How long to fall through a hole in Earth?

sonhouse
Science 26 Mar '15 15:14
1. sonhouse
Fast and Curious
26 Mar '15 15:14
Through a theoretical hole drilled all the way through the center of the Earth and back to the other surface on the opposite side of the world?

I have this piece saying the regular answer is 42 seconds but nay nay, should be 38 seconds. I did the numbers and found if you had a constant acceleration of 1 G, the time it takes to get to 6,371,000 meters (radius of Earth) is equal to 19 minutes and therefore times 2 or 38 minutes to get from surface to surface.

http://phys.org/news/2015-03-fall-hole-earth.html

I don't see how EITHER number is correct. If I assign a value of half of 9.8 M/S^2 or 4.9, it comes out at 26 minutes to get to one R of Earth, X 2 of 52 minutes surface to surface.

So what is up with all that?
2. 26 Mar '15 15:20
The acceleration is not constant but changes linearly as a function of position. To find the correct answer, solve Newton's second law assuming that the force obeys F = -kx, where the coefficient k can be determined through the boundary conditions.
3. sonhouse
Fast and Curious
26 Mar '15 15:33
Originally posted by KazetNagorra
The acceleration is not constant but changes linearly as a function of position. To find the correct answer, solve Newton's second law assuming that the force obeys F = -kx, where the coefficient k can be determined through the boundary conditions.
The student in this article says it is not linear but even so, I used the old formula T=(2S/A)^0.5 and used 9.8 m/s^2 CONSTANT accel and come up with 19 minutes to reach dead center. So how could HE come up with 19 minutes since we know full well that even if not linear at the surface you get 9.8 and center you get zero. Well, 19 minutes and 5.5 seconds. Even the number they use as the accepted number seems too close to 19 minutes, 21 minutes to dead center or 42 minutes surface to surface.

So what gives?
4. 26 Mar '15 15:49
Originally posted by sonhouse
The student in this article says it is not linear but even so, I used the old formula T=(2S/A)^0.5 and used 9.8 m/s^2 CONSTANT accel and come up with 19 minutes to reach dead center. So how could HE come up with 19 minutes since we know full well that even if not linear at the surface you get 9.8 and center you get zero. Well, 19 minutes and 5.5 seconds. Ev ...[text shortened]... ose to 19 minutes, 21 minutes to dead center or 42 minutes surface to surface.

So what gives?
The object will perform harmonic motion with a period T = 2*pi*\sqrt{m/k}, assuming homogeneous density (the article mentions that a varying density gives a small correction). If that doesn't give the right answer your choice for k is inaccurate (note that the mass m will drop out as k is proportional to it).
5. sonhouse
Fast and Curious
26 Mar '15 15:56
Originally posted by KazetNagorra
The object will perform harmonic motion with a period T = 2*pi*\sqrt{m/k}, assuming homogeneous density (the article mentions that a varying density gives a small correction). If that doesn't give the right answer your choice for k is inaccurate (note that the mass m will drop out as k is proportional to it).
So you are saying this is equivalent to a pendulum, where end of swing to bottom of swing = surface to center of earth. And the full swing one end to the other = surface to surface of Earth.
6. 26 Mar '15 16:17
Originally posted by sonhouse
So you are saying this is equivalent to a pendulum, where end of swing to bottom of swing = surface to center of earth. And the full swing one end to the other = surface to surface of Earth.
Exactly. You can prove the two are equivalent by showing that the force is indeed a linear function of displacement with respect to the center of the Earth.

Of course, there are a bunch of assumptions/idealizations in this picture, of which the homogeneity of the density of the Earth is one that is addressed in the article.
7. sonhouse
Fast and Curious
26 Mar '15 17:23
Originally posted by KazetNagorra
Exactly. You can prove the two are equivalent by showing that the force is indeed a linear function of displacement with respect to the center of the Earth.

Of course, there are a bunch of assumptions/idealizations in this picture, of which the homogeneity of the density of the Earth is one that is addressed in the article.
So how can EITHER answer be right? If it takes 19 minutes to go the distance of one Earth Radii and double that to decel at the same rate so you now are 2 radii away from the starting point, presumably going zero velocity at both ends of travel, how could that kid possibly be coming up with 38 minutes for the whole journey when I already showed it comes out at 38 minute under 1 g of constant acceleration?
8. 26 Mar '15 17:50
Originally posted by sonhouse
So how can EITHER answer be right? If it takes 19 minutes to go the distance of one Earth Radii and double that to decel at the same rate so you now are 2 radii away from the starting point, presumably going zero velocity at both ends of travel, how could that kid possibly be coming up with 38 minutes for the whole journey when I already showed it comes out at 38 minute under 1 g of constant acceleration?
Your numbers are wrong. I just did the calculation and it comes out at just over 42 minutes, the same as the figure mentioned in the article.
9. sonhouse
Fast and Curious
26 Mar '15 18:041 edit
Originally posted by KazetNagorra
Your numbers are wrong. I just did the calculation and it comes out at just over 42 minutes, the same as the figure mentioned in the article.
Ok, I used 1 radii as 6371 km or 6,371,000 meters and 9.8 M/S^2.

I used the formula T=(2S/A)^1/2 In other words, the square root of twice the distance divided by the acceleration.

I come up with 1140 and change seconds, or 19 minutes to fall to the center of the Earth with that constant accel. Am I not using the right formula?

I see your formula is the harmonic motion of a mass on a spring. Is that the same thing as falling through the Earth? I got 84 minutes for one whole swing from one end of travel to the other, twice the 42 minutes you mentioned.
10. DeepThought
26 Mar '15 18:45
Originally posted by sonhouse
The student in this article says it is not linear but even so, I used the old formula T=(2S/A)^0.5 and used 9.8 m/s^2 CONSTANT accel and come up with 19 minutes to reach dead center. So how could HE come up with 19 minutes since we know full well that even if not linear at the surface you get 9.8 and center you get zero. Well, 19 minutes and 5.5 seconds. Ev ...[text shortened]... ose to 19 minutes, 21 minutes to dead center or 42 minutes surface to surface.

So what gives?
The formula you used δx = 1/2 at² assumes constant acceleration. As Kazet said, the total field at a point a distance r from the centre is due to the mass contained within the sphere. This is the same as Gauss's law for elecromagnetism - that the total flux emitted from a fixed volume is proportional to the contained charge. The mass contained within a sphere goes as r³ assuming constant density and the force goes as 1/r², so the force increases linearly with r.

The acceleration is -GM/r², the minus sign indicates it's an inwards force,

M= 4 ρπ r³ /3

a = -4Gρπr/3

If R is the radius of the earth, and g the magnitude of the acceleration due to gravity:

g = 4GρπR/3

So:

a = - gr/R

which you can also get just by noticing it's linear and knowing the force due to gravity at the surface. The formula for simple harmonic motion is:

F = -kr,
or:
a = -(k/m)r

Comparing we have: k = mg/R

The period of simple harmonic motion is T = 2π√m/k = 2π√R/g

For the Earth R = 6,371,000 metres, g = 9.81 m/s, which gives T = 5060 seconds = 84 minutes 34 seconds. So to get from one side to the other we need half of this which is 42 minutes 17 second.
11. 26 Mar '15 18:53
Originally posted by sonhouse
Ok, I used 1 radii as 6371 km or 6,371,000 meters and 9.8 M/S^2.

I used the formula T=(2S/A)^1/2 In other words, the square root of twice the distance divided by the acceleration.

I come up with 1140 and change seconds, or 19 minutes to fall to the center of the Earth with that constant accel. Am I not using the right formula?

I see your formula ...[text shortened]... tes for one whole swing from one end of travel to the other, twice the 42 minutes you mentioned.
I used the formula T=(2S/A)^1/2 In other words, the square root of twice the distance divided by the acceleration.

I come up with 1140 and change seconds, or 19 minutes to fall to the center of the Earth with that constant accel. Am I not using the right formula?

Ah, I see, I didn't realize you also assumed that the acceleration changes sign at the center of the Earth. In this case you would indeed get about 38 minutes. Of course, the assumption that acceleration is constant is wrong.

I see your formula is the harmonic motion of a mass on a spring. Is that the same thing as falling through the Earth?

The point is that you always get this formula if the force can be written as F(x) = -kx. A mass on an (idealized) spring is just an archetypal example.

When I was a TA in the autumn we posed a similar problem to the students, except that the hole through the Earth was at a certain angle. Try to figure out how long it would take to get to the other side in that case.
12. 26 Mar '15 19:13
The Earth rotates. The reasoning above presupposes that the hole is to be drilled from the exact north pole to the exact south pole.

Because if we drill a hole from a point of the equator to another point also at the equator, then we will not have a free fall. We will quite immediately touch the sides of the drilled hole and the path will not be straight, but a slight spiral to the center and then another spiral to the surface again.

How does this fact complicate the formulae?
13. DeepThought
26 Mar '15 19:301 edit
Originally posted by sonhouse
Ok, I used 1 radii as 6371 km or 6,371,000 meters and 9.8 M/S^2.

I used the formula T=(2S/A)^1/2 In other words, the square root of twice the distance divided by the acceleration.

I come up with 1140 and change seconds, or 19 minutes to fall to the center of the Earth with that constant accel. Am I not using the right formula?

I see your formula ...[text shortened]... tes for one whole swing from one end of travel to the other, twice the 42 minutes you mentioned.
I got 84 minutes for one whole swing from one end of travel to the other, twice the 42 minutes you mentioned.
The period is the time taken to return to it's starting point. We're calculating half a period.

There are three idealisations being used. One is that the earth has constant density. This isn't true and we'd need some information about the functional dependence of density with depth. I haven't looked at the article, but from what Kazet said above that seems to be what the calculation described in the article corrected for.

The second idealisation is that it's a sphere. This isn't true either, and it produces corrections to the force. Correcting for that's a bit of a pain.

The third problem is angular momentum, the earth is spinning, dropping pole to pole that's not a problem, but at the equator the thing dropping to an antipode had better be able to shed angular momentum...

Edit: Oh yeah, and there's the minor matter of air-resistance...
14. 26 Mar '15 20:10
Originally posted by DeepThought
I got 84 minutes for one whole swing from one end of travel to the other, twice the 42 minutes you mentioned.
The period is the time taken to return to it's starting point. We're calculating half a period.

There are three idealisations being used. One is that the earth has constant density. This isn't true and we'd need some informati ...[text shortened]... to shed angular momentum...

Edit: Oh yeah, and there's the minor matter of air-resistance...
If you are really being nitpicky, you also have to take into account a small correction due to being obliterated by a huge mass of molten rock.
15. joe shmo
Strange Egg
26 Mar '15 20:56
Originally posted by FabianFnas
The Earth rotates. The reasoning above presupposes that the hole is to be drilled from the exact north pole to the exact south pole.

Because if we drill a hole from a point of the equator to another point also at the equator, then we will not have a free fall. We will quite immediately touch the sides of the drilled hole and the path will not be straig ...[text shortened]... ter and then another spiral to the surface again.

How does this fact complicate the formulae?
I believe you have to include the "ficticious" Coriolis Force and Centrifugal Force for a rotating frame of reference. Which certainly complicates things.