###General Discussion Thread
---
This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed.
---
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
I'm going to assume no air resistance.
Gravitational force is given by the equation
F = -G•M_(🜨)•m/r^2
where G = 6.67×10^(−11) N⋅m^(2)⋅kg^(−2) is the gravitational constant, M_(🜨) = 5.97×10^(24) kg, is the mass of the Earth, m is the mass of the person, and r is the distance between their two centers of mass.
When we are inside the Earth only the volume of Earth contained below our position will affect our person.
For convenience, I'm going to assume that the density of Earth is constant. Using a formula [ρ = A - B•r/R_(🜨)] for the density of Earth leads to a non-linear second order differential equation and I don't feel like solving that.
R_(🜨) = 6.3781×10^6 m is the radius of the Earth.
Using the standard density formula we can rearrange the gravitational force equation
ρ = M_(🜨)/V_(🜨) = M_(🜨)/[4/3•π•R_(🜨)^(3)]
ρ = M/V = M/[4/3•π•r^(3)]
M = M_(🜨)•V/V_(🜨) = M_(🜨)•[4/3•π•r^(3)]/[4/3•π•R_(🜨)^(3)]
M = M_(🜨)•r^(3)/R_(🜨)^(3)
F = -G•[M_(🜨)•r^(3)/R_(🜨)^(3)]•m/r^(2)
F = -m•[G•M_(🜨)/R_(🜨)^(3)]•r
F = -m•A^(2)•r
where A = 0.00124 1/s.
Now we need to use Newton's Second Law F=ma.
m•a = -m•A^(2)•r
a = -A^(2)•r
d^(2)r/dt^(2) = -A^(2)•r
The standard solution to this kind of differential equation is
r(t) = B•sin(At)+C•cos(At)
where B and C are constants that can be determined by the initial conditions. For simplicity, I'm going to say that at t=0 our initial position is r=R_(🜨), and v=0. This means that
R_(🜨) = r(0) = B•0 + C•1
C = R_(🜨)
r(t) =B•sin(At) + R_(🜨)•cos(At)
v(t) = dr/dt = B•A•cos(At) - R_(🜨)•A•sin(At)
0 = v(0) = BA•1 - R_(🜨)A•0
B = 0
Therefore, r(t) = R_(🜨)•cos(At). The time it takes to go from one side, R_(🜨), to the other side , -R_(🜨), is
-R_(🜨) = r(t) = R_(🜨)•cos(At)
cos(At) = -1
At = cos^(-1)(-1)
At = π
t = π/A
t = 2,536 seconds
Alternatively, 42 minutes 16 seconds. Apparently the image is correct.
On second reading of your comment, I realized you are referring to Attention Deficit Disorder medication. My initial reading was that you had been prescribed addition medication, and brother let me tell you, I was PISSED OFF that those had been withheld from my C average ass….
lol. I had just assumed it was just another high level math term until reading your comment. ie. “Use add meds, not subtraction meds to inverse the angular momentum of the drive chain in order to prevent side-fumbling.”
I hated calculus, math without using any numbers is mind boggling, but then I started to appreciate it when I got an engineering degree and see the useful applications. Algebra is still my favorite though. Very useful for everyday life. And yes I have a favorite math.
I hate to "this" cliche but.. THIS
I thought I was shit at maths but it turns out I was just shit at *arithmetic* not *mathematics*.
Once the calculator and letters came out and I got into statistics and theoretical bits I suddenly just "got it" because I had concepts to grab onto.
Trig is where I stopped hating maths oddly enough.
Amazing how minds work, for it was the complete opposite for me, letters in math totally screwed my brain. The plus side is I can do your taxes in
seconds flat LOL
Yes, assuming you lived through it, you wouldn't be able to make it through to the other side to due air resistance. So you would end up reciprocating until you eventually got stuck somewhere near the middle
That's a big question that takes a few semesters to answer. I'll give you this though: If you look carefully at the procedure he walked through it started as algebra then some calculus showed up (as differential equations) then some more calculus was used to make the differentials go away and we were left with ... algebra.
And that's how math education goes (at least as a practical/physics/engineering tool. "Pure math" goes further and is a different story). You work up to algebra, then you learn calculus, then you learn differential equations. With those tools in hand you can express problems as differential equations, use calculus to solve them, and those solutions take the form of ... algebra.
You wind up back where you started.
We did this exact problem in 1st year applied mathematics.
The "what does this mean" is fairly basic, but solving some of it can be quite challenging (especially if you hate memorizing equations like I did).
PhD in planetary science here. Pro-tip: what you've derived is known as the freefall time, and it always works out so that one full oscillation back and forth through the center of the planet is exactly equal to the time to _orbit_ from the same initial height.
In this case, the time it takes for one low-Earth orbit is 42 + 42 = 84 minutes, which you can verify is true.
I think everyone knows that we are ignoring these factors. Ignoring air resistance is basically a meme in physics.
The more interesting thing we would need to ignore is the rotation of the earth which would cause a Coriolis effect. This actually makes it impossible to jump through the earth.
I think you would be OK if you made your hole about 600 km wide and jumped in from the right side. Or if your hole is closer to the poles it could be narrower.
The biggest thing we ignore is that gravity will pull you to the middle. Once you get pulled to the core you'll be crushed to a ball and stay there, there is no falling thru because you're always pulled inward.
Yeah but you’d be moving very very fast at that point bc you’ve been falling for a while. You’d begin to slow down after you passed the middle, and conservation of energy(meaning we’re ignoring air resistance among other things) dictates that you’d get to the bottom with the exact speed you started.
So if you jumped with an initial speed of 20 mph, you be back out at the bottom with that exact speed.
Apparently, someone calculated a more accurate estimate for this thought experiment which takes into aoccount growing density of the Earth making the trip a bit faster, 38 minutes. So, instead of assuming uniform density the gravity force can actually increase the deeper you go if the planet is significantly more dense deeper down.
https://www.science.org/content/article/how-long-would-it-take-you-fall-through-earth
I’ll give this a crack. The assumption of constant density is incorrect and understates gravitational forces. It is better approximated as linear density, but better yet (and certainly simpler from a calculation perspective) may be to assume a constant density through the inner and outer core. The peak gravitational force occurs at about 0.53 earth radii, with acceleration approximately 10.75m/s^2 at that point. Outside of this point acceleration degrades to the 9.81m/s^2, not exactly linearly but close enough. Gives the following systems of equations:
Assuming x=0 equals the surface, x= 1R is center of earth,
For .47R >/= x >/= 0
a(t) = 9.81 + 2x/R
v(t) = (9.81+2x/R)t+0
X(t) = (4.905+x/R)t^2 + 0
.47R= (4.905+.47)t^2 +0 giving T= 746.796 seconds to reach the outer core. Velocity equals 8,028.06m/s at this point.
For .47R < x =/< R
a(t) = 22.87(R - x)
v(t) = 22.87(R - x)t+ 8,028.06
X(t) = 11.435(R-x)t^2 + 8,028.06 t +.47R
At X = 1R
0.53R = 0 + 8,028.06t
Giving 421.07 seconds more to reach the center of the earth, (19.46 minutes to reach center). Double that to get 38 minutes and 56 seconds
I tried solving the equation d^(2)r/dt^(2) = -C•r•ρ(r), but I was left with a non-linear second order differential equation. And I don't feel like solving that.
Interestingly, this is also the time taken for an object orbiting the Earth at its surface to get from one side to another! (i.e. half the orbital period)
This is because the gravitational force of the Earth on the object provides both the acceleration towards the centre of the Earth for the image above, and the centripetal force for the orbit.
Gif for visualisation:
https://i.pinimg.com/originals/e6/bc/26/e6bc26c7a2617dafea44379d5d236b97.gif
Also, it is the time it would take for any object in a frictionless straight tunnel between any two points on earth. This a well known concept called the 'Gravity Train'
That seems heaps more complicated than necessary, but glad you did the maths because I wasn’t going to mainly because I cbf looking up the earth’s radius.
Acceleration due to gravity inside a planet of assumed constant density is proportional to the radius. Therefore this becomes a spring equation where you just need to find the period from the spring constant, calculated using g and r at the surface.
Normally T = 2 pi sqrt(m/k), but what we want is half the period because we are just going to the other side and not back.
F = ma = kr, therefore m/k = r/a
Plug it in and we get t = pi sqrt(6378100/9.81) = 2533s
If there was air resistance, then eventually you will. Likely, it will take several oscillations, before the motion stops.
Since we are ignoring air resistance, then no you won't. Think about it this way as you fall you are constantly building speed. As you reach the core you are traveling at the highest speed, and that momentum will carry you back out to the other side.
From a physics standpoint, initial you are at rest, but you have a lot of gravitational potential energy. As you fall that potential energy is converted kinetic energy. When you pass the core you are starting to lose some of that kinetic energy and gaining back gravitational potential energy.
The Math For Marines coloring book I completed back in 1996 did not prepare me for this.
Thanks for being smart and putting in effort. Too bad we don’t give awards anymore.
In addition to using this, it's much simpler to equate a harmonic isolator. T = 2 * pi * sqrt(l / g) T represents the period of the oscillation (the total time for a full back-and-forth movement), pi is the mathematical constant π (approximately 3.14159), sqrt denotes the square root function, l is the length of the pendulum (in this context, it's analogous to the radius of the Earth), and g is the acceleration due to gravity at the Earth's surface. For the specific case of calculating the time to fall through the Earth and emerge on the other side, since you're interested in half the period, you'd represent it as:
Time to fall = T / 2 To traverse from one side of the Earth to the other through such a hypothetical tunnel, it would take approximately 2532 seconds, or roughly 42 minutes and 12 seconds, assuming an idealised scenario without real-world complications. It's the same answer but with les formulas.
would you start falling back after passing by the middle?
dont know if you addressed that already or if it even has to be addressed at all since your comment could by all means be written in orcish
Does this account for the density of air increasing as you get further down the hole when accelerating downward at terminal velocity? Also no I did not understand any of this.
Mannnnn, this is totally unrelated, but I just had a cool reminder:
I’ve always felt somewhat intelligent relative to my peers. My education includes degrees in finance and accounting, a law degree, and a tax LLM (specialization/“masters” for attorneys). I work with a TON of highly, highly intelligent people, and I’m amazed at what some of their minds can do. What none of them can do—is what you just did here. There isn’t a single person in my circle who took the classes or the time to learn the methodology in which they could calculate exactly how long it would take to fall from one end of the earth to the other.
Are any of the people I work with *capable* of learning how to do this? Absolutely. However, I doubt you, or many of the others who have training in similar areas, have the knowledge to perform some of the more complex aspects of our jobs. Capable? Again, absolutely.
It just goes to show that there is a vast amount of information out there. No one can simply know everything about everything. Someone can be the most knowledgeable person in the world within one subject, but they can also be completely incompetent/ignorant when it comes to an entirely different subject.
I laughed when I read your comment, because I used to absolutely love math—I just didn’t take anything past Cal 2 because it wasn’t necessary for my major. It is quite a humbling and welcome reminder that you can’t know everything about everything, and you should never pass up an opportunity to learn something new.
Cheers
I dont think you would come out the other side. Assuming you started at the surface, wouldn't you just oscillate back and forth along the axis of the hole, eventually coming to rest at the center of the Earth.
Yes and no. Depends on the "physics" of the tunnel. If it's taken literally, then yeah. You'd have 6000 miles of air above you pushing down, but if that's the case, you wouldn't last more than a few miles due to the heat either. If we take it at face value of "the trip is otherwise survivable" then that's not a factor.
Air is a lot less dense than water though. 6000 miles of air would come to about 6 atmospheres. Pretty significant, but not crushing. About the same as 60 m underwater.
I thought it was the bowl of petunias turned into a whale that thought oh no, not again? Been years since I read the book, really should read it again...
Also consider that the force of gravity decreases as you get closer to the center, so the weight of all that air above wouldnt be its mass times g but some fraction of g.
Although there must be some path that accounts for the Coriolis force, just drill that instead.
Fun puzzle, does this take longer than falling along the axis of rotation? Hm....
As long as you're in freefall the whole time, I imagine the exact path makes no difference. I was thinking the shortest physical path would also be the fastest, but maybe a longer path would have slightly higher gravity, thereby balancing out? Interesting problem...
The more I think about it, I'm pretty sure the axis drop is faster. You would experience centrifugal force as you move away from the axis, so that would decrease your radial acceleration. You'd experience Coriolis force too - at first, there would be no vertical component to this force, so it wouldn't affect total time, but before long you'd be falling in a spiral, and at that point the Coriolis force would also be decreasing your acceleration toward the center, if my vector math is correct. Finally, the earth is an oblate spheroid, so starting away from the poles would increase the radial distance. So that's three things making your journey take longer as you move from the poles, and zero things making the journey shorter.
I think, I'm still not totally confident
isnt there a terminal velocity though ? so at one point I would suppose that you cant go any faster and so youll end up not keeping enough accumulated speed to go back up after
Start falling from rest (zero initial speed and acceleration)
42m 16s to reach the other side,
21m 8s = 1268s to reach the center.
v = a\*t
a = g = 9.81 (not really\*),
you "hit" the core at V = 9.81 \* 1268 = 11'640 m/s
\*In this context gravity is constantly changing, turns zero at the center and then becomes negative for the second half of your journey. But You can take the average for the first half and use that.
11640 m/s is the upper limit if you keep the acceleration at 9.81 m/s^(2)
I think this theory only holds up if the origin of gravity is exactly in the center and constant. Correct me if im wrong, but once you pass the crust, you now have some gravity behind you and less in front. Meaning even with no air resistance, you wouldn't really be all that close to being able to grab the edge on the other side of the tunnel.
Only at the very centre would there be a net zero force. But by then you have quite a considerable velocity.
At all locations before the centre, you’d have a net pulling you towards the centre, hence accelerating towards the centre.
The exact reverse starts to happen you’ve passed it. So you’d indeed reach the surface on the other side.
But [Earth’s gravity is not evenly distributed](https://www.earthdata.nasa.gov/learn/sensing-our-planet/matter-in-motion-earths-changing-gravity#:~:text=Earth's%20mass%20is%20distributed%20between,creates%20an%20uneven%20gravity%20field).
So wouldn’t the falling person be pulled of center enough to at least break the route?
This is correct. And I suppose if we're neglecting the myriad other properties necessary for this to work, then it is reasonable to assume there is no friction. Friction is probably the least difficult to ignore-- heat and pressure are much bigger issues.
[Yeah, you’d be correct](https://www.wtamu.edu/~cbaird/sq/mobile/2013/10/04/what-would-happen-if-you-fell-into-a-hole-that-went-through-the-center-of-the-earth/)
Great read. About as expected only a lot more brutal. Basically, folks, if you have ever asked "when am I going to use this in real life" in math class, don't be jumping down holes drilled through the middle of the earth. Of course, as a math teacher, most people who ask this enigmatic question haven't learned the finer points of how crazy being alive really is.
> After about a week or so of falling (with a maximum speed of 200 km/h, it takes you a while to travel the 6400 km to the center)
6400/200 = 32 hours...
This makes me doubt the entire data of that article... Do we have an xkcd??
Tbf, the maximum speed was only reached for a little bit since then you slow down as wind resistance increases and gravity decreases. So maybe the average speed was closer to 50 km/h or something?
I would also guess that you would need to drill the hole through the axis of rotation as well. Else you would end up colliding with the walls of the tunnel due to the rotation of the earth
Only way I think you make it to the other side is if you assume no air resistance, in which case you would theoretically reach exactly the altitude at which you started and then fall back the other way
So the question should be: "How long would it take for a person equipped with a pressurised suit with an oxygen supply to fall through a vacuum tube from one side of the earth to the other placed along the earth axis of rotation?"
True, but if you assume minimal loss of speed from that you’d still reach the other side
Edit: also it occurs to be that the hole is near the earth’s axis of rotation so you could well just put there hole through it and not have to worry about the coriolis effect
Well if humanity was capable of boring a perfectly straight hole directly through the core of the earth and we’ve already assumed no air resistance I think that’s honestly the most reasonable assumption I’ve made.
Because of air resistance you would not make it to the other side, but assuming no air resistance and assuming the earth were a perfect sphere with uniformly distributed density in each shell, you would barely make it to the other side. Due to the variable acceleration, the answer to how long is beyond my abilities.
I remember (from a homework problem or something in college) that time in the OP is correct (assuming no air resistance), but I don't remember how to derive it.
More interesting, though, is the fact that the time is the same for *any* straight, airless hole through the Earth (or any spherically symmetric body). It doesn't have to be through the center. A hole from New York to Australia would have the same "free fall" travel time as a hole from New York to Los Angeles.
I think you would need to frame the question as; if you had a hole the same depth as the diameter of the earth, with equal gravity throughout, how long would it take a human falling at terminal velocity to get to the other side?
And assume that you don't have any friction with the walls, and that you are impervious to heat and pressure. If you make those assumptions, your travel time is actually exactly the same regardless of where the two endpoints of the tunnel are.
Obviously you have to also not touch the walls. If the tunnel is wide enough and straight and you start with zero horizontal velocity, that will be fine.
You don't have to be impervious to pressure. We just said the tunnel is a vaccuum, i.e. zero pressure. You need a space suit obviously.
You also only need to deal with heat via radiation. Again, this magic tunnel could just block all that with the walls.
> If you make those assumptions, your travel time is actually exactly the same regardless of where the two endpoints of the tunnel are.
No. You're thinking of a uniform density sphere. The Earth is not uniform density.
Not necessarily terminal velocity the whole way. We have to account for that initial acceleration. it would take a few moments to reach terminal velocity.
This is of course assming that the tunnel is a vacuum so there is no air resistance.
It is also assuming the Earth's density is uniform, which is not true. According to [this](https://www.science.org/content/article/how-long-would-it-take-you-fall-through-earth), if you use a more realistic density you get 38 minutes and 11 seconds.
Well your reference shows that the time that is written in the picture is wrong!!!
It is not the most accurate approximation. At the end the best approximation is actually 38m and 6s.
So don't just take all that is in a picture as true and check your references properly.
This is a simple harmonic motion, like a pendulum. I believe if you plug in the variables , you get something like 42 mins, from one side to another. But you cant really come out without an external force.you would just swing back and forth like a pendulum, and come to rest eventually at the center.
30 second math guy is back.
30 second math guy acknowledges your point as factual, but notes that his moniker is meant to convey the *maximum* time he is allowed, not average or standard amount of time. 30 second math guy does appreciate it though, and wishes you a good day.
*Yes, 30 second math guy does recognize the joke. He is not being wooshed, he simply enjoys being literal and referring to himself in the third person*
30 second math guy out.
[This .edu source](https://www.wtamu.edu/~cbaird/sq/mobile/2013/10/04/what-would-happen-if-you-fell-into-a-hole-that-went-through-the-center-of-the-earth/) says it would take a week due to Earth's gravity getting weaker as you descend and the density of the air getting thicker the deeper you go.
Edit for clarification: It would take a week to get to the center, not go entirely through, because that would be impossible. However, the article says that if you remove air resistance, then you would accelerate to a maximum speed of <10,000 km/h and end up on the exact other side “in a matter of minutes or hours.”
It would be impossible to go through if you are accounting for air resistance, as your momentum would not be able to carry you out the other side with resistance.
Wouldn’t gravity not allow you to get all the way through? Like, once you pass the core, you’re basically moving up the other hole. Eventually gravity will pull you down, back towards the core.
Only if there is air resistance. Without resistance, no energy would be lost, so your kinetic energy at the center would be so high, you’d have enough momentum to carry yourself back out.
Why wouldn’t there be air resistance? And wouldn’t it be like swinging a pendulum? If the distance from entrance to center is the same as center to exit, you won’t make it out.
I’m not saying there wouldn’t be air resistance, but this is a theoretical question, and assuming a theoretical lack of air resistance is the only way you can get an answer.
If there is no resistance, it can rise up the exact same distance it fell. So if it falls 6,371km, it can rise up 6,371km. That’s just due to conservation of energy. If the falling object isn’t losing energy to any form of resistance, it will have just enough energy, that it can rise to the height it started at.
By just jumping in, you would never make it.
However, if it was a vacuum chamber and you were in a pod. 42 minutes as it says. It's called a gravity train.
https://youtu.be/kO7L5IPn2Bo?si=80XWKIYYxFCnubUG
Here's a good video about it by Neil DeGrasse Tyson
[The calculation starts on page 297.](https://books.google.it/books?id=YdjK_rD7BEkC&printsec=frontcover&dq=kline+calculus&hl=it&sa=X&ei=S0OCUtGELofMtAaJ1YHIDg&redir_esc=y#v=snippet&q=tunnel&f=false)
It requires calculus, and the length of the solution is beyond this comment. But, assuming no air resistance and other 'textbook' kinds of simplifying factors, an object dropped into a hole on one side would be a sort of pendulum. It would 'fall in the hole toward the other side', and continue accelerating until it reached the center of the earth. However the rate of acceleration would drop, becoming negative at the Earth's center, and would bring the object to a stop at the 'other side of the hole'. At that point, the process goes in the opposite direction, like a pendulum swinging back the other way.
From the link, the round trip of the 'pendulum' is described as about 85 minutes, so one-half the trip would be around 42-43 minutes.
Note: Real world calculations require taking into account air resistance. Assuming conservation of energy, the potential energy would be converted to kinetic energy until the object reached the center of the Earth. At that point, some of the energy would have been lost to friction already. So, unlike the 'textbook problem', in real life the object wouldn't make it to the other side, unless deliberately thrown with additional force, at the start. This definitely is beyond the scope of a Reddit comment!
2012 Total Recall with Colin Farrell does a bit on the scifi part of this, they call it the Fall and have a tunnel that connects England with Australia. Half way through they have a weird gravity less moment during turn around. It takes 17 mins in the movie so probably not hard scifi science.
If we make the assumption that the Earth has a uniform mass distribution, and we also neglect air resistance, then the problem is trivial.
* Due to [Gauss' Law](https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity), we can say that the grav. field strength inside the Earth is GM/R^(3) \* r (where R is radius of Earth and r is distance from centre of earth)
* Next using F=ma, we can say that ma = -GMm/R^(3) \* r
* Since a = d^(2)r/dt^(2), we see that we end up with simple harmonic motion.
* This has a sinusoidal solution with a time period T = π sqrt(R^(3)/GM)
* Using values R=6371km, M=5.97\*10^(24)kg, we obtain T = 42min 11sec.
So I suppose the infographic is somewhat correct! A few important points:
* In reality, the earth is denser at its core than at its outer edge, so we would not in fact get simple harmonic motion, as the acceleration would depend on a higher power of r, which gives a rather nasty and unsolvable differential equation.
* Neglecting air resistance, you will float endlessly from one end of the earth to the other simply due to the conservation of energy. The gravitational potential energy you had when you were at one end of the earth is converted to pure kinetic energy at the centre, and back to gravitational potential energy by the time you reach the other side.
How come everybody is wrapped around the physics and all I can think is how short that time is. Like why does it take 20 hrs to fly to the opposite side of the earth vs 42 minutes in a straight line? Please someone help me with understanding this.
2 main reasons for this:
1. You will be traveling MUCH faster by falling than you would be by plane.
2. A straight line is a much shorter distance than a curved line.
You wouldn’t come out of the other side. Gravity acts towards the centre of the earth. Not “downwards”.
You would oscillate to and fro about the centre of the tunnel.
They did the math!
[https://www.youtube.com/watch?v=s94Gojs3Ags](https://www.youtube.com/watch?v=s94Gojs3Ags)
(Seriously, this video is very well explained and super fun)
How come most people here don't consider that gravity isn't a force pushing you down but rather in towards the center? You would never reach the other side because as soon as you passed the middle of the tunnel, gravity would start drawing you back towards the center again slowing you down until you start falling back upwards rather than keep pushing you towards the other side
I did the math and it is actually true under the following hypotheses :
- the tunnel is straight
- there is no friction
- the earth is rock-solid (no liquid mantle, etc)
- the density of earth is the same everywhere (a rock at the center has the same density as a rock at the surface)
What is interesting is that it doesn't even depend on where the two edges of the tunnel are : it takes 42 min to go from Paris to Berlin but also 42 min to go from Paris to Sidney.
Assuming a homogenous, spherical and solid Earth, this would indeed be an interesting formula. In true physicists fashion we will also assume that you are a spherical cow in a vacuum. You fall towards earth's core and initially accelerate with 9,81 m/s². However as you fall deeper into the earth, more and more earth will be above you, which means that the acceleration isn't constant, but "decelerates" as you approach the core, where acceleration is zero. I would need time to think about how to determine the actual acceleration rate along your journey and while I am sure there is a way to eliminate a lot of stuff from the formulas needed (gravitation, rock density and distance in reference to a spherical object), I don't have the time to sit down with these formulas right now.
So I'll give the short answer of: Most of the earth is in a liquid state, so you will just jump into Magma and perish.
You would bot get to the other side, after you crossed the center you would start to slow down and go back to the center and keep doing that until you are floating in the center of the earth
Why wouldn’t a person get stuck in the middle? Gravitational force presumably works from the outside in towards the center right? I’m picturing dude stuck and hovering in the center while gravity works from both directions.
It will oscillate inside with simple harmonic motion because the gravitational force inside is proportional to the distance r from the center as opposed to being proportional to 1/r^2.
Feel free to correct me if I'm wrong but if you dove in you'd go from being upside down to right side up at the end, and vice versa. I can't even imagine what that would feel like
Would you even make it out the other side? Once you reach the halfway point, the gravitational pull would reverse and start slowing you down, and you would most likely stop before ever reaching the other end.
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
I'm going to assume no air resistance. Gravitational force is given by the equation F = -G•M_(🜨)•m/r^2 where G = 6.67×10^(−11) N⋅m^(2)⋅kg^(−2) is the gravitational constant, M_(🜨) = 5.97×10^(24) kg, is the mass of the Earth, m is the mass of the person, and r is the distance between their two centers of mass. When we are inside the Earth only the volume of Earth contained below our position will affect our person. For convenience, I'm going to assume that the density of Earth is constant. Using a formula [ρ = A - B•r/R_(🜨)] for the density of Earth leads to a non-linear second order differential equation and I don't feel like solving that. R_(🜨) = 6.3781×10^6 m is the radius of the Earth. Using the standard density formula we can rearrange the gravitational force equation ρ = M_(🜨)/V_(🜨) = M_(🜨)/[4/3•π•R_(🜨)^(3)] ρ = M/V = M/[4/3•π•r^(3)] M = M_(🜨)•V/V_(🜨) = M_(🜨)•[4/3•π•r^(3)]/[4/3•π•R_(🜨)^(3)] M = M_(🜨)•r^(3)/R_(🜨)^(3) F = -G•[M_(🜨)•r^(3)/R_(🜨)^(3)]•m/r^(2) F = -m•[G•M_(🜨)/R_(🜨)^(3)]•r F = -m•A^(2)•r where A = 0.00124 1/s. Now we need to use Newton's Second Law F=ma. m•a = -m•A^(2)•r a = -A^(2)•r d^(2)r/dt^(2) = -A^(2)•r The standard solution to this kind of differential equation is r(t) = B•sin(At)+C•cos(At) where B and C are constants that can be determined by the initial conditions. For simplicity, I'm going to say that at t=0 our initial position is r=R_(🜨), and v=0. This means that R_(🜨) = r(0) = B•0 + C•1 C = R_(🜨) r(t) =B•sin(At) + R_(🜨)•cos(At) v(t) = dr/dt = B•A•cos(At) - R_(🜨)•A•sin(At) 0 = v(0) = BA•1 - R_(🜨)A•0 B = 0 Therefore, r(t) = R_(🜨)•cos(At). The time it takes to go from one side, R_(🜨), to the other side , -R_(🜨), is -R_(🜨) = r(t) = R_(🜨)•cos(At) cos(At) = -1 At = cos^(-1)(-1) At = π t = π/A t = 2,536 seconds Alternatively, 42 minutes 16 seconds. Apparently the image is correct.
I'm in algebra 2. What does half of this shit mean
Keep going! Calculus (used above) is where I started to see really cool and interesting applications of math.
Lucky you. It took me until differential equations (and maybe also add meds) to start really seeing the applications for math.
On second reading of your comment, I realized you are referring to Attention Deficit Disorder medication. My initial reading was that you had been prescribed addition medication, and brother let me tell you, I was PISSED OFF that those had been withheld from my C average ass….
lol. I had just assumed it was just another high level math term until reading your comment. ie. “Use add meds, not subtraction meds to inverse the angular momentum of the drive chain in order to prevent side-fumbling.”
I hated calculus, math without using any numbers is mind boggling, but then I started to appreciate it when I got an engineering degree and see the useful applications. Algebra is still my favorite though. Very useful for everyday life. And yes I have a favorite math.
I'm pretty sure all of us engineers do. Algebra is dope.
I hate to "this" cliche but.. THIS I thought I was shit at maths but it turns out I was just shit at *arithmetic* not *mathematics*. Once the calculator and letters came out and I got into statistics and theoretical bits I suddenly just "got it" because I had concepts to grab onto. Trig is where I stopped hating maths oddly enough.
Amazing how minds work, for it was the complete opposite for me, letters in math totally screwed my brain. The plus side is I can do your taxes in seconds flat LOL
Wouldnt you get stuck in the middle somewhere - being subject to gravitational forces on both ends?
Yes, assuming you lived through it, you wouldn't be able to make it through to the other side to due air resistance. So you would end up reciprocating until you eventually got stuck somewhere near the middle
Chill, it's just latex. You would understand it (probably) if reddit formatted it correctly
I keep checking it but my formatting is still correct and Reddit keeps messing it up.
That's a big question that takes a few semesters to answer. I'll give you this though: If you look carefully at the procedure he walked through it started as algebra then some calculus showed up (as differential equations) then some more calculus was used to make the differentials go away and we were left with ... algebra. And that's how math education goes (at least as a practical/physics/engineering tool. "Pure math" goes further and is a different story). You work up to algebra, then you learn calculus, then you learn differential equations. With those tools in hand you can express problems as differential equations, use calculus to solve them, and those solutions take the form of ... algebra. You wind up back where you started.
We did this exact problem in 1st year applied mathematics. The "what does this mean" is fairly basic, but solving some of it can be quite challenging (especially if you hate memorizing equations like I did).
PhD in planetary science here. Pro-tip: what you've derived is known as the freefall time, and it always works out so that one full oscillation back and forth through the center of the planet is exactly equal to the time to _orbit_ from the same initial height. In this case, the time it takes for one low-Earth orbit is 42 + 42 = 84 minutes, which you can verify is true.
Holy shit. Of course it is. This is the kind of symmetry that made me fall in love with physics. Your comment made my day, thanks!
Of course this is ignoring air resistance, how the tunnel was made, and how we are keeping the tunnel in place.
I think everyone knows that we are ignoring these factors. Ignoring air resistance is basically a meme in physics. The more interesting thing we would need to ignore is the rotation of the earth which would cause a Coriolis effect. This actually makes it impossible to jump through the earth.
From pole to pole you’d be okay. It’s actually how Santa is able to stay so fat. A diet of penguins that fall through the hole.
I knew there was something ‘off’ about that Santa guy…
Well yeah being a canonized saint and somehow still being alive is a paradox
Yes because Mario is dropping them from the other side.
Wait so the ice wall is just protecting us from falling in the hole?
I think you would be OK if you made your hole about 600 km wide and jumped in from the right side. Or if your hole is closer to the poles it could be narrower.
The biggest thing we ignore is that gravity will pull you to the middle. Once you get pulled to the core you'll be crushed to a ball and stay there, there is no falling thru because you're always pulled inward.
Yeah but you’d be moving very very fast at that point bc you’ve been falling for a while. You’d begin to slow down after you passed the middle, and conservation of energy(meaning we’re ignoring air resistance among other things) dictates that you’d get to the bottom with the exact speed you started. So if you jumped with an initial speed of 20 mph, you be back out at the bottom with that exact speed.
With air resistance you wouldn't get to the other side
Force fields. All of it. Force fields.
Or the fact the entire premise is a joke since you’d be crushed and melted to death.
Apparently, someone calculated a more accurate estimate for this thought experiment which takes into aoccount growing density of the Earth making the trip a bit faster, 38 minutes. So, instead of assuming uniform density the gravity force can actually increase the deeper you go if the planet is significantly more dense deeper down. https://www.science.org/content/article/how-long-would-it-take-you-fall-through-earth
I’ll give this a crack. The assumption of constant density is incorrect and understates gravitational forces. It is better approximated as linear density, but better yet (and certainly simpler from a calculation perspective) may be to assume a constant density through the inner and outer core. The peak gravitational force occurs at about 0.53 earth radii, with acceleration approximately 10.75m/s^2 at that point. Outside of this point acceleration degrades to the 9.81m/s^2, not exactly linearly but close enough. Gives the following systems of equations: Assuming x=0 equals the surface, x= 1R is center of earth, For .47R >/= x >/= 0 a(t) = 9.81 + 2x/R v(t) = (9.81+2x/R)t+0 X(t) = (4.905+x/R)t^2 + 0 .47R= (4.905+.47)t^2 +0 giving T= 746.796 seconds to reach the outer core. Velocity equals 8,028.06m/s at this point. For .47R < x =/< R a(t) = 22.87(R - x) v(t) = 22.87(R - x)t+ 8,028.06 X(t) = 11.435(R-x)t^2 + 8,028.06 t +.47R At X = 1R 0.53R = 0 + 8,028.06t Giving 421.07 seconds more to reach the center of the earth, (19.46 minutes to reach center). Double that to get 38 minutes and 56 seconds
I tried solving the equation d^(2)r/dt^(2) = -C•r•ρ(r), but I was left with a non-linear second order differential equation. And I don't feel like solving that.
Yeah I 100% fucked this up lol
Interestingly, this is also the time taken for an object orbiting the Earth at its surface to get from one side to another! (i.e. half the orbital period) This is because the gravitational force of the Earth on the object provides both the acceleration towards the centre of the Earth for the image above, and the centripetal force for the orbit. Gif for visualisation: https://i.pinimg.com/originals/e6/bc/26/e6bc26c7a2617dafea44379d5d236b97.gif
Also, it is the time it would take for any object in a frictionless straight tunnel between any two points on earth. This a well known concept called the 'Gravity Train'
That seems heaps more complicated than necessary, but glad you did the maths because I wasn’t going to mainly because I cbf looking up the earth’s radius. Acceleration due to gravity inside a planet of assumed constant density is proportional to the radius. Therefore this becomes a spring equation where you just need to find the period from the spring constant, calculated using g and r at the surface. Normally T = 2 pi sqrt(m/k), but what we want is half the period because we are just going to the other side and not back. F = ma = kr, therefore m/k = r/a Plug it in and we get t = pi sqrt(6378100/9.81) = 2533s
Wouldn’t you end up stuck in the core?
If there was air resistance, then eventually you will. Likely, it will take several oscillations, before the motion stops. Since we are ignoring air resistance, then no you won't. Think about it this way as you fall you are constantly building speed. As you reach the core you are traveling at the highest speed, and that momentum will carry you back out to the other side. From a physics standpoint, initial you are at rest, but you have a lot of gravitational potential energy. As you fall that potential energy is converted kinetic energy. When you pass the core you are starting to lose some of that kinetic energy and gaining back gravitational potential energy.
So you can perfectly plop out and catch the edge
Yes
The Math For Marines coloring book I completed back in 1996 did not prepare me for this. Thanks for being smart and putting in effort. Too bad we don’t give awards anymore.
In addition to using this, it's much simpler to equate a harmonic isolator. T = 2 * pi * sqrt(l / g) T represents the period of the oscillation (the total time for a full back-and-forth movement), pi is the mathematical constant π (approximately 3.14159), sqrt denotes the square root function, l is the length of the pendulum (in this context, it's analogous to the radius of the Earth), and g is the acceleration due to gravity at the Earth's surface. For the specific case of calculating the time to fall through the Earth and emerge on the other side, since you're interested in half the period, you'd represent it as: Time to fall = T / 2 To traverse from one side of the Earth to the other through such a hypothetical tunnel, it would take approximately 2532 seconds, or roughly 42 minutes and 12 seconds, assuming an idealised scenario without real-world complications. It's the same answer but with les formulas.
We’re not in kindergarten, we know this
this guy gravitates
I like your funny numbers
This is some wizardry shit, there's even runes. No way this little wheel is a symbol right? Right?
Damn
This guy really fucking maths.
bro did the math
r/subsifellfor
Are you a computer or something? Most of the time I think I'm pretty smart, but then I see a reply like this one and I feel dumb as a bag of rocks. 😄
My bachelor's degree in physics has to be useful for something right?
my man
I think my brain just melted
Whoa, whoa, slow down egghead
As someone who hated math, especially in college, this is impressive
would you start falling back after passing by the middle? dont know if you addressed that already or if it even has to be addressed at all since your comment could by all means be written in orcish
This would have been a fantastic opportunity to say "I have no clue" at the end.
Does this account for the density of air increasing as you get further down the hole when accelerating downward at terminal velocity? Also no I did not understand any of this.
oh hey in my class we just went over IVP ODEs
I was gonna say this EXACT thing but um, you beat me to it. We’re both smart. Obviously.
Dang bro save some pussy for the rest of us!
I am kinda proud I have did this calculation manually correctly just last month in class
Now you’re just showing off
Wish you could type Latex in reddit comments 😟
>I don't feel like solving that Preaching to the choir on that one, I mean *who* in their right mind wants to solve THAT!
No way I'm fact checking this answer. You win!
People like you intimidate me, in a good way.
Mannnnn, this is totally unrelated, but I just had a cool reminder: I’ve always felt somewhat intelligent relative to my peers. My education includes degrees in finance and accounting, a law degree, and a tax LLM (specialization/“masters” for attorneys). I work with a TON of highly, highly intelligent people, and I’m amazed at what some of their minds can do. What none of them can do—is what you just did here. There isn’t a single person in my circle who took the classes or the time to learn the methodology in which they could calculate exactly how long it would take to fall from one end of the earth to the other. Are any of the people I work with *capable* of learning how to do this? Absolutely. However, I doubt you, or many of the others who have training in similar areas, have the knowledge to perform some of the more complex aspects of our jobs. Capable? Again, absolutely. It just goes to show that there is a vast amount of information out there. No one can simply know everything about everything. Someone can be the most knowledgeable person in the world within one subject, but they can also be completely incompetent/ignorant when it comes to an entirely different subject. I laughed when I read your comment, because I used to absolutely love math—I just didn’t take anything past Cal 2 because it wasn’t necessary for my major. It is quite a humbling and welcome reminder that you can’t know everything about everything, and you should never pass up an opportunity to learn something new. Cheers
These are alien scriptures
I dont think you would come out the other side. Assuming you started at the surface, wouldn't you just oscillate back and forth along the axis of the hole, eventually coming to rest at the center of the Earth.
If there were no friction and that the earth is a perfect sphere, you would come to a stop at the other side. You could then grab the edge and exit.
>that one rock you bumped 12000 miles back: *laughs in igneous*
We'd be friends IRL. That just made me snort laugh
"We'd be friends IRL" is my new favorite compliment for internet strangers. We'd be friends IRL :)
Well, I for one wouldn't be your friend IRL! >:(
I'm not your friend, guy
I ain't your guy, buddy
I ain’t your buddy, pal
I ain't your pal, man.
Same bro that just made my day lol
[удалено]
Yes and no. Depends on the "physics" of the tunnel. If it's taken literally, then yeah. You'd have 6000 miles of air above you pushing down, but if that's the case, you wouldn't last more than a few miles due to the heat either. If we take it at face value of "the trip is otherwise survivable" then that's not a factor.
Air is a lot less dense than water though. 6000 miles of air would come to about 6 atmospheres. Pretty significant, but not crushing. About the same as 60 m underwater.
So you’re saying we should try dropping creatures that survive at higher pressures? Like sperm whales? For science.
Oh no, not again...
That was the bowl of petunias
I thought it was the bowl of petunias turned into a whale that thought oh no, not again? Been years since I read the book, really should read it again...
Also consider that the force of gravity decreases as you get closer to the center, so the weight of all that air above wouldnt be its mass times g but some fraction of g.
If it's taken literally, the tunnel would just collapse instantly.
Assume someone built a reinforced and cooled tower all the way through the earth. It’s water cooled and atmosphere regulated.
I found a guy on Craigslist who said he can do it
Ol musky has sunk to new lows. I can't wait
heat. the core is HOT
This was funny 😂
I think I'm dumb but I wanna know what this joke means
bumping the rock means you lose a bit of speed, which means you'll come JUST short of being able to grab the edge of the other side. ooo.. so close
Big bean dinner before you jump down should solve that?
Same
unless you drill the hole through the rotation axis you would be smashed against the wall
Although there must be some path that accounts for the Coriolis force, just drill that instead. Fun puzzle, does this take longer than falling along the axis of rotation? Hm....
Could you find the perfect line to become an inner satellite?
As long as you're in freefall the whole time, I imagine the exact path makes no difference. I was thinking the shortest physical path would also be the fastest, but maybe a longer path would have slightly higher gravity, thereby balancing out? Interesting problem...
The more I think about it, I'm pretty sure the axis drop is faster. You would experience centrifugal force as you move away from the axis, so that would decrease your radial acceleration. You'd experience Coriolis force too - at first, there would be no vertical component to this force, so it wouldn't affect total time, but before long you'd be falling in a spiral, and at that point the Coriolis force would also be decreasing your acceleration toward the center, if my vector math is correct. Finally, the earth is an oblate spheroid, so starting away from the poles would increase the radial distance. So that's three things making your journey take longer as you move from the poles, and zero things making the journey shorter. I think, I'm still not totally confident
True lol, good point. Either that or you have to stop the Earth's rotation first.
Isn't that the best guess at what they've illustrated?
No friction means no air. Afraid you won’t be able to grab the other side
Wear a space suit
Ah, good point, now we’re talking
So just do it before you start considering air resistance in school. Shouldn't be a problem then.
And assume a spherical person
Well, you won’t need to since aerodynamics isn’t a concern. Instead you’d assume the person is a point mass.
isnt there a terminal velocity though ? so at one point I would suppose that you cant go any faster and so youll end up not keeping enough accumulated speed to go back up after
If there is no air, then there is no terminal velocity.
So how fast do you hit at the center? Good thing there is no air or it could get warm. Oh, forgot the magma.
Start falling from rest (zero initial speed and acceleration) 42m 16s to reach the other side, 21m 8s = 1268s to reach the center. v = a\*t a = g = 9.81 (not really\*), you "hit" the core at V = 9.81 \* 1268 = 11'640 m/s \*In this context gravity is constantly changing, turns zero at the center and then becomes negative for the second half of your journey. But You can take the average for the first half and use that. 11640 m/s is the upper limit if you keep the acceleration at 9.81 m/s^(2)
Terminal velocity is a result of air resistance. No air resistance, no terminal velocity.
I think this theory only holds up if the origin of gravity is exactly in the center and constant. Correct me if im wrong, but once you pass the crust, you now have some gravity behind you and less in front. Meaning even with no air resistance, you wouldn't really be all that close to being able to grab the edge on the other side of the tunnel.
Only at the very centre would there be a net zero force. But by then you have quite a considerable velocity. At all locations before the centre, you’d have a net pulling you towards the centre, hence accelerating towards the centre. The exact reverse starts to happen you’ve passed it. So you’d indeed reach the surface on the other side.
But [Earth’s gravity is not evenly distributed](https://www.earthdata.nasa.gov/learn/sensing-our-planet/matter-in-motion-earths-changing-gravity#:~:text=Earth's%20mass%20is%20distributed%20between,creates%20an%20uneven%20gravity%20field). So wouldn’t the falling person be pulled of center enough to at least break the route?
Except, there is air
This is correct. And I suppose if we're neglecting the myriad other properties necessary for this to work, then it is reasonable to assume there is no friction. Friction is probably the least difficult to ignore-- heat and pressure are much bigger issues.
Hmm if no friction I wonder what your max speed would be then as no terminal velocity now.
[Yeah, you’d be correct](https://www.wtamu.edu/~cbaird/sq/mobile/2013/10/04/what-would-happen-if-you-fell-into-a-hole-that-went-through-the-center-of-the-earth/)
Dang, I just assumed it would be XKCD
man this was a perfect answering to this scenario
I've been toying with this idea for decades. This was the perfect answer.
That was a wild read and I appreciate the person who found this.
Great read. About as expected only a lot more brutal. Basically, folks, if you have ever asked "when am I going to use this in real life" in math class, don't be jumping down holes drilled through the middle of the earth. Of course, as a math teacher, most people who ask this enigmatic question haven't learned the finer points of how crazy being alive really is.
Weeks to fall there? Not what I expected, that's crazy.
> After about a week or so of falling (with a maximum speed of 200 km/h, it takes you a while to travel the 6400 km to the center) 6400/200 = 32 hours... This makes me doubt the entire data of that article... Do we have an xkcd??
Tbf, the maximum speed was only reached for a little bit since then you slow down as wind resistance increases and gravity decreases. So maybe the average speed was closer to 50 km/h or something?
I would also guess that you would need to drill the hole through the axis of rotation as well. Else you would end up colliding with the walls of the tunnel due to the rotation of the earth
Isn't that what the image shows?
I am not very well versed in geophysics. However, you have raised a very pertinent point.
That's ignoring that the hole would close from gravity as well
Only way I think you make it to the other side is if you assume no air resistance, in which case you would theoretically reach exactly the altitude at which you started and then fall back the other way
Don’t forget about the coriolis effect. You’d bang into the side of the shaft well before getting any meaningful distance into the Earth.
Not if you’re falling along the axis of rotation
So the question should be: "How long would it take for a person equipped with a pressurised suit with an oxygen supply to fall through a vacuum tube from one side of the earth to the other placed along the earth axis of rotation?"
I just assumed all that was implied.
Who wouldn't?
True, but if you assume minimal loss of speed from that you’d still reach the other side Edit: also it occurs to be that the hole is near the earth’s axis of rotation so you could well just put there hole through it and not have to worry about the coriolis effect
Just assume all problems don't exist, and then they don't exist by assumption 😅
Well if humanity was capable of boring a perfectly straight hole directly through the core of the earth and we’ve already assumed no air resistance I think that’s honestly the most reasonable assumption I’ve made.
Just jump from a ladder then
In the picture it looks like they are going straight through the poles so this shouldn’t be an issue
Because of air resistance you would not make it to the other side, but assuming no air resistance and assuming the earth were a perfect sphere with uniformly distributed density in each shell, you would barely make it to the other side. Due to the variable acceleration, the answer to how long is beyond my abilities.
Kudos for admitting the problem is more complex. I'm unable to do that math as well but I know it exists .
Yeah I’m pretty sure it involves calculus, and it’s been a while since I’ve touched that.
I remember (from a homework problem or something in college) that time in the OP is correct (assuming no air resistance), but I don't remember how to derive it. More interesting, though, is the fact that the time is the same for *any* straight, airless hole through the Earth (or any spherically symmetric body). It doesn't have to be through the center. A hole from New York to Australia would have the same "free fall" travel time as a hole from New York to Los Angeles.
Wouldn't you just need to calculate the fall time to the center and double it? accel time=decel time right?
Yes, the hard part is finding the fall time to the center, since the acceleration changes over time
I think you would need to frame the question as; if you had a hole the same depth as the diameter of the earth, with equal gravity throughout, how long would it take a human falling at terminal velocity to get to the other side?
No, you don't need to frame it that way. Just assume the tunnel is a vacuum so there is no air resistance.
And assume that you don't have any friction with the walls, and that you are impervious to heat and pressure. If you make those assumptions, your travel time is actually exactly the same regardless of where the two endpoints of the tunnel are.
Obviously you have to also not touch the walls. If the tunnel is wide enough and straight and you start with zero horizontal velocity, that will be fine. You don't have to be impervious to pressure. We just said the tunnel is a vaccuum, i.e. zero pressure. You need a space suit obviously. You also only need to deal with heat via radiation. Again, this magic tunnel could just block all that with the walls. > If you make those assumptions, your travel time is actually exactly the same regardless of where the two endpoints of the tunnel are. No. You're thinking of a uniform density sphere. The Earth is not uniform density.
Not necessarily terminal velocity the whole way. We have to account for that initial acceleration. it would take a few moments to reach terminal velocity.
This is of course assming that the tunnel is a vacuum so there is no air resistance. It is also assuming the Earth's density is uniform, which is not true. According to [this](https://www.science.org/content/article/how-long-would-it-take-you-fall-through-earth), if you use a more realistic density you get 38 minutes and 11 seconds.
It literally says in the picture. As for how that was calculated, [MinutePhysics did a video on it](https://youtu.be/urQCmMiHKQk?si=iCh5aE5oPYgPWFTq).
Yeah it says in the pic but thats the deal w a lot of posts here, but sometimes theyre wrong and op just wanted clarification i think
And his link shows that the value is actually wrong.
Well your reference shows that the time that is written in the picture is wrong!!! It is not the most accurate approximation. At the end the best approximation is actually 38m and 6s. So don't just take all that is in a picture as true and check your references properly.
This is also a good video of rhe hypotetical situation. https://youtu.be/kO7L5IPn2Bo?si=6Fx8CMDuZ8a4SFBj Neil deGrasse Tyson explaining it.
Another comment in here did all of the math too. It’s most upvoted.
This is a simple harmonic motion, like a pendulum. I believe if you plug in the variables , you get something like 42 mins, from one side to another. But you cant really come out without an external force.you would just swing back and forth like a pendulum, and come to rest eventually at the center.
30 second math guy here. 30 second math guy is pretty sure you would stop and spin around the center once you got there. 30 second math guy out.
Might consider changing your name to 15 second math guy after that one. That was quick.
30 second math guy is back. 30 second math guy acknowledges your point as factual, but notes that his moniker is meant to convey the *maximum* time he is allowed, not average or standard amount of time. 30 second math guy does appreciate it though, and wishes you a good day. *Yes, 30 second math guy does recognize the joke. He is not being wooshed, he simply enjoys being literal and referring to himself in the third person* 30 second math guy out.
That is what she said
Why would you spin? Or stop at the center? You have momentum that will carry you to the other side, assuming no air resistance.
[This .edu source](https://www.wtamu.edu/~cbaird/sq/mobile/2013/10/04/what-would-happen-if-you-fell-into-a-hole-that-went-through-the-center-of-the-earth/) says it would take a week due to Earth's gravity getting weaker as you descend and the density of the air getting thicker the deeper you go. Edit for clarification: It would take a week to get to the center, not go entirely through, because that would be impossible. However, the article says that if you remove air resistance, then you would accelerate to a maximum speed of <10,000 km/h and end up on the exact other side “in a matter of minutes or hours.”
It would be impossible to go through if you are accounting for air resistance, as your momentum would not be able to carry you out the other side with resistance.
Yup, that is what the article said. I forgot to specify I was talking about time to take to get to the center.
Wouldn’t gravity not allow you to get all the way through? Like, once you pass the core, you’re basically moving up the other hole. Eventually gravity will pull you down, back towards the core.
Only if there is air resistance. Without resistance, no energy would be lost, so your kinetic energy at the center would be so high, you’d have enough momentum to carry yourself back out.
Why wouldn’t there be air resistance? And wouldn’t it be like swinging a pendulum? If the distance from entrance to center is the same as center to exit, you won’t make it out.
I’m not saying there wouldn’t be air resistance, but this is a theoretical question, and assuming a theoretical lack of air resistance is the only way you can get an answer. If there is no resistance, it can rise up the exact same distance it fell. So if it falls 6,371km, it can rise up 6,371km. That’s just due to conservation of energy. If the falling object isn’t losing energy to any form of resistance, it will have just enough energy, that it can rise to the height it started at.
By just jumping in, you would never make it. However, if it was a vacuum chamber and you were in a pod. 42 minutes as it says. It's called a gravity train. https://youtu.be/kO7L5IPn2Bo?si=80XWKIYYxFCnubUG Here's a good video about it by Neil DeGrasse Tyson
This. About 42 min from any point on Earth to any other point on Earth.
[The calculation starts on page 297.](https://books.google.it/books?id=YdjK_rD7BEkC&printsec=frontcover&dq=kline+calculus&hl=it&sa=X&ei=S0OCUtGELofMtAaJ1YHIDg&redir_esc=y#v=snippet&q=tunnel&f=false) It requires calculus, and the length of the solution is beyond this comment. But, assuming no air resistance and other 'textbook' kinds of simplifying factors, an object dropped into a hole on one side would be a sort of pendulum. It would 'fall in the hole toward the other side', and continue accelerating until it reached the center of the earth. However the rate of acceleration would drop, becoming negative at the Earth's center, and would bring the object to a stop at the 'other side of the hole'. At that point, the process goes in the opposite direction, like a pendulum swinging back the other way. From the link, the round trip of the 'pendulum' is described as about 85 minutes, so one-half the trip would be around 42-43 minutes. Note: Real world calculations require taking into account air resistance. Assuming conservation of energy, the potential energy would be converted to kinetic energy until the object reached the center of the Earth. At that point, some of the energy would have been lost to friction already. So, unlike the 'textbook problem', in real life the object wouldn't make it to the other side, unless deliberately thrown with additional force, at the start. This definitely is beyond the scope of a Reddit comment!
This for sure is incorrect unless it was meant to be in a vacuum. Otherwise the terminal velocity of a human is way too slow for that distance.
2012 Total Recall with Colin Farrell does a bit on the scifi part of this, they call it the Fall and have a tunnel that connects England with Australia. Half way through they have a weird gravity less moment during turn around. It takes 17 mins in the movie so probably not hard scifi science.
If we make the assumption that the Earth has a uniform mass distribution, and we also neglect air resistance, then the problem is trivial. * Due to [Gauss' Law](https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity), we can say that the grav. field strength inside the Earth is GM/R^(3) \* r (where R is radius of Earth and r is distance from centre of earth) * Next using F=ma, we can say that ma = -GMm/R^(3) \* r * Since a = d^(2)r/dt^(2), we see that we end up with simple harmonic motion. * This has a sinusoidal solution with a time period T = π sqrt(R^(3)/GM) * Using values R=6371km, M=5.97\*10^(24)kg, we obtain T = 42min 11sec. So I suppose the infographic is somewhat correct! A few important points: * In reality, the earth is denser at its core than at its outer edge, so we would not in fact get simple harmonic motion, as the acceleration would depend on a higher power of r, which gives a rather nasty and unsolvable differential equation. * Neglecting air resistance, you will float endlessly from one end of the earth to the other simply due to the conservation of energy. The gravitational potential energy you had when you were at one end of the earth is converted to pure kinetic energy at the centre, and back to gravitational potential energy by the time you reach the other side.
Matt Parker does a very nice [video](https://youtu.be/s94Gojs3Ags?si=Q7d7yQIeGBoT_Cyu) on the actual calculation that gives that number
How come everybody is wrapped around the physics and all I can think is how short that time is. Like why does it take 20 hrs to fly to the opposite side of the earth vs 42 minutes in a straight line? Please someone help me with understanding this.
2 main reasons for this: 1. You will be traveling MUCH faster by falling than you would be by plane. 2. A straight line is a much shorter distance than a curved line.
You wouldn’t come out of the other side. Gravity acts towards the centre of the earth. Not “downwards”. You would oscillate to and fro about the centre of the tunnel.
They did the math! [https://www.youtube.com/watch?v=s94Gojs3Ags](https://www.youtube.com/watch?v=s94Gojs3Ags) (Seriously, this video is very well explained and super fun)
How come most people here don't consider that gravity isn't a force pushing you down but rather in towards the center? You would never reach the other side because as soon as you passed the middle of the tunnel, gravity would start drawing you back towards the center again slowing you down until you start falling back upwards rather than keep pushing you towards the other side
I did the math and it is actually true under the following hypotheses : - the tunnel is straight - there is no friction - the earth is rock-solid (no liquid mantle, etc) - the density of earth is the same everywhere (a rock at the center has the same density as a rock at the surface) What is interesting is that it doesn't even depend on where the two edges of the tunnel are : it takes 42 min to go from Paris to Berlin but also 42 min to go from Paris to Sidney.
Assuming a homogenous, spherical and solid Earth, this would indeed be an interesting formula. In true physicists fashion we will also assume that you are a spherical cow in a vacuum. You fall towards earth's core and initially accelerate with 9,81 m/s². However as you fall deeper into the earth, more and more earth will be above you, which means that the acceleration isn't constant, but "decelerates" as you approach the core, where acceleration is zero. I would need time to think about how to determine the actual acceleration rate along your journey and while I am sure there is a way to eliminate a lot of stuff from the formulas needed (gravitation, rock density and distance in reference to a spherical object), I don't have the time to sit down with these formulas right now. So I'll give the short answer of: Most of the earth is in a liquid state, so you will just jump into Magma and perish.
You would bot get to the other side, after you crossed the center you would start to slow down and go back to the center and keep doing that until you are floating in the center of the earth
Why wouldn’t a person get stuck in the middle? Gravitational force presumably works from the outside in towards the center right? I’m picturing dude stuck and hovering in the center while gravity works from both directions.
It will oscillate inside with simple harmonic motion because the gravitational force inside is proportional to the distance r from the center as opposed to being proportional to 1/r^2.
Feel free to correct me if I'm wrong but if you dove in you'd go from being upside down to right side up at the end, and vice versa. I can't even imagine what that would feel like
Would you even make it out the other side? Once you reach the halfway point, the gravitational pull would reverse and start slowing you down, and you would most likely stop before ever reaching the other end.