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Author Topic: How does an object behave in a theoretical tunnel through the Earth's centre?  (Read 18622 times)

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.
 

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 »
 

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)
 

Offline JP

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

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 :)
 

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!

 

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 :)
 

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


 

Offline JP

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What's a factor of 10 among friends?
 

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.
 

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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.............
 

Offline yor_on

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How do you go through Mercury?
 

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 »
 

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.
 

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...
 

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.
 

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
 

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.)
 

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 !
 

Offline syhprum

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

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.
 

Offline Chemistry4me

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

Offline Chemistry4me

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I got around 4.4 meters per second
 

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.
 

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 :)
 

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