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!
