New way to calculate how long it would take to fall through a hole in the Earth

March 26, 2015 by Bob Yirka weblog
A composite image of the Western hemisphere of the Earth. Credit: NASA

(Phys.org)—Alexander Klotz a student at McGill University in Canada has calculated a new answer to the commonly asked physics question, how long would it take a person to fall all the way through the Earth? Instead of the commonly accepted 42 minutes, he claims it is 38. He has published his reasoning, math and conclusions in a paper published in The American Journal of Physics.

If someone were to drill a hole all the way through the planet, and then somehow manage to fall into it, how long would it take them to arrive on the other side? That is a physics question put to students every year, and those who give it expect the answer to be 42 minutes. But is that answer correct? Klotz says no and has the math to prove it, Science reported.

The accepted answer of 42 minutes takes into account the constantly changing impact that gravity will have (and ignoring drag due to the presence of air) on the person falling, becoming less and less of a factor as the center of the Earth is approached then growing stronger and stronger as the person heads "up" against gravity on the other side. It is accepted that the speed attained during the descent on the first half of the journey would be significant enough to cause the person to continue moving against gravity on the other side of the planet, right up until the surface is reached.

But Klotz argues that it is time to start taking the different densities of the Earth's layers into consideration—after all, a lot of research has shown that our planet is a lot denser at the center than at the crust for example—and that of course would have an impact on the person falling through. He has used to calculate the different densities at different depths and has used that data to give a more accurate answer to the falling man question, stating that it would in fact, take just 38 minutes (and 11 seconds) to fall all the way through, not 42 and (12 seconds).

Interestingly, Klotz also notes that if were to be assumed to be at a surface level constant throughout the duration of the trip, the math shows it would take just 38 minutes as well.

Explore further: How long does it take sunlight to reach the Earth?

More information: The gravity tunnel in a non-uniform Earth, Am. J. Phys. 83, 231 (2015); dx.doi.org/10.1119/1.4898780 . On Arxiv: arxiv.org/abs/1308.1342

Abstract
This paper examines the gravity tunnel using the internal structure of Earth as ascertained from seismic data. Numerically, it is found that the time taken to fall along the diameter is 38 min, compared to 42 min for a planet with uniform density. The time taken to fall along a straight line between any two points is no longer independent of distance but interpolates between 42 min for short trips and 38 min for long trips. The brachistochrone path (minimizing the time between any two points) is similar in shape to the uniform-density solution but tends to reach a greater maximum depth and takes less time to traverse. Although the assumption of uniform density works well in many cases, the simpler assumption of a constant gravitational field serves as a better approximation to the true results.

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billpress11
3 / 5 (2) Mar 26, 2015
I certainly could be wrong about this but why would one even need to calculate the different densities of the interior of the earth? Wouldn't the acceleration toward the center and the deceleration back toward the surface be exactly or very nearly balanced?
Expiorer
5 / 5 (3) Mar 26, 2015
I certainly could be wrong about this but why would one even need to calculate the different densities of the interior of the earth? Wouldn't the acceleration toward the center and the deceleration back toward the surface be exactly or very nearly balanced?

You will be exposed to larger mass at center longer time and reach center faster.
billpress11
3 / 5 (2) Mar 26, 2015
I certainly could be wrong about this but why would one even need to calculate the different densities of the interior of the earth? Wouldn't the acceleration toward the center and the deceleration back toward the surface be exactly or very nearly balanced?

You will be exposed to larger mass at center longer time and reach center faster.

I would agree with that, but wouldn't the deceleration, once beyond the center, be equal to the acceleration gained approaching the center?
Taterbug
1 / 5 (4) Mar 26, 2015
The person or thing would slow down and slide against the side of the hole the closer they came to the center of the earth. at some point that person would stop and would be able to walk around the walls of the hole. and would weigh much less .
Science Officer
3 / 5 (4) Mar 26, 2015
The path to Hell is paved with good calculations......
TheGhostofOtto1923
2 / 5 (4) Mar 26, 2015
Reaching an equivalent radius on the other side implies perpetual motion. A pendulum swinging in a vacuum still loses energy with each swing - am I right?

If the person were allowed to oscillate back and forth, eventually he would come to rest at the center.
Sonhouse
1 / 5 (1) Mar 26, 2015
Well I get 38 minutes just using 9.8 m/s^2 and 6371000 meters as the radius of Earth using T=(2S/A) * 2 where 2S is twice the radius or 12,742,000 and A = 9.8 M/S^2

Time to center, 19 minutes, 1140 seconds and double that to get back to the surface, 2280 seconds or 38 minutes. What's the big deal and where did they get 42 minutes anyway? I get 38 minutes using a constant 9.8 M/s^2 but it can't be just that because at dead center of the Earth you would have zero gravity so it must be something like half that as an average but that is probably not right but just an idea as to the proper number, but that will give you 38 minutes to just get to dead center which would be 76 minutes surface to surface.
TheGhostofOtto1923
3 / 5 (2) Mar 26, 2015
Apparently I am wrong.

"this is a perfect case of a so called perpetuum mobile. It would respresent a perfect (ideal) non-dissipative system where entropy production diS/dt=0, in accordance with the 2nd law of thermodynamics. Indeed the first law of thermodynamics (energy conservation) does not say much about this, except that no term for energy loss included."

-But in a less perfect case I would think that the movement would somehow affect the earths rotation and thus be slowed.
Sonhouse
not rated yet Mar 26, 2015
Forgot the square root deal, so if we use an average of half of 9.8 or 4.9 M/S^2 it comes out at 26 minutes to reach dead center and 52 minutes to go surface to surface. The thing that strikes me is it takes 19 minutes to free fall a distance = Earth Radius in a constant field of 9.8 and using half as an average so the real number would have to be closer to 9.8 to come up with 42 minutes. I don't see how this kid comes up with 38 minutes since I was using a constant acceleration and we know the actual acceleration is a variable, going to zero at dead center. I don't see how EITHER answer is correct.
gkam
1 / 5 (1) Mar 26, 2015
"-But in a less perfect case I would think that the movement would somehow affect the earths rotation and thus be slowed."
---------------------------------------

Wouldn't it then be boosted by the approximately the same amount when it reversed direction?
adam_russell_9615
5 / 5 (1) Mar 26, 2015
I certainly could be wrong about this but why would one even need to calculate the different densities of the interior of the earth? Wouldn't the acceleration toward the center and the deceleration back toward the surface be exactly or very nearly balanced?


True, so consider only how long it takes to reach center then multiply by 2.
As you proceed toward center the rate of acceleration diminishes until it reaches zero at the center. Consider a case where most of the mass is very close to the center. Then your acceleration wont diminish much until you are almost all the way to the center. Your trip from surface to center will be much quicker.
billpress11
5 / 5 (2) Mar 26, 2015
I would think that your rate of acceleration would start to decrease as soon as you drop below the surface until you reach dead center where your acceleration rate would drop to zero. Then from the center on it would be the exact opposite, you would very slowly decelerate just beyond the center with increasing deceleration until you reached opposite surface.
Raygunner
not rated yet Mar 26, 2015
A falling person through air would be subject to terminal velocity and have a stabilized maximum speed of around 120 mph at about 1 atmosphere. Factoring this in with increasing air density and lessening gravity you would not fall "up" past the center of the earth. Air resistance would be more of a factor as you neared the center of the earth and you approach weightlessness, slowing the body falling more and more. So you would not fall "up" more than a few miles past the zero gravity point IMO. Assuming a vacuum all the way through and this works - 38 minutes. The fact is that terminal velocity limits the speed that can be obtained, thus the ability to have enough speed to make it out of the other side. Unless I'm missing something obvious here.
walter_marvin
2 / 5 (1) Mar 26, 2015
Actually, simple physics shows you actually can't "fall" all the way through the earth. If you start at zero velocity, the forces at you going down equally balance the forces against you coming back up. If the hole is air filled, then you have air friction against you as well, and you will osculate, eventually ending up at the centre. In a vacuum you oscillate forever. Now if you start with enough initial velocity or use a vacuum tube and pierce the Earth in a path such that the return path is shorter than the downward path, then you might be able to do it. Basically, the question is ill formed, since the initial conditions are now known.
Mayday
not rated yet Mar 26, 2015
The deeper you would fall, more mass would be above & beside you, and less mass would be below you. Your rate of descent should therefore decrease based on these combined masses. But, if the tube is a vacuum, your momentum at your highest speed (a relatively short way down) would need to be accounted for. I don't believe you would get very far past the center.

The pencil on my desk is drawn toward the Earth's center of mass with a force equivalent to the mass "below" it minus any other commingled mass attractions(the Moon comes to mind, for one), not toward some magical point at the Earth's center that exerts a continuous attractive force no matter how close it is to it.

If there is air in the tube, all bets are off. Then even the diameter of the tube plays a part.

A better thought experiment might be to let the tube fill with air and calculate the depth at which the air reaches maximum density. And what might that air density be? Any takers?
gkam
3 / 5 (2) Mar 26, 2015
Decent discussions, but I hope to not take the trip.
billpress11
not rated yet Mar 26, 2015
The 42 minutes and 12 seconds is certainly figured for a perfect vacuum, if it wasn't you would never reach the other side and end up in the center eventually. If I were a betting person my money would be on the 42 minute 12 time period.
adam_russell_9615
not rated yet Mar 26, 2015
Just remember if you try this that you need to do it from the geographic north pole because - Foucault's pendulum.
:-)
shavera
not rated yet Mar 26, 2015
Sonhouse: addressed in other media is the coincidence that assuming constant acceleration is roughly the same as the calculation with densities.

More-or-less, since so much of the mass is in the very dense center, and one travels fairly quickly through that center, the gravitation near the surface (on either end) has the longest time to act (and thus a larger impulse on the momentum), and so the acceleration being *roughly* the same as surface acceleration is a fairly decent estimate.
Victor G_
not rated yet Mar 26, 2015
Something does not make sense in this calculation... We get the same time of travel (38 min) for (1) a multilayer model of Earth, and (2) for the constant-acceleration case. The latter case corresponds to an idealized situation when the whole mass of the Earth is concentrated in a single point its center. So how can the multilayer model produce the same result as the point-mass model? It would only be possible if the outer layers had negligible density, which is not true.
baby-panda
not rated yet Mar 26, 2015
(and ignoring drag due to the presence of air) - where's the fun in that? They would never make it to other side - gravity would pull the falling person up against wall of the hole and slide them to the center. In a vacuum what would prevent the drift to one side of the hole?
RayInLv
not rated yet Mar 26, 2015
Funny thing is if you were able to get to the "center" of the earth, you would weigh nothing. Gravity would be balanced in all directions.
Frank99
not rated yet Mar 26, 2015
I don't think you'd make it through the Earth. With no mass directly under you, you'd float until the mass of the Earth on the sides of the hole started attracting you. Gravity pulls you in straight lines towards the greatest concentration of mass. Since the straight line of mass in all directions is considerably less than the direct line mass of an Earth-width you would be slowly pulled to the line of next greatest mass which would be a line at a slight angle off of your hole's centerline. You'd be pulled to the wall, bounce off and be pulled to the next greatest mass until you hit it. Basically you'd ricochet off of the walls going slower and slower as you descended.

As you fell, more and more mass would be above you countering any pull 'down' until you eventually came to a stop at the center, completely weightless because you're being pulled equally in all directions.
TopCat22
not rated yet Mar 27, 2015
Interestingly, Klotz also notes that if gravity were to be assumed to be at a surface level constant throughout the duration of the trip, the math shows it would take just 38 minutes as well.

... based on this observation it would be obvious that if using the gravity at the surface of both ends whatever exists in between the ends would be averaged out to be zero effect since whatever is happening going down happen coming back up the other side thus the 38 minutes would be shown to be more correct than the official 42.
ilya_simkhovich
not rated yet Mar 27, 2015
i hope everyone understands that the earth isn't spherical and because it's like a oblong egg-shaped thing that expands at its fat belly because it's spinning and not fully solid, that means that the different layers inside may very well be pretty different at different places. a true cross-section would probably show pockets of different densities. i don't quite understand this calculation posed to students; does it assume a pure, solid mass that is also perfectly spherical? either way, this is idiotic and was a terrible premise for the "total recall" reboot.
ROBTHEGOB
not rated yet Mar 27, 2015
Does the weight of the man make any difference? Would the time and speed of fall change?
Lex Talonis
not rated yet Mar 27, 2015
It's a stupid question about acceleration and braking due to gravity....

And leaving out air friction? Like WTF?

THAT is not science.... it's mental masturbation.

Z99
5 / 5 (2) Mar 27, 2015
What a waste. No air resistance? But resistance scales approx. proportional to v², so ignoring that is a very good example of how NOT to do the calculation. Aside from nonsymmetric density (and shape) distributions (the oceans come to mind), the calculation must also ignore the TEMPERATURE during the trip. Even assuming vacuum, the radiant energy would be a problem. Then we have spin. We have evidence that the spin isn't uniform - that the Earth, big surprize, isn't a rigid rotating spheroid (ellipsoid). Finally, it isn't a question of how you dig a hold through the Earth so much as how you could possibly keep it open. You say potato, I saye potatoe...We haven't even got to 15 km, for gosh sakes! And we talk about 12,500 km?? Oh, the hubris.
impatricko
1 / 5 (1) Mar 27, 2015
I'm not a physicist, but how is it that all of the geniuses here don't get the idea that once you pass the center you'd be falling "up", start decellerating, and then start falling "down" again. You would then ocillate back and forth closer and closer to the center, (similar to the effect gravity has on a bouncing ball) until you stopped in the center

I'm dying for someone to tell me how this train of thought is wrong.
OceanDeep
not rated yet Mar 28, 2015
I'm glad someone narrowed down the time. Next maybe he can refine the angels on the head of a pin calculation.

Seriously, though, it's good to see someone take a problem and think it through in this much detail. It's a nice counterpoint to the typical thoughtlessness of most behavior these days.
OceanDeep
not rated yet Mar 28, 2015
PS I need to read the paper, because I also wonder whether it make a difference in falling through the rotating Earth from the poles, in between, or at the equator. Does it?
loneislander
not rated yet Mar 28, 2015
[Question for the boffins] Taking the two results, one which accurately maps density changes and one which simply ignores mass 'above your head' and uses surface gravity for the entire calculation, am I misled when I feel there is a natural connection between the stratification of densities (of whatever materials happen to be available) when building an 'earth' and the force of gravity being what determines that distribution and so we ought to expect that the surface gravity ought to be -more or less- some average of all the densities?

If one were to mathematically 'build' another planet with an arbitrary availability of material densities, it seems somehow intuitive that they ought to show a similar correlation. One ought to be able to prove an upper limit in the variance between the two. (a ball of pure hydrogen would leave the two equations with precisely the same result)
jonesdenson
not rated yet Mar 28, 2015
AS you fall wouldn't acceleration reach a cap? Once you reach the max speed where friction resistence and gravitational pull are equalized, how could you continue to accelerate?
To assume that you could reach the other side of the planet is an argument for perpetual motion, which is impossible.
Also in the case of falling down the first half, friction is working to resists the fall, while gravity is working to propell you to the other end; but after the first half both forces are working to keep you down.
vidyunmaya
1 / 5 (1) Mar 29, 2015
Sub; Wastage of Talents
Appeal : Please do no dig a Graveyard hole in the earth to verify which calculations are right or wrong.
Please put Energies on "How to save earth planet and life Support'
adam_russell_9615
not rated yet Mar 29, 2015
PS I need to read the paper, because I also wonder whether it make a difference in falling through the rotating Earth from the poles, in between, or at the equator. Does it?


If you tried it at the equator you would hit the side of the hole as the earth rotates and you do not.
Mayday
not rated yet Mar 29, 2015
My calculations indicate that no matter the care taken, some hole construction debris will inevitably be left at the center. I would strongly advise against jumping into said hole.
Ducklet
not rated yet Mar 29, 2015
My calculations indicate air resistance can't be disregarded. The terminal velocity will decrease during the fall towards center, after quickly having reached near-surface terminal velocity. Because everything will settle at the center the hole will immediately begin to fill up with airborne dust and any debris. A person leaping into the hole will hit this debris and add to it. Even if it were kept hypothetically clean and cool and structurally sound, reaching the opposite surface would still be a physical impossibility.

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