if you fell into a tube through the earth

Cecil's Columns/Staff Reports
21

newbie with a question/comment or two.

About the Coriolis effect? Which pole are we talking about when we mention poles, the earth has many different kinds. North, true north (?), magnetic north. However I’ll assume for purposes of the Coriolis effect we are placing the tunnel through earth’s axis of rotation. So here is my potentially dumb question, doesn’t the earth have a little wobble to it? For some reason, i remember something about the earth’s rotation not being pure, but I’m probably mistaken.

As far as any concerns about perpetual motion, I would imagine that such minute effects such as earth’s magnetic field and the gravitational pull of the moon would not allow such a thing to occur.

22

I think the better question would be, at what point does deceleration due to air resistance overcome acceleration due to gravity? (Assuming uniform air density) After this transition point, the falling body would begin to slow down, and depending upon where the transition point occurrs, you may either stop at the center, or fly past it to some point, then fall back.

But I should think that, until you reach that transition point, you would continue at the same terminal velocity (whatever that is) as above the surface of the earth, just as if you’d jumped out of a perfectly good airplane.

So yeah.

Matthew Baillie
ace22@geologist.com
“Hurrah, we are all
free now” - REM, 1986

23

Suppose the outside of this perfectly smooth tube was wound with copper wire. And suppose we dropped a 10 ton magnet down the hole. Would this be a source of perpetual free electricity?

24

Not if I owned it, natch! :smiley:

But seriously, I think that much copper wire would seriously tax the world’s copper reserves. I just can’t see past that.

25

wouldn’t the magnet under go a force to make the electricty and thus slow down?

I mean it does take work to move the magnett through thus introducing friction-like slowing to the magnet.

26

I suspect yes, but does the moving magnet in a generator receive any sort of drag by so moving through the wires? If so, my 10 ton magnet would eventually slow down to a stop.

27

If you really want to see the effect, replace the coil with a thick-walled copper pipe, and use a good strong rare earth magnet. Set the pipe upright, and drop the magnet in the top end. Then wait about two minutes for it to fall out the bottom.

Always remember, There Ain’t No Such Thing As A Free Lunch. If you get energy out of a system, the system has to be losing energy. Eventually, it’ll be all used up.

28

The chief wobble of the Earth has a period of about 26,000 years. The next biggest has a period of about 19 years. I don’t think they need to be factored in.

29

On the magnet idea:

To make things more intesting lets ditch the idea of the earth altogether and imagine a large uniform sphere far, far in space where gravity from planets, stars, etc. wont interfere. Drill two holes along a diameter of the sphere and place a uniform tube of copper wire along that diameter. Place a uniform magnetic sphere outside the large uniform sphere along a diameter line, and the fun begins.

The magnetic sphere accelerates toward the shell due to gravitation force (F = GmM/r^2). When the sphere is inside the shell - anywhere in the shell - net force is zero, and the magnet does not accelerate (although, it does have velocity). The magnet passes through the center of the shell with that same velocity and mantains that velocity until it reaches the other side of the diameter it is travelling. Upon exiting the shell, the magnet experiences an acceleration pulling it back towards the shell. This acceleration slows down the magnet until it reaches a distance equal to the starting distance from the shell’s surface. At that distance, velocity is zero and acceleration continues. Thus the magnet begins travelling into the opposite direction. The cycle continues.

Now here’s the good news: This motion will continue uninterrupted - so yes, we have perpetual motion. We have accounted for all given forces and can double check using conservation of energy (there is no thermal loss, because there is no friction, because we are sufficiently far in space, and the given bodies never touch). An alternating current is induced and therefore we have inexhaustible current.

And the bad news: First, this is just your stock textbook problem with a slew of impossible givens. A truly uniform solid is in fact unachievable - ANY loss of mass through transportation or production ruins our unifrom body. That means we gotta label all the atoms and account for lost ones… Second, placing a mass sufficiently far from any other mass that we don’t want interfering would take quite some time and energy to transport. Third, setting up the system requires other masses to be present - thus tainting the system. Fouth, using the current provided by circuit would introduce loss of thermal energy - again, tainting the system. Imagine running a wire (massless for sake of discussion) around the shell from one end to another end of the tube. Place a light bulb (again, massless) along this wire and you lose energy due to heat and light.

So in the end: Sadly, perpetual motion and an inexhaustable source of usable energy eludes us…

30

They may be thinking of the Chandler wobble, with a period of about 14 months, or the forced annual wobble which is about half of the Chandler. Still wouldn’t make much of a difference in a 42 minute trip.

31

moox you forgot to account for random quantum fluctuations that occur even in vacuums. Heisenburg would never allow you to have a system devoid of outside forces, there is no such thing as ground state energy 0, right?

BTW, two diff ppl have mentioned that the force of gravity is net zero inside the sphere/earth regardless of location within the tunnel. Is there some ready explanation of this i can take a gander at?

32

I think what they said was that the net zero gravity was inside a hollow shell. In other words, the force of gravity inside the tube is only due to the part of the Earth below that same distance to the center, there is no influence from the shell above that point.

That’s a fairly straightforward calculus calculation, using Newton’s law of gravitation.

As a heuristic, Chronos has used the idea that two objects of the same density that look the same size will have the same gravitational influence–since angular area is inverse square, and gravity is also. So, if you’re inside a shell, every visible piece is balanced by another piece directly opposite.

33

We haven’t yet accounted for all given forces, because there’s still the magnetic force. And we still have thermal losses, because the copper wire isn’t a perfect conductor. If you increase the conductivity, you’ll just make the magnetic forces that much more significant. And even if you did have “free current”, so what? You can get that easily enough in a loop of superconductor: Any current that’s in it will stay in it forever. What you would want is a current that can maintain a load.

34

I don’t see why air resistance would cause an equlibrium. (it’s misleading to call it friction, because friction does not depend on velocity, and air resisitance does). Seems to me it would asymtoptically approach equilibrium, but never reach it. Even if there is loss from gravity waves (and if there is, why are the planets still orbiting?) that would still be an asymtoptical loss.

And “perpetual motion” is a misnomer; strictly speaking there’s nothing impossible about perpetual motion; just put an electron in orbit around a proton.

RM Mentock

I think you could also prove it using Stoke’s theorem, but that’s not exactly a “ready” explanation.

35

Assumptions: The Earth has a core in the center, in which the density is uniform. The core extends up to a radius aR, where R is the radius of the Earth, and a is a parameter between 0 and 1. Outside the core, the acceleration toward the center is uniform.

Find: Travel time through a Cecilian tube.

Solution:

T = T[sub]0[/sub] × f(a)

where:

T[sub]0[/sub] = pi (R[sup]3[/sup] / GM)[sup]1/2[/sup] = 42m 14s
f(a) = 2/pi × ((2(1-a))[sup]1/2[/sup] + a[sup]1/2[/sup] arctan((2(1-a)/a)[sup]1/2[/sup]))

For the case of the core extending all the way to the surface, a = 1 and f(1) = 1, so we get the normal result. For the core to be a point in the center, a = 0, and f(0) = 8[sup]1/2[/sup] / pi = 0.9003, so the trip will only be 90.03% as long. A more realistic picture will have a somewhere in the middle, like a = 0.5464. Since f(0.5464) = 0.917, the trip will be 91.7% as long, or about 38m44s.

36

Because for the planets (and for our particle falling through the Earth, for that matter) general relativistic effects are very small, so energy loss from gravitational waves is very slow. I do agree that, given the usual models for air resistance, you would only asymptotically approach stationary equilibrium. But eventually, you’re going to get to a point where the usual models break down, and things like Brownian motion become significant. Of course, Brownian motion could bring our test mass to rest, so I think we should consider velocities comparable to Brownian motion to be “at rest for all practical purposes”.

And I’m not sure about Stoke’s theorem, but you can definitely prove “no gravity inside a shell” using Gauss’ Law. Put the Gaussian surface at your radius, you have spherical symmetry so the field must be uniform, and you have no mass enclosed. So the gravitational field is zero.

37

Yeah, that was my point. Relativistic effects would be too small a term to worry about.

Ooops! I was trying to remember which theorem to use, and the only ones I could remember were Green’s theorem and Stoke’s theorem; I was trying to figure out which one proves this. In fact, it’s the diveregence theorem, also known as Gauss’ Theorem. Gauss’ Law is a special case of Gauss’ Theorem (although when applied to point sources, it requires Dirac delta functions).

38

for adding in air resistance, we can model the situation with a linear homogenous second-order differential equation:
m y’’(t) + lambda y’(t) + k y(t) = 0

m is the mass of the object, lambda is the damping constant, and k is the spring constant. k is obtained from the equation f = - k y, and since gravity varies linearly with height under the earth’s surface you can use the gravity at the earth’s surface to determine k: k = - m g / (radius of earth), where m is the same mass of the object.

lambda represents the effects of air friction. air friction can be modeled in a variety of ways, with the friction force being proportional to the velocity, velocity squared, or other functions, however to have a linear equation we will model the air friction force as being proportional to the velocity. lambda is obtained from the terminal velocity equation: m g = lambda v (i think). also note that the units in each term of the first equation are of force, which serves as a check for the lambda equation.

now you can solve the equation by either plugging in the numbers or leaving the constants there. trying a solution of the form y(t) = exp[r t] (standard way of solving this de equation) yields:
(r^2 m + r lambda + k) exp[r t] = 0
solve for the roots of r and you have the two solutions.

the value of lambda affects how damped the system is. as ski and ace22 noticed, there are different cases depending on the magnitude of lambda. if lambda is very large, the system is overdamped and the object does not cross the axis after release. think of the object falling through a thick liquid, it will just slow down approaching the center. other cases are when the system is critically damped, passes the axis just once, and underdamped, passes the axis many times. also, as the ryan noticed, the object approaches the center but doesn’t reach it, as shown by the exponential solutions. finally, also notice as citybadger pointed out the density of the atmosphere will not be constant and will vary with height. you could integrate to find out how the density would vary with y under the earth’s surface, but then lambda would be a function of y and the de would be non-linear. for this calculation you could choose lambda as a value midway between the surface and the center.

whew long post! anyways, if anyone finds a suitable value for lambda, i’ll complete the calculation. hope this helps explain this interesting question!

39

Actrually, in the critically damped case the object never reaches the axis.

40

If I’m remembering my DiffEq correctly, Elemental is right on this one. Overdamp, and you monotonically and asymptotically approach equilibrium; underdamp, and you cross equilibrium an infinite number of times; critically damp, and you cross equilibrium exactly once, and then approach it asymptotically.