Kill Time With Recreational Math: Journey Through the Centre of Psyche

There is a classic physics thought experiment in which you suppose that you could drill a hole right through the centre of the Earth then jump in and fall through it and pop out the other side. There is even an anitpode website that lets you find out what is on the other side of the Earth from you here.

In this scenario you evacuate all of the air in the hole to create a vacuum and then drop an object down the hole. How long does it take for the object to fall through and reach the other side?

In this post we are going to calculate just that. Then we are going to go a bit further and also do it for the asteroid 16 Psyche which is said to be worth $10,000 quadrillion because it is said to be made entirely of nickel-iron and other valuable minerals. I wrote a post about this asteroid here.

The reason that I want to do the calculation for Psyche is because a hole through the Earth is too fanciful and will never actually happen but a hole through Psyche might one day just happen as a sort of asteroid subway system (metro system for the European readers).

The Spreadsheet

Below is a screenshot of the spreadsheet. I have set up the relevant parameters:

  • G for the Universal Gravitational constant
  • radius is the radius of the Earth
  • density is the average density of the Earth. It changes as you go deeper from the crust to the core but that doesn't affect the answer too much (more on that later).

The Columns are set up as:

Column A

This column calculates the time 't' in seconds. It is a simple column in which you start at 0 and then just add 1 to each cell below it. You may use any time increment you feel, I chose 1 second for this calc and it works well enough.

Column B

For convenience this column calculates the time 't' in minutes. It is simply the cell to the immediate left divided by 60.

Column C

This is the distance, or radius 'r', that the object is from the centre of the Earth. The Earth has a radius of 6378 km at the equator. Each cell will use the calculation show in the screenshot below:

Cell C8 "=6378000-I7"

Cell C8 calculates the radius of the Earth minus the distance fallen so far which is held in Column I. To avoid a circular calculation reference we have to offset the cell by one. This is why we are subtracting off I7 and not I8. This introduces a small off-by-one-second error. Not a big deal.

Column D

Column D merely stores the average density of the Earth and in this example it never changes. To be honest, you could actually keep it in a single cell as shown for Cell B3 but I want to retain the option of having a density that increases as you go deeper which I may do in a later post. That is why I did it this way.

Column E

Column E calculates the volume of the planet that is below the falling object as shown below.

According to Shell Theorem you only feel the pull of gravity of the matter that is below you as you travel down into a asteroid, planet or star. The matter above you pulls up and is perfectly balanced out by the entire spherical shell of matter that is further away from the core that you are.

I know this is true because it was a homework question I had to work out back when I was studying physics in University a long time ago.

This equation you can see in the screenshot above calculates the volume of a sphere which is:

Column F

Column F calculates the mass of the volume calculated in Column E. It multiplies the density in column D times the volume in column E as shown below.

Column G

Column G calculates the force due to gravity on a object at a particular distance above the core of the Earth according to the equation:

The calculation is shown in the screenshot below.

Cell B$1 contains G the universal gravitational constant and is set to be an absolute cell reference.

The other cells are the mass of the volume still below the object and the radius the object is still above the core. These are relative cell references because they will be changing constantly as the object falls down through the Earth.

Column H

Column H calculates the velocity of the object. It is falling through a vacuum so we do not need to worry about air drag.

The equation is thus just uses the simple equation "v = at" where,

v is the velocity (always increasing as we drop deeper and deeper),
a is the gravitational acceleration (always decreasing as we drop deeper and deeper) and
t is the time increment.

The cell's equation is shown below:

Column I

Column I calculates the distance that the object has fallen so far and it used the classic kinematics equation:

where d0 is the distance fallen so far,
v is the velocity the object is currently falling at,
a is the current acceleration, and
t is the time increment.

The equation is shown below:

The Result and Verification

There are a number of sources that calculate this answer (cited below) and they generally provide an answer of 42 minutes and 12 seconds to fall through the Earth and pop out the other side.

My spreadsheet gives me this:

It take 1265 seconds to fall from the surface to the center. The trip back up will be a mirror image of the first half of the trip so we double the answer to get 1265 x 2 = 2530 seconds or 42 minutes and 10 seconds.

Not bad. My spreadsheet is only 2 seconds different from the other 'official' calculations.

You might also notice that the object's velocity as it passes through the centre is about 7.9 km/s. That sucker is really moving at a fast clip. Hope it doesn't hit any walls.

This high speed is also a partial reason why the constant density assumption works in this calculation. The object just flies through the denser parts of the Earth and so it spends less time in the high density regions of the planet.

The short time means less of a build up in error for the real answer if you use the actual density as a function of depth.

The Asteroid 16 Psyche

If humans ever get into space in a big way then this asteroid is going to be the centre for a lot of mining and industry. A subway system that drops through the middle of this asteroid is actually a feasible idea.

So we copy the spreadsheet above for Earth into another tab, and replace the critical parameters as shown here:

You will see that the radius has been changed to 113 km (the average radius of the asteroid) and the density has been changed to 4500 kg/m3 the average density of the asteroid.

The Result for Psyche

So after plugging in the new numbers for 16 Psyche we find that it takes 1400 seconds to fall through to the centre of the asteroid meaning that it takes 1400 seconds x 2 = 2800 seconds or about 47 minutes to make it the 113 km through the asteroid.

The journey is very slow at first because gravity acceleration on this asteroid is small. By the time you get to the centre though you will be traveling at 127 m/s which is 457 km/h just on gravity alone.

The trip can be made faster if you give it a bit of a boost at the start.

If we accelerate to 100 km/h at the start it takes 40 minutes to go through the asteroid, a 7 minute time savings. Not really worth the effort actually.

Closing Words

Traveling through the centre of the Earth is a silly idea and it is never going to happen. You might as well take the SpaceX BFR transport instead.

However traveling through the centre of a largish asteroid just might be feasible and this calculation shows that it would take less than an hour to do so on Psyche!

Thank you for reading my post.

Post Sources

Shell Theorem:
https://en.wikipedia.org/wiki/Shell_theorem

How Long to Fall Through the Earth?
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/earthole.html
http://www.physicscentral.com/explore/poster-earth.cfm
https://www.sciencemag.org/news/2015/03/how-long-would-it-take-you-fall-through-earth

Earth Data:
https://www.livescience.com/50312-how-long-to-fall-through-earth.html
https://en.wikipedia.org/wiki/Earth

16 Psyche Data:
https://en.wikipedia.org/wiki/16_Psyche

Sort:  

Interesting.
If there was no air friction would you end up oscillating forever around the center?

Yes. You would just keep cycling back and forth.

However there is no such thing as a perfect vacuum so in reality you would eventually settle at the centre and be stuck. Not a good day.

On a positive note the centre is at zero g so that would be fun.

Hi, I found some acronyms/abbreviations in this post. This is how they expand:

AcronymExplanation
BFRBig Falcon Rocket (2017 enshrinkened edition),Yes, the F stands for something else; no, you're not the first to notice
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