Science
What If You Fell Through the Earth?
How long is a gravity-train ride through a tunnel along a diameter?
A straight tunnel through Earth's centre is a seventeenth-century daydream with a clean answer if you stack the right lies. Uniform density, vacuum tunnel, no Coriolis, no magma: the gravity inside a sphere is proportional to radius, the motion is simple harmonic, and the one-way trip is half a period — about 42 minutes, the same for a chord through Earth, famously. Takeoff speed at the centre is orbital-ish, near 8 km/s. Turn friction on, even as a toggle, and the oscillation dies; you do not pop out in Australia. Real Earth is denser in the core, so the true vacuum-tunnel time would shift a little. This page uses the uniform-density formula and says so. It will not drill the hole.
Change this
Results update as you move the controls.
Off: you do not oscillate, you do not exit the far side.
Your result
42.2 min
one-way through a diameter
uniform density, no friction
- One-way
- 42.2 minutes
- Round trip
- 84.4 minutes
- Peak speed at centre
- 7,904 m/s
- g used
- 9.81 m/s²
- Friction
- ignored
42 minutes 12 seconds
28,456 km/h
What does that mean?
In the uniform-density vacuum tunnel, one-way time is 42.2 minutes (T/2 = π sqrt(R/g)). Peak speed at the centre is 7,904 m/s. At 1× surface g the clock scales as 1/sqrt(g). Turn friction on and this harmonic story ends: you would not come out the other side. Real Earth's denser core would tweak the 42 minutes, not replace them with a Hollywood core of lava-surfing.
At a glance
- One-way42.2 min
- Round trip84.4 min
Timeline
0
Step in
Weight is still mg. The tunnel is the fiction.
21.1 min
Centre
Speed peaks at 7,904 m/s. g_local is zero at the exact centre in this model, but you are moving too fast to linger.
42.2 min
Far side
If the other hatch is open and the vacuum held, you arrive at rest — then fall back, forever, in the ideal oscillator.
Compare scenarios
Why 42 minutes
Inside a uniform sphere, g(r) = g_surface × (r / R). That is Hooke's law. Period is 2π sqrt(R/g), one-way is π sqrt(R/g). Plug in mean radius and standard g and you are near 42 minutes. Raise g with the multiple slider and the trip shortens as 1/sqrt(g). The same period shows up as a low-Earth-orbit period for a grazing circle, which is not a coincidence: both are sqrt(R³/GM) in disguise.
The middle is fast
Energy conservation puts your peak speed at the centre: v = sqrt(g R) for this model, around 7.9 km/s on Earth. That is not a sightseeing pace. It is also why 'just jump in' is a thought experiment. Air would turn the tunnel into a furnace; rock would not permit a hole; Coriolis would slam you into a wall on a rotating planet. Those are the reasons the toggle labelled ignore friction is on by default.
Friction ends the joke
With drag or with a medium in the way, you dissipate the potential difference between surface and centre. You do not climb the other side. In a cartoon of strong friction you stop somewhere short of the far exit, perhaps near the centre if you leave enough energy on the table. This page does not integrate a drag coefficient. It changes the story from 'oscillator' to 'you do not come out.' That is the scientifically honest punchline once vacuum is refused.
How we calculated this
One-way time is physics.gravityTrainSeconds(gMultiple): π sqrt(R / (g n × gn)) with mean Earth radius. Peak speed is sqrt(g R) in the frictionless uniform model. gMultiple scales surface g. If ignoreFriction is false, the result reports no oscillation and no far-side arrival instead of Infinity or a fake damping time. Not a PREM density profile, not a rotating-Earth Coriolis integrate.
Go further
A curated rabbit hole from this question. Each link is a real experiment, not a random suggestion.
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