How does an object behave in a theoretical tunnel through the Earth's centre?

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Offline Foolosophy

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Let's assume that is is practically possible to dig a tunnel through the center of the Earth.

How will a body behave if it fell into the tunnel from rest?

« Last Edit: 09/12/2010 22:31:15 by chris »

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Offline QuantumClue

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You would fall towards the center at an increasing rate, slow down as you pass the center and maybe reach the other end due to your kinetic energy resulting in your speed... You would fall back if you did not grab hold of anything.

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Offline JP

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This gets asked here a lot and the answer is what QC said--you would oscillate back and forth around the center.  Without air resistance you would keep going forever, popping up to the surface on one side, falling back to the other side and so on.

Looking at the picture I have another question: if the hole were narrow enough, would you hit the wall since you'd be rotating along with the earth's surface when you jumped in?

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Offline QuantumClue

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That I don't have an answer to. I would imagine, the fall would not be too bumpy.

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Offline yor_on

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I will now make the sound..

 Boiiiing -----> <---Boiiing---> <--Boiing--> <-boing etc.

And then a lonely "Allloh, anyone here?"
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Offline syhprum

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Your transit time down to the other side and back is the same as the orbit time of a ground level satellite about 84 minutes.
This subject has been discussed many times on this forum.
syhprum

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Offline Foolosophy

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Your transit time down to the other side and back is the same as the orbit time of a ground level satellite about 84 minutes.
This subject has been discussed many times on this forum.
82.7 minutes to be precise

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Offline JP

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I thought about it a bit, and I suspect you'd smack into the wall on the way down.  When you're on the surface, you're traveling in a circle at a constant velocity given by the earth's rotation.  Further down, the velocity that you'd have to move to stay in the hole would be smaller than at the surface, but since I don't think you have any way of "braking" yourself, you'd probably smack into the wall...

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Offline maffsolo

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If gravity decreases as you descend to the center, your weight become less, your rate of decent will approach zero.
Also the affects like being in a space craft orbiting the earth.
 You will not hit the sides of the hole walls, just like the astronauts are not pinned to the aft section of the bulkhead during orbit.
What about how a plumb-bob works.
You will decelerate until you reach the center and stop and float indefinitely, or maybe when you reach the point where the earths radius, revolution per second  rate, matches the rate of descent.
« Last Edit: 07/12/2010 06:35:15 by maffsolo »

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Offline JP

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I'm not sure I understand fully.  The tangential velocity of any section of the earth varies with radius, so it's highest at the surface and lowest near the center.  For you not to hit the walls, wouldn't your tangential velocity have to decrease as you fell?  I feel like I'm missing something obvious here.

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Offline yor_on

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Maff?

If I'm free falling towards the middle my velocity should diminish like you think, I think :)But I would expect that you still would have a momentum taking you past that middle point? And I think JP is right, the earth is like a carousel, rotating. The fastest rotation at its surface just like the rim of that carousel. If you laid a path 'in' that carousel, from its rim to its center, and then let a ball roll to the center, would you expect it to keep to the middle of your path with the carousel spinning?
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Offline Foolosophy

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....is the path or trajectory parabolic?

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Offline yor_on

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In which frame :)
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Offline yor_on

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 No, I don't think so? The trajectory should slow down as gravity becomes less but it should work from the middle of the 'shaft' to the wall, shouldn't it?

What would one see if one was a ant on the wall? (With binoculars)

Yes, ants have eyes too.
==

small ones.

The eyes I mean.
Blue?

Coriolis force too? Would it be noticeable?

« Last Edit: 07/12/2010 20:21:07 by yor_on »
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Offline maffsolo

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Maff?

If I'm free falling towards the middle my velocity should diminish like you think, I think :)But I would expect that you still would have a momentum taking you past that middle point? And I think JP is right, the earth is like a carousel, rotating. The fastest rotation at its surface just like the rim of that carousel. If you laid a path 'in' that carousel, from its rim to its center, and then let a ball roll to the center, would you expect it to keep to the middle of your path with the carousel spinning?

yep I am having trouble wrapping my head around it.
I am visualizing revolution of a small wheel compared to a large wheel.
I see the small wheel spinning more frequently and big wheel slowly spinning
I know I am looking at this incorrectly, the same wheel speed slows to zero as you approach the absolute center, as if the hole was a wedge or rather cone shape... Thanks JP..


Since we are looking at a hole shaft with  parallel sides, that do not converge as the descend to the center!  The revolving speed is constant, wouldn't it take the same time to travel across the diameter of the hole, no matter the depth?  So wouldn't the rotational velocity in this case be the same with respect to the hole? Until the radius of the hole matches the distance to the center.

----
As far as having inertia and the momentum while approaching the center of the earth, acceleration becomes less and less by per second squared, your weight becomes less and less, so inertia will become less and less it will be like putting the breaks on
« Last Edit: 08/12/2010 01:50:37 by maffsolo »

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SteveFish

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If the tunnel were between the exact north and south pole, and all air was removed from the tunnel, it would be as Quantum Que and JP's first posts just after the 0P. If the tunnel were from equator to equator, as an object fell it would have to decelerate in a direction parallel to the surface from 1,000mph at the surface to 0 at the center, and then accelerate back to surface speed again as it approached the opposite surface of the earth. I believe that this is a pure example of Coriolis force in action. This situation would require some sort of frictionless slide, such as maglev, for the object to make it all the way back to the surface.
« Last Edit: 08/12/2010 00:39:03 by SteveFish »

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Offline Geezer

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If the tunnel were between the exact north and south pole, and all air was removed from the tunnel, it would be as Quantum Que and JP's first posts just after the 0P. If the tunnel were from equator to equator, as an object fell it would have to decelerate in a direction parallel to the surface from 1,000mph at the surface to 0 at the center, and then accelerate back to surface speed again as it approached the opposite surface of the earth. I believe that this is a pure example of Coriolis force in action. This situation would require some sort of frictionless slide, such as maglev, for the object to make it all the way back to the surface.

Good point about the pole to pole trip Steve.

In the equator case, does anyone know of a shortcut to compute the trajectory? It's a bit tricky as the accelerating force is not constant. Based on absolutely nothing at all, I have a hunch that the trajectory will coincide with the center of the shaft as the Earth rotates, but I'm trying to avoid doing the calculus required to find out.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Foolosophy

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No, I don't think so? The trajectory should slow down as gravity becomes less but it should work from the middle of the 'shaft' to the wall, shouldn't it?



I am refering to its trajectory through space, wouldnt it be some sort of parabolic curve?

And if the entry point to the tunnel is not directly along the earths axis the trajectory would be 3D helical in nature too? Wouldnt it?

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Offline Geezer

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I am refering to its trajectory through space, wouldnt it be some sort of parabolic curve?

And if the entry point to the tunnel is not directly along the earths axis the trajectory would be 3D helical in nature too? Wouldnt it?

Yes. As Steve pointed out, along the axis of rotation, the trajectory would be a straight line.

Equator to equator will be a bit more complicated due to the variation in gravity. The Coriolis forces will cancel, so the trajectory will be in a single plane. Can you figure out an equation for the trajectory that takes account of the reduction in gravitational attraction?

It's beyond me.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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SteveFish

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Just to be clear. The pole to pole tunnel completely avoids the Coriolis force problem because surface speed due to rotation is 0. Speed of falling to the center will increase all the way but the rate of acceleration of this speed will taper off as the center is approached. The trip back out will be a mirror image of this.

On the equator to equator tunnel, Coriolis force will be maximal because the speed due to the earth's rotation is maximum there (about 1K mph I think) and it must decrease to 0 at the center and increase back to the original speed as it approaches the opposite surface. If the tunnel is somewhere between the pole and equator the surface speed from the earth's rotation will be something between the equator and poles depending upon latitude, but the center is obviously still 0.

The Coriolis force is a problem because it would be applied to the object free falling in the tunnel by the walls of the tunnel.


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Offline Geezer

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On the equator to equator tunnel, Coriolis force will be maximal because the speed due to the earth's rotation is maximum there

(

That's true, but precisely at the Equator, the Coriolis force will consist of two components. Due to symmetry, these components will exactly cancel each other.
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Offline syhprum

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There seems to be some confusion between acceleration and velocity ones acceleration would be at its highest at the Earths surface but the velocity would be at it highest as you passed thru the center, (you would have enough momentum to carry you up to the surface).
syhprum

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Offline Geezer

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There seems to be some confusion between acceleration and velocity ones acceleration would be at its highest at the Earths surface but the velocity would be at it highest as you passed thru the center, (you would have enough momentum to carry you up to the surface).

Yes. I entirely agree with you.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Foolosophy

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I am refering to its trajectory through space, wouldnt it be some sort of parabolic curve?

And if the entry point to the tunnel is not directly along the earths axis the trajectory would be 3D helical in nature too? Wouldnt it?

Yes. As Steve pointed out, along the axis of rotation, the trajectory would be a straight line.

Equator to equator will be a bit more complicated due to the variation in gravity. The Coriolis forces will cancel, so the trajectory will be in a single plane. Can you figure out an equation for the trajectory that takes account of the reduction in gravitational attraction?

It's beyond me.

If an obeserver outside the earths boundaries tracked the bodies trajectory through the tunnel, that tracjectory would be dependent on where the tunnel is drilled.

THe earth is spinning through on its axis and moving in an orbit around the sun

The trajectory is not linear from the observers point of view - very complex when the tunnel isnt through the earths axis.

Its a complex spiral towards the earths center in some cases

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Offline JP

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You're right.  It depends where you drill the hole.  I guess that's probably the answer to my question about hitting the walls.  You clearly won't hit them if you drill it between the poles.

I don't think it's necessary to worry about orbiting the sun unless you really want to complicate the problem.  For this problem, it should be a very good approximation that the earth is stationary, but rotating.

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Offline Geezer

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I am refering to its trajectory through space, wouldnt it be some sort of parabolic curve?

And if the entry point to the tunnel is not directly along the earths axis the trajectory would be 3D helical in nature too? Wouldnt it?

Yes. As Steve pointed out, along the axis of rotation, the trajectory would be a straight line.

Equator to equator will be a bit more complicated due to the variation in gravity. The Coriolis forces will cancel, so the trajectory will be in a single plane. Can you figure out an equation for the trajectory that takes account of the reduction in gravitational attraction?

It's beyond me.

If an obeserver outside the earths boundaries tracked the bodies trajectory through the tunnel, that tracjectory would be dependent on where the tunnel is drilled.

THe earth is spinning through on its axis and moving in an orbit around the sun

The trajectory is not linear from the observers point of view - very complex when the tunnel isnt through the earths axis.

Its a complex spiral towards the earths center in some cases

The gravitational effect of the Sun bends the space/time of all the components in the experiment equally. Consequently, you can treat the Earth and the falling object as if the Earth is travelling in a straight line.

When you eliminate the Coriolis effect, which you can at the poles and at the equator, the problem is simply two dimensional. In the Equator case, all we need is an equation of motion that takes into account the diminishing effect of gravity as the object approaches the centre of the Earth.

It should be a fairly trivial exercise for a mathematician to tell us what the equation of motion is. We even know that it should give us an answer for the period of the oscillation that is a bit less than 168 seconds.
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Offline JP

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Here's how you do it:

Force = -m M G r-2,
where m is your mass, M is the mass contained within a spherical shell under your feet, at any given depth, and r is how far you are from the center.

You can compute M=4/3πr3ρ,
where ρ is the mass density of the earth (you assume it to be uniform, but whatever... this is a back-of-the-envelope computation). 

ρ=Me/(4/3π R3),

where Me is the earth's mass.

Plugging all this back in

F=-mMeGR-3r.

All that junk to the left of r is constant.  This is identical to the force exerted by a spring, F=-kx, where k is given by all that constant stuff.  The period of a spring is 2π(k/m)-1/2, so the period of your oscillations through the center of the earth is
(MeGR-3)1/2

I get ~250 seconds transit time.  I might have made a slight mistake, but your motion through the earth is just like a spring.
« Last Edit: 08/12/2010 12:19:44 by JP »

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Offline maffsolo

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Here's how you do it:

Force = -m M G r-2,
where m is your mass, M is the mass contained within a spherical shell under your feet, at any given depth, and r is how far you are from the center.

You can compute M=4/3πr3ρ,
where ρ is the mass density of the earth (you assume it to be uniform, but whatever... this is a back-of-the-envelope computation). 

ρ=Me/(4/3π R3),

where Me is the earth's mass.

Plugging all this back in

F=-mMeGR-3r.

All that junk to the left of r is constant.  This is identical to the force exerted by a spring, F=-kx, where k is given by all that constant stuff.  The period of a spring is (k/m)1/2, so the period of your oscillations through the center of the earth is
(MeGR-3)-1/2

I get ~250 seconds transit time.  I might have made a slight mistake, but your motion through the earth is just like a spring.


I have a math question about what I am interpreting in this literal equation…

ρ=Me/(4/3π R3)
Am I looking at?

Density = mass / volume

Volume of a sphere = 4(pi*R3) / 3

ρ=Me/(4 π R3/3)

ρ=3Me/(4 π R3)

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Offline JP

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That's mass divided by volume in my equation, which is the same as your bottom two equations.

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Offline yor_on

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Ok guys, we're on our way..

Now on to the financing, I suggest you all send me your money so I can oversee the project.. We all are good on something. I'm good at spending money :)
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Offline imatfaal

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JP & Geezer!  Transit/oscilation times of 168 seconds and 250 seconds - for 12000 kilometres!?!  Numerically show that is impossible (ie basis constant acceleration)
s = ut +1/2at2
u=0 a=10m/s2 t=250s
s = 1/2.10.250.250 
s = 312,500 m
312,500m << 12,000 km

Again using constant acceleration to set an lower bound for time
s = ut +1/2at2
u=0 s=12,000km a=10m/s2
12*106 = 0.t + 1/2at2
t2 = 24*105
t=1549

Now I am gonna go away and do my sums for what actual time of passage will be - I can't believe I have just told Geezer AND JP they are wrong!

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Offline yor_on

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heh, as long as you remember me..
Don't worry, we don't have to be over-precise, as far as my bank account in Zürich goes, any hole will do.

Shovels are on me :)
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Offline imatfaal

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F=-mMeGR-3r.

All that junk to the left of r is constant.  This is identical to the force exerted by a spring, F=-kx, where k is given by all that constant stuff.  The period of a spring is (k/m)1/2, so the period of your oscillations through the center of the earth is
(MeGR-3)-1/2

I get ~250 seconds transit time.  I might have made a slight mistake, but your motion through the earth is just like a spring.

I thought the easiest way was to work out where JP had gone wrong - cos his calc looked theoretically spot on, apart from the answer :-) I redid his sums and got to 809 seconds (ie if T= MeGR-3)-1/2)  - still too low - see lower bound previously calculated

But of course the period of a harmonic oscillator is given by
T = 2π √(m/k)
we want half a period so we get
Travel time = π.(MeGR-3)-1/2= 2534 seconds

Suddenly making it clear that JP just lost a factor of ten


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Offline JP

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What's a factor of 10 among friends?

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Offline imatfaal

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What's a factor of 10 among friends?
  Exactly!  [;D

It's a marvellously cute result that regardless of the inclination of the tunnel, the length of the tunnel and the size of the planet - all it comes down to is the Gravitational Constant and the Earth's density.  It would take about the same time to get through mercury or venus as it would to get through earth!

Leonard Susskind said in one of his theoretical minimum lectures that Newton coming up with shell theorem from no real basis was one of his most amazing feats.
There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

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Offline Foolosophy

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interesting how the gravitational force decreases as the body falls through the tunnel towards the center of the earth and yet its weight is equal to zero once free fall conditions apply.

Weightlessness is a state that is independent of the mass of the body and the gravitational field it is free falling in.

And yet some cannot accept the direct connection of this state to the earth orbiting the sun.

interesting.............

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Offline yor_on

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How do you go through Mercury?
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SteveFish

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Geezer:

Please explain how Coriolis force cancels out on the equator-to-equator tunnel. What I am talking about is the problem that the speed, due to the earths rotation, is maximum at the surface along the equator, and down the tunnel half way to the center of the earth the speed has been cut in half. This would require deceleration of the falling object, probably by the spinward side of the tunnel. The deceleration would continue till speed was zero at the center, and then acceleration back to surface speed ensues as the object approaches the opposite side of the earth on the equator.  I don't see what would cancel this.

This is a similar situation where surface winds at the equator travel in a poleward direction and this requires deceleration, which in turn changes the direction of the wind. Coriolis.

Steve
« Last Edit: 08/12/2010 16:49:22 by SteveFish »

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Offline syhprum

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I would be interested to calculate the speed of the body as it zips thru the chamber at the centre of the Earth, it is easy to calculate the speed of a satellite by assuming the potential and kinetic energy are always in balance but the case is different here the potential energy is always diminishing as it approaches the centre so the speed must increase to compensate also the distance traveled is shorter and the trajectory could be described as an ellipse with one axis zero.
syhprum

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Offline rosy

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If we ignore the question of the rotation of the earth being discussed upthread, and assume there's no air resistance, the potential energy at the surface should all be converted to kinetic energy at the centre, and so:
Ep = G x m1 x m2 / r^2

Where r is the distance between the centres of mass of the 'zipping' body and the earth, and m1 and m2 are the masses of the body and the earth respectively.

So if all that's converted to kinetic energy:
Ek = m1 x v^2 / 2

m1 cancels out

v^2 / 2 = G x m2 / r^2

v = ( 2 x G m2 / r^2 )^(1/2)

I can't be bothered to do the actual calculation, but I think the algebra's right...

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Offline Geezer

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JP & Geezer!  Transit/oscilation times of 168 seconds and 250 seconds - for 12000 kilometres!?!  Numerically show that is impossible (ie basis constant acceleration)

Hey! It wasn't my number. I was just parroting what Syphrum and Foolosophy said.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Geezer

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Geezer:

Please explain how Coriolis force cancels out on the equator-to-equator tunnel. What I am talking about is the problem that the speed, due to the earths rotation, is maximum at the surface along the equator, and down the tunnel half way to the center of the earth the speed has been cut in half. This would require deceleration of the falling object, probably by the spinward side of the tunnel. The deceleration would continue till speed was zero at the center, and then acceleration back to surface speed ensues as the object approaches the opposite side of the earth on the equator.  I don't see what would cancel this.

This is a similar situation where surface winds at the equator travel in a poleward direction and this requires deceleration, which in turn changes the direction of the wind. Coriolis.

Steve

Steve,

We may only have a terminology difference. I agree with your description of the dynamics of the object descending into the hole, but I'm not thinking of that as a Coriolis effect, although, perhaps I should.

The cancelling out I was refering to would be the tendency for the object to descend along a helical trajectory while it curved towards the center of the Earth. If that is true, the trajectory is three dimensional. I believe this would happen if the axis of the tunnel did not lie on the plane of the equator (or the axis of the poles).

G
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Offline Geezer

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OK - So, all we need to do now is plot the locus of the falling body at various angles of rotation of the tunnel to find out if the body always lies in the tunnel or not.  Any takers?

(I'd simulate it myself, but my simulator can't handle variable gravity.)
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline syhprum

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Putting some numbers into Rosy,s formula give this result
G=9.81
M=6*10^24 Kg
r=6366,620 meters
r^2=4.0528*10^13

((9.81*2*6*10^24)/4.0528*10^13)*10*.5 = 1704.2959 Km/s
This seem much to High !
syhprum

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Offline syhprum

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I have 'Mathmatica' available if anyone would like me to enter their calculations into it !
syhprum

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Offline Geezer

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Putting some numbers into Rosy,s formula give this result
G=9.81
M=6*10^24 Kg
r=6366,620 meters
r^2=4.0528*10^13

((9.81*2*6*10^24)/4.0528*10^13)*10*.5 = 1704.2959 Km/s
This seem much to High !


It does seem just a tad too quick.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline Chemistry4me

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Isn't 'G' the universal gravitational constant 6.673 × 10-11

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Offline Chemistry4me

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I got around 4.4 meters per second

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Offline Geezer

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Isn't 'G' the universal gravitational constant 6.673 × 10-11

Yes. I think we should be using G, not g. But 4.4 m/s seems a bit on the slow side.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline yor_on

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I've been thinking of this, a lot, ever since I started my bank account..
Now I will add, as a first time limited offer, gold tipped shovels for those of you joining our escape tunnel. And Mr Chem, see you soon :)
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."