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An oft-discussed question in school physics classes is the notion of a hypothetical tunnel through the earth, that you could jump into and, ignoring friction/clipping the walls on the journey etc, would take you to the exact opposite side of the globe from where you are now.
My take on this is that it would only take you to the centre, whereupon you'd stop. Now, I understand the theory is that you'd accelerate at 9.8m/s/s due to gravity, and would overshoot the centre at peak velocity and then start to decelerate at the same rate that you accelerated, but my thinking is that, as you near the exact centre of the sphere, there's more mass 'above' you which would reduce the gravitational effect. Consequently en-route to the exact centre you'd start to decelerate at some point, as all gravity at the centre must be negative, ie pulling 'away/outwards' from the centre.
Do any Friday afternoon physicists have a view? Flat-earthers are also welcome, but only for a roasting.
Energy equivalence. Your gravitational potential energy on the one side would be converted entirely into kinetic energy as you hit the centre (assuming the tunnel is a vacuum tube) and be converted into the exact same amount of gravitational potential energy as you come out of the other side. Assuming you haven't lost any mass on the way through e.g by shitting yourself.
It's the same as a pendulum swinging.
Would it make a difference how fast you entered the hole? Is it like firing stuff into space or being a newly formed planet - if you go fast enough you can break through the gravitational field and pass out the other side? So maybe if you were fired out of a canon you'd not stop.
We did a similar prob way back in uni for thermodynamics (studying the entropy of the earths core), You’d only get so down the hole before the coriolis effect would throw you against the walls resulting in your death, you could do it at the poles as the coriolis effect is negated and from memory you’d be doing about 5/6 miles a second and if you could slow down you’d be weightless at the core but you’d be going so fast that you’d zip through and pop out the other side into the atmosphere before falling back down to earth
Consequently en-route to the exact centre you’d start to decelerate at some point, as all gravity at the centre must be negative,
No - at the centre the gravity is exactly zero: the same mass of the earth above and below you. So you no longer accelerate but haven’t started to decelerate.
as you come out of the other side.
Well, not so much out the other side as just reaching the surface before falling back in again.
You’d oscillate.
Yes.
Would it make a difference how fast you entered the hole?
Yes you would leave the hole on the far side with the same kinetic energy as you had when you entered on the near side.
But as you're whistling down through the tunnel (ignoring earth rotation and Coriolis effect) would the amount of earth's mass 'above' (behind?) you that you've already passed through not start to have a cancelling-out effect? Theoretically (presumably, in my head anyway) the gravitational force (pulling to the centre) at the exact centre is zero, because everything that causes gravity, ie earth's mass, is all around you in all directions, so it must have peaked and then diminished before you reach the centre. If it's diminishing then you'd start to slow down, decelerated by that amount of mass you've already passed through. If you're slowing down then you'd never make your stop at the other side.
For context, I know I'm almost certainly wrong, but I don't know why.
but my thinking is that, as you near the exact centre of the sphere, there’s more mass ‘above’ you which would reduce the gravitational effect. Consequently en-route to the exact centre you’d start to decelerate at some point, as all gravity at the centre must be negative, ie pulling ‘away/outwards’ from the centre.
No, because until you get to the center, you're getting closer to more of the other side than you are getting further away from what's above you. Gravity doesn't act from the surface, it acts from the center of mass. All the earth surface is doing is holding you up from falling down to the core.
F = GMm/R²
R is the distance between the center of mass of the two objects.
G is always 6.674x10^-11 m^3/ks.s^2
So the answer is 9.81 N/kg bodymass at the surface. But when you get to the core your bodyweight increases.
So on the one hand you would be 'weightless' floating in the tunnel. But on the other you would be very much stuck there unable to lift yourself out.
At 1m from the center you would weigh about 4x10^16N
Why 1m from the center? Well if you're ~2m tall, imagine half that pushing down on your head, half that from your feet. You'd be a floating meatball.
So really it's less of a thought experiment, and more like something from a horror story.
The gravitational acceleration would reduce from 9.81m/s2, that it is at the earth's surface, towards zero as you approach the centre but you'd still be accelerating and velocity increasing just by a reducing amount (assuming no frictional losses). Then the same the other side of the centre except rather than increasing your velocity, gravity would be accelerating in the opposite direction to your velocity slowing you down until you reach the same radius.
If you start with some velocity (being fired out a canon) then you'll have the same velocity at the same radius the other side. The velocity from being fired out of a canon is the same as jumping from higher up in terms of energy as you'd accelerate to that canon speed if you jumped from the correct height.
The gravitational acceleration would reduce from 9.81m/s2, that it is at the earth’s surface, towards zero as you approach the centre
Nope, and it's observable on the earths surface.
It's something lie 9.83 at the poles and 9.80 somewhere in the Andes because centripetal force pushes out the equator a bit so pole to core is a shorter distance than core to equator (plus a mountain).
In fact even without the heat from the core, the ever increasing acceleration and air density in the tunnel would probably cook the human meatball as it fell.
At 1m from the center you would weigh about 4×10^16N
If you consider the earth as two separate hemispheres placed together then not sure this makes sense. I would suggest being weightless similar to in space?
I can add a couple of insites
There is a thing called Stokes theorem. Basically this says if you are 3000km from the the centre of the Earth your pull to the centre is just the same as it would be from a 3000km sphere of the same mass as the middle 3000km of the Earth. The effect of all the rest of the planet cancels out to no effect. This is because there is mass above you which is closer and pulling you up. There is mass below you, a good 6000km away. But there is more mass the other side, at any given angle. So it all cancels out.
You’ll often see graphs based on this that suggest as you go down the hole “g” or acceleration drops linearly. So 3000km down the hole you’d expect your acceleration to have dropped to about 5m/s^2. But This isn’t what would happen. The middle bit of the Earth is all very iron rich and much denser than the edges. So actually your acceleration stays at about 10m/s^2 for a huge chunk of the way to the centre. It does of course eventually drop to zero at the centre of course
9.81 being what taught as average value to use. As you say the change is measurable at different places on the earths surface.
Value of g surely is effectively negligible at very large distance, maximum at the earth's surface, then reducing to zero as you approach the centre
But yep think mostly you'd come out BBQ'd as TINAS says so probably doesn't matter too much
Inertia, innit.
In a frictionless vacuum, you'd arrive (dead) on the other side at the same altitude as you set off, before reversing course and returning from whence you came in an infinite loop.
In the real world, in so far as this can be considered real at any rate, other forces would slowly ebb you of energy until you stopped oscillating and came to rest at the core.
It's like, pull a swing back, let go, what happens? Gravity is Bottom Dead Centre but the swing doesn't slam to a halt when the chains become vertical.
As an aside to this I was reading about atomic clocks which the boffins have got down to such a mathematical art form that they are able to differentiate the difference in aging of the human body between steps on a staircase, bonkers really.
I think I'm confusing velocity v with acceleration a. The a would diminish (although probably not linearly as we're a sphere) but that wouldn't reduce v. So at the centre a=0 but velocity=28000kmh.
There is a thing called Stokes theorem. Basically this says if you are 3000km from the the centre of the Earth your pull to the centre is just the same as it would be from a 3000km sphere of the same mass as the middle 3000km of the Earth. The effect of all the rest of the planet cancels out to no effect. This is because there is mass above you which is closer and pulling you up. There is mass below you, a good 6000km away. But there is more mass the other side, at any given angle. So it all cancels out.
Ahhh dammit, I was forgetting that bit.
No baked human meatballs.
It kind of depends on the tunnel diameter, will it be 26", 27.5" or 29"?
Really makes the trail come a- ... ah, no, quite the opposite.
I think I’m confusing velocity v with acceleration a. The a would diminish (although probably not linearly as we’re a sphere) but that wouldn’t reduce v. So at the centre a=0 but velocity=28000kmh.
No, F=ma, so as a drops, so does F, so your terminal velocity would reduce, you'd slow down as you approached the center.
Air density would be increasing too, so drag would be going up.
Would there be a conveyor belt in this tunnel?
you’d slow down as you approached the center.
What force would be decelerating you?
What force would be decelerating you?
Air resistance.
How long to get through is slightly mind blowing. About forty minutes IIRC.
And if you build a tangential tunnel from London to Manchester, also frictionless, it would behave in the same way.
No, F=ma, so as a drops, so does F, so your terminal velocity would reduce, you’d slow down as you approached the center.
Air density would be increasing too, so drag would be going up.
Was about to ask you to remind me, you're a science teacher right?
Till I read this:
Air resistance
Go back and read the question. 😉
All aside, I really like the thought that one day someone will cut through a dead star for exactly this experiment.
My kids watched this not long ago…
Some good episodes in the series!
If there was a hole the diameter of the diagonal across a football pitch passed all the way through the earth, from one side to the other, straight thorough the centre, what effect would that have to our planet and how many elephants giraffe halves could it be filled with?
The go straight through to the other side answer is correct (assuming no friction), but perhaps it's interesting to note that if the Earth was a hollow shell there would be no gravitational acceleration in the interior and anyone on the inside would just float around aimlessly.
I think. (From memory, I haven't just worked it out.)
no gravitational acceleration in the interior and anyone on the inside would just float around aimlessly.
Interesting situation. But I don’t agree - except of course in the dead centre. Somebody falling in would experience gravitational deceleration once they had passed the thin shell because most of the nearby mass would now be above them. So they would slow down, stop and then accelerate back out. They would continue to oscillate in and out, assuming no friction or rotational effects.
Even somebody who starts randomly floating inside, but not the dead centre, would be attracted to the nearest piece of the shell and fall upwards towards it.
At which point do you think you would regret starting the experiment?
matt_outandabout
Full Member
Would there be a conveyor belt in this tunnel?
Yes, and also a 747, it's quite busy in the tunnel.
Interesting situation. But I don’t agree
Having googled it, my recollection was correct, and this calculation dates to Newton.
Interesting situation. But I don’t agree – except of course in the dead centre. Somebody falling in would experience gravitational deceleration once they had passed the thin shell because most of the nearby mass would now be above them. So they would slow down, stop and then accelerate back out. They would continue to oscillate in and out, assuming no friction or rotational effects.
Back to my previous post. It’s zero ever where in the she shell.
9.81 being what taught as average value to use. As you say the change is measurable at different places on the earths surface.
At university I used a gravity meter that could measure the difference in g between being on the table and on the floor. g also varies with latitude. We eventually measured g across a landfill site to estimate the density of the rubbish
you’d slow down as you approached the center.
No, your acceleration would slow but you'd still be accelerating.
You're all over thinking it. Without air resistance, the energy just changes from potential to kinetic and back, so you end up on the surface at the other side. The energy can't go anywhere else. Conservation of energy can't be violated. Any other analysis you want to do must agree with this.
With air resistance, you'd lose some, so you'd go through the centre and part of the way back up the hole, then fall back down as some of.your energy has been lost to heat. In practice, you'd lose so much to air resistance you'd be lucky to get a few hundred metres from the centre.
But I don’t agree
You'd better write in to my university Physics dept with your mathematical proof, because they proved this in a first year lecture and if it were wrong I'm sure a lot of mathematicians would like to know...
In practice, [s] you’d lose so much to air resistance [/s] you’d be lucky to get a few hundred metres [s] from the centre[/s] .
FTFY
If it's anything like Alton Towers Oblivion you'd pop out into daylight thinking - "that wasn't worth a two hour queue"
As said, you'd oscillate. It's effectively an elliptical orbit with a zero minor axis.