I recently received this question via email:
Hey Henry,
A while ago you did a video... called "What if the Earth were Hollow?" where you showed
how long it would take to fall through the earth. I was simply wondering how that was
even calculated since the force of gravity would constantly be changing due to the growing
amount of mass above you.
Peter
Ok, so we've got a hole through the earth, and it goes from the north pole to the south
pole – that way we don't have to worry about the Coriolis effect from the earth's spinning.
There's also no air in the hole, otherwise you'd reach terminal velocity pretty quickly
and your trip would be slow and boring. There are a few different ways to figure out how
long it'll take for you to reach the other side.
One is to jump in, but it'll be faster – and probably a greater chance of survival – if
we calculate using math & physics.
First, more simplifications: assume the earth is perfectly spherical and has the same density
everywhere. It turns out that the gravitational attraction from any spherically symmetric
object is the same as if all its mass were concentrated at the center of that object
– the closer parts attract more than average, the far away parts attract less, but over
the whole sphere it averages out. In a similar vein, if you're _inside_ a spherical shell,
then the gravitational pulls from all the different parts cancel out and you experience
zero effect from the shell.
This means that, inside the earth, any parts that are farther away from the center than
you are cancel out and have no effect – almost like they've been trimmed off and you're temporarily
on the surface of a smaller, shaved earth. Since we assumed the same density everywhere,
the shaved-earth's mass is simply proportional to its volume, which is proportional to its
radius cubed. And because it's a sphere we get to pretend all that mass is actually concentrated
at a single point in the middle.
So how much does the shaved-earth-point pull on you? Well, the gravitational attraction
between two objects is proportional to their masses but inversely proportional to the distance
between them, squared, so we have to divide the mass of the shaved-earth by the square
of the distance you are from the center – which is just the radius of the shaved earth. R
cubed divided by r squared is r, so the force on you is simply F equals some constant stuff
times r, your distance from the center.
Essentially, as you fall the mass beneath you decreases, while the **average gravitational
pull on you** from any bit of that mass increases, but the mass decreases **more than the average
pull** of gravity increases.
So as you approach the earth's center, you go faster and faster but the force pulling
you towards the middle gets smaller and smaller. Exactly in the middle you experience zero
net force because the earth is pulling you equally in all directions, though since you're
going so fast you'll continue to speed towards the other side, gradually slowed by the now
increasing force pulling you back towards the middle. F equals some constant stuff times
distance.
The exact same equation – some constant stuff times a distance – also describes
a mass on a spring, or simple pendulum, or a cat in a parabola. And from studying _those_
equations we know that **the time taken by the moving object – whatever it is –
to go from** one side to the other has a simple formula: pi times the square root of the mass
divided by the "constant stuff". In the case of falling through the earth, your mass cancels
out of the equation so we just need to put in numbers for the density of the earth and
the gravitational constant to get the answer - 42 minutes to fall through the earth.
This turns out to be exactly the same as the time it takes to fall _around_ the earth to
the other side, and it's the number you'll find commonly mentioned on the internet. Even
more surprising, the radius of the earth didn't factor into the time calculation – it predicts
you'll take 42 minutes to fall through or orbit around to the other side of ANY sphere
with the same density as the earth.
But the earth isn't exactly the same density throughout – we know from seismology that
the earth's core is much denser than its mantle and crust. So as you begin to fall, most of
the mass is still below you, pulling, so the pull of gravity doesn't decrease as much as
our simple model predicted. In fact, the force is actually pretty constant until about halfway
to the center, at which point it starts quickly decreasing as more and more of the earth is
"above" you.
The calculations here are a bit more annoying because we have to piece together two different
parts – the falling with constant acceleration part, which is easy, and the falling with
decreasing gravity proportional to your radius part, which is the same thing we did before,
except now you're starting out halfway to the middle of the earth with a speed of 17
thousand miles per hour, instead of on the surface with no speed. Once our mathemagical
dust settles, we combine the two parts and multiply by two to get the total time back
to the surface on the other side: 37 minutes.
Of course, this is still just an approximation – slightly more realistic than before, but
far from perfect. If you carefully piece together the time for a falling-through-the-earth trip
based on a more detailed density profile of the earth, like maybe the Preliminary Reference
Earth Model, you can can be even more precise – 38 minutes and 6 seconds from pole to
pole.
But either way, if instead of calculating you jumped into the hole at the start of this
video, you still have a long ways to go before reaching the other side of the earth. Safe travels!