$$

Newton's Law of Gravitation states that the force on a test mass $m$ at this point is $F=G\frac{m{M}_{enc}}{r^2}$. The gravitational field strength $g$ is force per unit mass ($F/m$). Therefore,

$$g=\frac{G{M}_{enc}}{r^2}=\frac{G}{r^2}\left[\rho \cdot \frac{4}{3}\pi r^3\right]$$

Simplifying, we arrive at the core result:

$$g=\frac{4}{3}\pi G\rho r$$

This linear equation, $g\propto r$, reveals a profound fact: inside a uniform sphere, gravitational field strength increases linearly from zero at the center to a maximum of $g_{surface}=GM/R^2$ at the surface. The force is a restoring force, always directed toward the center and proportional to the displacement from it.

Application: Simple Harmonic Motion in a Planet-Wide Tunnel

A classic application of this result is analyzing the motion of an object dropped into a frictionless tunnel drilled straight through a uniform Earth (or any planet). This is not just a fun thought experiment—it demonstrates how a non-constant force can produce perfectly simple harmonic motion (SHM).

Let the tunnel pass through the center. At any point in the tunnel, a distance $r$ from the center, the net force on an object of mass $m$ is given by the field we just derived: $F=m\cdot g=m\cdot (\frac{4}{3}\pi G\rho r)$. Since the force is opposite the displacement (e.g., if the object is to the right of center, force points left), we write Hooke's Law form:

$$F_{net}=-\left(\frac{4}{3}\pi G\rho m\right)r$$

This is the defining condition for SHM: $F=-kx$, where the effective spring constant is $k=\frac{4}{3}\pi G\rho m$. The angular frequency $\omega$ for any SHM system is $\omega =\sqrt{k/m}$. Substituting our $k$:

$$\omega =\sqrt{\frac{\frac{4}{3}\pi G\rho m}{m}}=\sqrt{\frac{4}{3}\pi G\rho }$$

Crucially, the mass of the object $m$ cancels out, meaning all objects take the same time to complete an oscillation, regardless of mass. The period $T$ of the oscillation is:

$$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{3}{4\pi G\rho }}=\sqrt{\frac{3\pi }{G\rho }}$$

For a uniform Earth with an average density of approximately 5500 kg/m³, this period is about 84 minutes. This means a trip from one side of the Earth to the other would take roughly 42 minutes. In reality, Earth's density is not uniform (it increases toward the core), which would modify the period and the motion, but the uniform model provides an excellent foundational understanding and a very close approximation.

Common Pitfalls

1. Misapplying the Shell Theorem: The most common error is to use the sphere's total mass $M$ in the formula $g=GM/r^2$ for an interior point. Remember, for $r<R$, you must first calculate the enclosed mass $M_{enc}$ using the density and interior volume. The shell theorem explicitly tells you to ignore all mass at radii greater than $r$.

1. Incorrect Force Direction in the Tunnel Problem: When setting up the SHM equation, students sometimes forget the negative sign indicating a restoring force. The gravitational force $F=(4/3)\pi G\rho mr$ is magnitude only. You must manually apply the direction: acceleration is always toward the center ($-\hat{r}$), so $a=-(4/3)\pi G\rho r$. Omitting this sign yields an exponential, not oscillatory, solution.

1. Assuming Constant Gravity: A quick, intuitive guess might be that gravity is constant inside the Earth. The derivation shows this is false for a uniform sphere—gravity decreases linearly to zero as you approach the center. This linear relationship is the direct cause of the SHM.

1. Confusing Field Strength with Potential: The gravitational field $g$ is a vector (force/unit mass), while potential is a scalar (energy/unit mass). Inside the sphere, the field magnitude is proportional to $r$, but the potential is proportional to $r^2$ (a parabolic well), which is consistent with the SHM potential energy function $U=\frac{1}{2}k{r}^2$.

Summary

• Newton's Shell Theorem is essential: inside a uniform spherical shell, the gravitational field is zero. Inside a solid sphere, only the mass interior to your radius contributes.
• For a uniform solid sphere of density $\rho$, the gravitational field strength increases linearly with distance from the center: $g=\frac{4}{3}\pi G\rho r$.
• This linear restoring force results in simple harmonic motion for an object moving in a frictionless tunnel through the planet's center. The motion's period $T=\sqrt{3\pi /(G\rho )}$ is independent of the object's mass and the size of the planet, depending only on its density.
• This model, while idealized, provides a foundational framework for understanding central forces and oscillations under non-constant acceleration, directly linking gravitational theory to the core mechanics of SHM.