Falling down a hole through the Earth.

  • #1
Okay, I have seen the calculation for how long it takes to fall through the Earth if you had an evacuated hole all the way through it from pole to pole. The problem is that they always assume a constant density for the Earth, and in reality it is far from that. In fact I just watched a YouTube video where a fame physicist, who shall remain nameless said that at halfway the force of gravity would be about half of what it was at the surface. He could not have been more wrong. At about half way you are at the outer core/mantle boundary and your acceleration is actually about 10.8 m/s^2 or 1m/s^2 higher than it is at the surface.

So my question is can any physics people give me an answer better than the 42 minutes estimated for a constant density Earth? Here is a link to a Wiki article on the structure of the Earth:http://en.wikipedia.org/wiki/Structure_of_the_Earth" [Broken]

They have a couple of helpful graphs there. They show density as you approach the center and g as you approach the center. I am estimating 35 minutes rather than the standard 42 minutes for a uniform Earth.
 
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Answers and Replies

  • #2
2,685
22
Why don't you use this graph from your link: http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg and do the calculation yourself?

It would be fairly straight forward. Just divide it into approximate segments and substitute the new value of g each time.

I assume we're ignoring air resistance here?

My first impression is that it's irrelevant so far as the increased gravity goes because whatever you gain whilst accelerating towards the centre is cancelled out as you decelerate once on the 'exit run'.

So although you'd get greater acceleration whilst descending, and as such a higher velocity, you'd lose the advantage because you'd have it acting against you on the way out.
 
  • #3
I was hoping that someone else would do the hard work for me. Sorry.

But I am certain that the transit time would be shorter with real world densities. The force of g is always higher in the real world example than in the constant density example as you can see from the graph. In other words you would accelerate faster to the center and decelerate faster on the way out.

Think of it this way. If you took a round trip from one planet to another at one g, and the same trip at two g, the second trip would take one over the square root of 2 as much time as the first ( the inverse square root is a figure that I remember from some better science fiction stories ). The same applies to our journey through the Earth. Since the value for g is higher through the whole trip with the actual densities, then the time has to be shorter.

I am thinking that to do this you would need to enter the data from either one of the graphs into a half decent computer program and have it analyze it, since we don;t have a mathematical function that describes the density or the value of g.
 
  • #4
2,685
22
But what you gain on the way down, you lose on the way back.

I suppose if g is constantly higher you would shave some time off that way.

I'll do a very basic analysis and post the result for you.
 
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  • #5
Thank you. As I already said, there is more than a two times difference in expected g force at the half way point. With a constant density Earth the force of gravity is exactly half, in reality it is slightly higher than at the surface. I will check this out periodically. Or I could turn my notifications on for this thread.
 
  • #6
2,685
22
OK, initial extremely basic analysis gave a time of 39.30 minutes.

I 'refined' the analysis and came out with a final time of 38.84 minutes.

If I had the full dataset for the graph I could get it significantly more accurate, but I can't find it.

Results file available on request. It's only a very simple excel file, nothing amazing.
 
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  • #7
14
0
If I had the full dataset for the graph I could get it significantly more accurate, but I can't find it.

Results file available on request. It's only a very simple excel file, nothing amazing.[/QUOTE]

Wouldn't the hole's position be the most important factor? We can probably make a pretty good model for a hole straight through the equator, but the further you get from it, the more our polar offset is going to mess with the gravity gradient.
 
  • #8
2,685
22
Wouldn't the hole's position be the most important factor? We can probably make a pretty good model for a hole straight through the equator, but the further you get from it, the more our polar offset is going to mess with the gravity gradient.

Which is why it's being constructed under perfect conditions. If you start involving all the other factors possible it gets far too complicated to do simply.

As I made very clear, it is only an extremely basic analysis based on that graph only.
 
  • #9
Thank you. I am too old for all of this computer stuff. I did learn some programming back in the stone age, but we were still using Fortran and Pascal in those days. So to round it off we can call it 39 minutes. I thought it would have been slightly faster than that.
 
  • #10
2,685
22
Thank you. I am too old for all of this computer stuff. I did learn some programming back in the stone age, but we were still using Fortran and Pascal in those days. So to round it off we can call it 39 minutes. I thought it would have been slightly faster than that.

No, I thought I'd added this but apparently not so I'll put it now:

As I improved the accuracy the speed increased. I'd expect that if you could do it as accurately as possible then the actual time would come out between 37.5 to 38.0 minutes.
 
  • #11
318
0
This data might help
http://geophysics.ou.edu/solid_earth/prem.html" [Broken]

Using this I get 37.25 minutes for a trip.
 
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  • #12
256
2
How can you fall through the earth? That makes no sense at all dudes.

EDIT:

Ok, maybe it does. I guess at the center your speed would be at its maximum. I don't really see how you could make such a calculations though.
 
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  • #13
318
0
This is a hypothetical question, when you drill a hole through the earth's core.
 
  • #14
256
2
It'd be funny if that actually happened and you couldn't grab on to anything when you got to the "exit hole"; you'd start oscillating :)
 
  • #15
uart
Science Advisor
2,776
9
Here's a related question.

If the moon was made of blue cheese, and an average sized field mouse started eating a hole through it. How long would it take to get right through to the other side?

Please include the fact the the mouse may start to suffer from obesity which would slow it's progress in your solution as I need the answer to be as accurate as possible.:bugeye:
 
  • #16
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
686
Using this I get 37.25 minutes for a trip.
I get 38.19 minutes using a 1000 step leapfrog integrator and using piecewise linear interpolation to determine gravitational acceleration at some point inside the Earth.
 
  • #17
2,685
22
I get 38.19 minutes using a 1000 step leapfrog integrator and using piecewise linear interpolation to determine gravitational acceleration at some point inside the Earth.

Hmm, I'm feeling quite happy with my 20 step giving 38.84 minutes. Only knocked it up quickly in excel to have a rough set of calcs to work with.

Plus I'm happy with my interpretation of between 37.5 to 38.0 for more accurate results.
 
  • #18
D H
Staff Emeritus
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I did this in Excel also. Leapfrog is easy to implement in Excel.
 
  • #19
2,685
22
I did this in Excel also. Leapfrog is easy to implement in Excel.

I'll take a look. Sounds useful.
 

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