How to travel halfway across the world for free

[kingjoshi]

To get to the end-surface from the core takes exactly the same time as it takes from the start-surface to the core. So it's that times 2. But I don't think acceleration is constant.

I'm pretty confident it changes. But I made an assumption for ease of calculation.

On the surface, we're attracted to the center of mass to the Earth. As we fall into the earth, we'd lose acceleration towards that center as we'd have opposing forces surrounding us attracting us towards it. And the simple calculations aren't handling for any friction.

This ignores the whole absurdity of the idea. But I'm sure they said the same thing when someone first proposed the cockamani (sp?) idea of flying, much less flying off the planet!

 

[friskymike]

Well, a tunnel through the centre of the earth is obviously impractical due to their being no material which can remain solid at the high temperatures involved.

However, it would be technically possible to build a tunnel through the crust at a shallow angle, for a example a straight tunnel joining two points on the earths surface going to a maximum depth of 1km can be shown to cover a distance on the surface of about 220 km (simple pythagorus with radius of the Earth ~6400km).

Evacuate this tunnel so there is no air friction and put a superconducting maglev train in it, and you have a transport system capable of truly astronomic speeds.

Of course, I don't think this will ever happen due to the astronomic investment costs required... a teleconference or VR will be much cheaper in the future

 

[Erik Mesoy]

Acceleration decreases towards 0 at the center and then increases in the opposite direction of your travel.

Ugly integrals will come, I'm sure. Since a sphere may be given by y=sqrt(1-x²) rotated around the x-axis, the volume behind you at any given point p (where p is from -1, the starting point, to 1, the other side) will be given by pi*Int[(1-x²), -1, P] and the volume ahead of you by pi*Int[(1-x²), P, 1]. These determine your acceleration.
The first one equals pi(x-x³/3)[-1, P] = pi* ((P-P³/3) + 2/3), and the second one pi* (2/3 - (P-P³/3)). Since the sum of these is 2pi/3, we need a correcting factor of about 4.7 to get the acceleration of 9.8 m/s².

If I give that your acceleration is -14.75*(P- P³/3) m/s², can someone work out the rest?

 

[kingjoshi]

anybody have Maple or Matlab? My brother has my TI-92 since I haven't needed it in a long time.

EDIT: this is where computers are better then humans..

 

[Erik Mesoy]

Computers still have trouble with symbolic logic, though. I have Maple at my university, but I'm not sure how to set up the equations.

 

[A'AbarachAmadan]

How to travel halfway across the world for free

Drill a hole? Nah. Just join the US Army. Worked for me, I'm about half-way (well, 11/24ths) around the world now.

 

[AL_DA_GREAT]

You would be pressed against the tunnel wall by the centrefugal (how do you spell it?) force. Some one figured out that it would pull you further than you can imagine.

 

[Erik Mesoy]

There is no such thing as centrifugal force, so you must be wrong. (Just like there is no luminiferous ether. That idea also took a while to die.)

 

[funxus]

If gravitation is analogous with electromagnetism in this sense (it is in many at least), and we figure the mass distribution is uniform (not very likely though...) the equation for the acceleration would be:

a= G*M*r/(R^3) where G is the gravitational constant, M the mass of earth, R the earth's total radius and r the distance from the center you're currently at. Integrating this would give:

v = sqrt(G*M*(R^2-r^2)/(R^3)) giving at the center a velocity of

v=sqrt(G*M/R) ~7920 m/s

Maybe I've done my calculations wrong or th assumptions are, but if not, then that's pretty fast

 

[Gelion]

Your world looks like a butt crack.

And as Ainwood said. Drilling is expensive. My father's company has spent millions trying to research technologies that would allow faster, cleaner and more accurate drilling.

Harvard application puzzle answered!!!!


To the thread point: ainwood covered all the difficulties. However what about drilling NOT through the center?

 

[Erik Mesoy]

If gravitation is analogous with electromagnetism in this sense (it is in many at least), and we figure the mass distribution is uniform (not very likely though...) the equation for the acceleration would be:

Actually, due to the composition of the earth's core and whatnot, this may be an underestimate. Not sure what analogy you mean with electromagnetism.

a= G*M*r/(R^3) where G is the gravitational constant, M the mass of earth, R the earth's total radius and r the distance from the center you're currently at.

r/(R^3) is probably wrong, seeing as we're looking at "slices" of Earth, hence the ugly integrals in my above post. Also, you're forgetting that the bits of the earth you've passed will be slowing you down.

 

[funxus]

Actually, due to the composition of the earth's core and whatnot, this may be an underestimate. Not sure what analogy you mean with electromagnetism.

I've been reading up a bit on it (part of my current studies anyway ), and I can't see if there's any error in my thinking, and further calculations just made me more confused

The analogy I was thinking of was

E = k*Q*r/(R^3)

where the charge Q is changed into mass, and E would give the acceleration. To get the formula for a non-uniform mass distribution, I figured you could derive the formula again and use a density dependent on r. If this analogy works then you'd get the acceleration at any distance from the center within the earth. But it involves Gaussian surfaces, and I'm not sure how this works with gravitation.

 

[Erik Mesoy]

Where is that equation from? It seems to function differently for R>1 and R<1, as R^3 either grows larger or smaller.

 

[pboily]

See why you need math, Tenochtitlan?

 

[funxus]

Where is that equation from? It seems to function differently for R>1 and R<1, as R^3 either grows larger or smaller.

The formula is derived from Gauss' law applied on a sphere with uniform charge. Gauss' law basically says the electric field on a surface integrated over the area of this surface will be constant if the charge remains the same. I assumed this would be true for a gravitational field as well, but I could be wrong there. A force between two particles is still a force though, no matter if it's electromagnetic or gravitational, which is why I thought they'd be similar in this case.

I don't see why the value of R would make any difference? You could pick any actual value on R you want by just changing the unit, as long as r is measured in this same unit. Only thing is that the formula only works as long as r < R.

 

[Tenochtitlan]

See why you need math, Tenochtitlan?

I was just mad that day because they wouldn't let us use a crib sheet or a calculator in my math class. I can get easy A's with a graphing calc, I don't see why it is not allowed.

 

[Erik Mesoy]

The formula is derived from Gauss' law applied on a sphere with uniform charge. Gauss' law basically says the electric field on a surface integrated over the area of this surface will be constant if the charge remains the same. I assumed this would be true for a gravitational field as well, but I could be wrong there. A force between two particles is still a force though, no matter if it's electromagnetic or gravitational, which is why I thought they'd be similar in this case.

Electric field over the area of the surface sounds like it's correct, but I don't see how your formula works inside the planet, because it doesn't seem to take into account the bits you've passed by.

I don't see why the value of R would make any difference? You could pick any actual value on R you want by just changing the unit, as long as r is measured in this same unit. Only thing is that the formula only works as long as r < R.

Well, it looks like r and R should be raised to the same power. You said:
"a = G*M*r/(R^3) where G is the gravitational constant, M the mass of earth, R the earth's total radius and r the distance from the center you're currently at."
If we had a planet 10 meters in radius (R=10) and a mass of 100 WhateverUnits (WU), then on the surface a = 9.8*100*10/1000 = 9.8 (with meters as unit)
The planet is also 0.01 kilometers in radius (R=0.01), so then a = 9.8*100*0.01/0.000001 = 98000000 (kms) = 98000 (with meters as unit).
That's four decimals that have been misplaced somewhere, and I say they're in the difference between r/(R^3) and (r/R)^3.

Second, the units on your equation are (meters / second&#178 * (kilograms) * (meters) / (meters^3). This ends up as "Kilograms per meter per second per second", which makes little sense. Putting r inside the above parenthesis would change it to kilogrammeters per second per second.

 

[funxus]

Well, it looks like r and R should be raised to the same power. You said:
"a = G*M*r/(R^3) where G is the gravitational constant, M the mass of earth, R the earth's total radius and r the distance from the center you're currently at."
If we had a planet 10 meters in radius (R=10) and a mass of 100 WhateverUnits (WU), then on the surface a = 9.8*100*10/1000 = 9.8 (with meters as unit)
The planet is also 0.01 kilometers in radius (R=0.01), so then a = 9.8*100*0.01/0.000001 = 98000000 (kms) = 98000 (with meters as unit).
That's four decimals that have been misplaced somewhere, and I say they're in the difference between r/(R^3) and (r/R)^3.

Second, the units on your equation are (meters / second²) * (kilograms) * (meters) / (meters^3). This ends up as "Kilograms per meter per second per second", which makes little sense. Putting r inside the above parenthesis would change it to kilogrammeters per second per second.

You seem to have mixed up the Gravitational constant (often written G) with the gravitational acceleration at sea level (often written g).

G = 6.6726 * 10^-11 Nm²/kg²
g ~ 9.81 m/s²

The formula for gravitation is F = G*M*m/r², and if you subsitute M and r with earth's dimension, you'd get F = mg, where g = G*M/r² ~ 9.86 m/s². The units in the formula should then even out and give you an acceleration.

Electric field over the area of the surface sounds like it's correct, but I don't see how your formula works inside the planet, because it doesn't seem to take into account the bits you've passed by.

The formula is for an electric field over a surface within a uniformly charged sphere. In this situation it has already passed by charges closer to the surface, just like you would have passed by mass that is closer to the surface. And the formula does exclude mass both ahead of you (all outside the radius r ahead of you), and behind you (all mass outside radius r behind you).

Both gravitation and the electromagnetic force are two forces that work between two single particles and added to give an effect on large scale, which is why I think the formula would work for gravitation as well. I can't see why it would matter whether the force comes from the attraction of charges or mass in this case, except for the magnitude of the force?

 

[Perfection]

About the transit time, I actually solved it in my physics class last semester, and the trip time of a falling body in frictionless gravity-powered tunnel through a perfectly-spherical non-rotating Earth of uniform density (note that Earth has significant variations in density) between any two points on the surface (not neccearily antipolar) is ~45 minutes. Of course given the nonuniform density it will be lower. But this sets up a good upper bound.

 

[ZiP!]

Your world looks like a butt crack.

lol I think that was the philosophical reflection of the day