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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.

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?

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.

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?

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

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).

Quote from: Foolosophy on 08/12/2010 02:10:38I 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.

Quote from: Geezer on 08/12/2010 02:42:32Quote from: Foolosophy on 08/12/2010 02:10:38I 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 sunThe 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

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=4/3πr^{3}ρ,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=M_{e}/(4/3π R^{3}),where M_{e} is the earth's mass.Plugging all this back in F=-mM_{e}GR^{-3}r.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(M_{e}GR^{-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.

F=-mM_{e}GR^{-3}r.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(M_{e}GR^{-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.

What's a factor of 10 among friends?

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

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

Putting some numbers into Rosy,s formula give this resultG=9.81M=6*10^24 Kgr=6366,620 metersr^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 !

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