Is it just me or did Cecil miss something important in his response to this question? Coriolis effect would smear you against the side of the tube as you descended/ascended.

[ul]

[li]This is in reference to What if you fell into a tube through the earth? [/li][li]That column is signed by Cecil, and so belongs in **Comments on Cecil’s Columns.**[/li][li]The original question specifies a hole through the poles. Consequently, coriolis force is not a factor.[/li][/ul]

If you’re not at a pole, you could have a curved tube which would return you to another point on the surface (if it was frictionless), without Mr. Coriolis smashing you against the side. I’ll leave solving for the exact shape of this tube as an excercise for the student.

Yeah… but what about the wobble? I guess it depends how big the hole is…

Does anyone know how much the Earth wobbles in it’s axis? That might smoosh you on the sides before you made it through… Also, I’m wondering about the effects of air resistance… would there even be any air in the earth’s core? What reason would it have to be there? I mean with gravity pulling equally in all directions outwards…

Perhaps instead of coming to a rest at the center as Cecil suggested, you’d end up bobbing back and forth between the ends of the airless zone… (well, the airless zone I’m supposing to exist… but maybe that’s all wrong…)

Oops… got carried away.

Ah, lessee, so we’re positing a smooth frictionless tube through the center of the earth, with uniform heat (so as not to be uncomfortable) … and you’re worried that it’s not realistic because Cecil didn’t consider the Coriolis effect?

What reason would air have for *not* being there? True, there’s no gravitational force keeping it there, but it still have pressure forces pushing down on it from all of the other air above it. In fact, the air pressure should increase to a maximum at the middle of the tube.

Assuming a frictionless tube, we get a simple differential equation. Why Cecil did not solve it in 20 minutes is beyond me, as it didn’t take me more than 5 minutes, and I am a mere mortal.

The radius of the earth ® is 6.378e6 m, the initial acceleration is 10m *s^-2 = g0. Initial position is at earth’s surface, so r0 = R

The acceleration decreases linarly with the distance ® from earth’s corner, so

g® = g0(1-r/R),

and the acceleration is also the second differential of the speed, so g® = d^2r/dt^2.

We get the simple differential equation

d^2r/dt^2 = g0(1-r/R) with the harmonic solution

r= R(1-sin(w*t)) where w (omega) = sqrt(g0/R), or about 1/800s
as w*T = 2

*Pi , the period T equals 2*Pi/ w = 5015s

This means that after a mere 2507.7 seconds of weightlessness, you reach the antipodes.

- Alex

As you can see from my last post, the position r is

r(t) = R(1-sin(wt)), where w = sqrt(g0/R)

If we look at the horizontal speed with which we’re spinning around the center of the earth, at the equator we get

v® = v0*r/R, v0 = 2*Pi*R/24h where r is the distance from the center

Using the equation above for r(t), we get

v(t) = v0*(1-sin(wt))

And the coriolis acceleration is

a = dv/dt = v0*w*cos(wt)

The greatest acceleration a0 = v0*w is at t = 0
As we have seen, v0 = 2*Pi

*R/86400s and , so*

a0 = 2Pi

a0 = 2

*sqrt(g0 * R) / 86400s = 0.5 m*s^-2, a twentieth g.

So even from equator to equator coriolis certainly wouldn’t smash me!

- Alex

More mortal. Cecil was trying to figure out how long it would take for a body to stop in the center, because of air resistance, no?

Lexel, the coriolis acceleration would certainly make you hit the side of the tube instead of passing striaght through the center! So maybe you wouldn’t be smashed, but you’d definitely lose some skin due to sliding down the east side of the tube.

However, we assumed a frictionless tube, so smashing into the side wouldn’t make you lose any skin, nor should it slow you down at all, right? If the tube is frictionless, we should be able to ignore the coriolis effect.