How Long To Fall Through The Earth?

TL;DR

Model Earth as a uniform-density sphere and assume an airless tunnel to avoid drag and simplify gravity via spherical symmetry.

Briefing Cornell Notes

Briefing

A fall straight through a frictionless, airless tunnel from the North Pole to the South Pole takes on the order of tens of minutes—about 42 minutes under a simple model, shorter (about 37 minutes) once Earth’s layered density is accounted for. The key insight is that gravity inside a uniformly dense spherical Earth behaves like a restoring force that depends linearly on distance from the center, turning the trip into a problem with a known “oscillation” time.

The calculation starts with major simplifications: Earth is treated as perfectly spherical with uniform density, and the tunnel contains no air (so there’s no terminal-velocity slowdown). With spherical symmetry, the gravitational pull from mass shells outside a point cancels out. Only the “shaved” portion of Earth beneath the falling object matters—mass inside the radius you’ve reached. Because the mass of that shaved sphere scales with its volume (proportional to radius cubed), and gravity scales like mass divided by distance squared, the net gravitational force becomes proportional to distance from the center: F ∝ r. That means the fall speeds up as it approaches the center, reaches zero net force at r = 0 (equal pulls in every direction), and then continues toward the far side while the force reverses direction and slows it.

This linear-in-distance force matches the mathematics of a mass on a spring or a simple harmonic oscillator. For such motion, the time to travel from one side to the other has a compact form involving π and the relevant “constant stuff” (set by Earth’s density and the gravitational constant G). In the uniform-density approximation, the mass of the falling person cancels out entirely, leaving a result of about 42 minutes. A striking consequence follows: Earth’s radius doesn’t enter the final time, so any sphere with the same average density would yield the same pole-to-pole fall time. The same number also matches the time to go around the Earth to the opposite side in an idealized orbit.

Real Earth isn’t uniform. Seismology indicates the core is denser than the mantle and crust, so as the fall begins, gravity doesn’t drop as quickly as the uniform model predicts. The force stays relatively steady until roughly halfway to the center, then declines more rapidly as more mass shifts to the “above you” side. Incorporating a two-part density behavior—first a constant-acceleration-like segment, then a region where gravity decreases with radius—yields a shorter total time. Combining the segments and doubling to return to the surface on the far side gives about 37 minutes. A more detailed density profile (using the Preliminary Reference Earth Model) refines the estimate to roughly 38 minutes and 6 seconds. Either way, the trip remains long enough that “jumping in” would be a serious commitment—though still far faster than any air-filled fall.

Cornell Notes

A pole-to-pole fall through an airless tunnel can be modeled as motion under a gravity force that depends on distance from Earth’s center. For a uniformly dense spherical Earth, mass outside the current radius cancels by symmetry, leaving only the “shaved” inner mass. That produces a net force F proportional to r, the same form as simple harmonic motion, so the travel time comes out to about 42 minutes. Earth’s real layered density changes the force profile: gravity decreases more slowly at first because the core is denser, then drops faster near the center. Accounting for this structure reduces the estimate to about 37 minutes, with a more realistic density model giving about 38 minutes and 6 seconds.

Why does only the mass “below you” matter for gravity inside a spherical Earth?

For a spherically symmetric mass distribution, the gravitational effect of any spherical shell outside your current radius cancels out. Inside a spherical shell, pulls from different parts of the shell balance to zero net force. So during the fall, mass farther from the center than the object’s current position contributes nothing; only the inner mass within that radius affects the net gravitational force.

How does the uniform-density assumption lead to a force that scales like F ∝ r?

With uniform density, the mass of the inner “shaved” sphere scales with volume, so M ∝ r^3. Gravity scales like M divided by distance squared, giving F ∝ (r^3)/(r^2) = r. This linear dependence means the object accelerates toward the center more strongly as it approaches the center, but the force goes to zero at the center because pulls balance in all directions.

Why does the trip time become a simple π-based formula, and why is the person’s mass irrelevant?

A force proportional to displacement (F ∝ r) produces simple harmonic motion, the same mathematical structure as a mass on a spring. For such motion, the time to go from one extreme to the opposite extreme is proportional to π times a square root of the relevant “constant stuff.” In the falling-through-Earth case, the falling object’s mass cancels out of the dynamics, so the time depends on Earth’s density and the gravitational constant G rather than on the jumper’s mass.

What surprising result follows from the uniform-density model?

The estimated time (~42 minutes) does not depend on Earth’s radius. That means any sphere with the same average density would give the same pole-to-pole fall time under the same ideal assumptions. The model also predicts the same time for an idealized trip “around” Earth to the opposite side, matching the commonly cited internet number.

How does Earth’s layered density change the estimate from 42 minutes?

Because the core is denser than the mantle and crust, gravity doesn’t decrease as quickly as the uniform model predicts. The net force stays relatively constant until about halfway to the center, then drops more rapidly as more of the Earth lies “above” the falling object. Treating the motion as two regimes—one with roughly constant acceleration and another where gravity decreases with radius—reduces the total time to about 37 minutes.

What does a more realistic density profile do to the final number?

Using a detailed density model such as the Preliminary Reference Earth Model refines the pole-to-pole travel time to about 38 minutes and 6 seconds. The correction is modest compared with the uniform-density estimate, but it reflects the real variation in density with depth.

Review Questions

In the uniform-density model, what symmetry argument makes gravitational contributions from outer shells cancel inside the Earth?
Explain why a force proportional to distance from the center implies simple harmonic motion and leads to a π-based travel time.
How does a denser core alter the force-vs-position curve compared with the uniform-density case, and why does that shorten the fall time?

Key Points

1
Model Earth as a uniform-density sphere and assume an airless tunnel to avoid drag and simplify gravity via spherical symmetry.
2
Inside a spherical shell, gravitational pulls cancel; only the inner mass within the object’s current radius affects the net force.
3
For uniform density, the net gravitational force inside Earth is proportional to distance from the center (F ∝ r), producing simple harmonic motion.
4
Under the uniform-density approximation, the pole-to-pole fall time is about 42 minutes and does not depend on Earth’s radius or the falling person’s mass.
5
Real Earth’s denser core keeps gravity from dropping as quickly early in the fall, shortening the trip relative to the uniform model.
6
Including Earth’s layered density reduces the estimate to about 37 minutes, while a more detailed density profile gives about 38 minutes and 6 seconds.

Highlights

Inside a uniformly dense Earth, gravity behaves like a restoring force: it grows linearly with distance from the center and vanishes at the center.

The idealized 42-minute result is independent of Earth’s radius—only the average density and G matter.

Accounting for Earth’s denser core keeps the gravitational pull stronger for longer, cutting the time to roughly 37 minutes (or ~38 minutes 6 seconds with a detailed model).

Topics

Gravity Inside Earth
Simple Harmonic Motion
Spherical Symmetry
Earth Density Profile
Pole-to-Pole Fall