# Weightlessness on Earth

Would the following work?

Imagine one would

1) create a straight 4000 km long tube (at ground level), following the curvature of the Earth, wide enough to hold a pod the size of an aircraft cabin;

2) remove the air in front of the pod and make sure it accelerates to a speed of 28000 kph.

Would one achieve weightlessness inside the pod for ten minutes? If not, what are the obstacles?

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The sudden stop at the end of the tube might be disconcerting but otherwise it's exactly the same as an orbit. If you made the tube 40,000km and arrange for it to join up the trip might be more pleasant – Martin Beckett Feb 7 '13 at 23:57
I agree, but I lack the budget for the 40K model. :) The sudden stop, of course, could easily be avoided. – Domus Feb 7 '13 at 23:58
but you have funding for the transcontinental prototype? – Martin Beckett Feb 7 '13 at 23:59
I wish.:) Just curious why I couldn't find any reference to any such theory/project on the net. – Domus Feb 8 '13 at 0:04
My son once asked me if you had to go to space to be weightless. I had my car keys in my hand and told him to watch them while I gently lofted them up, holding my hand about an inch below them, and catching them on the way down. See? I said. They were weightless. It's a cheaper demo than a ride in the Vomit Comet. – Mike Dunlavey Feb 8 '13 at 1:07

Yes, equating the acceleration due to gravity with centripetal acceleration, I get the following required speed (which differs slightly from your rough figure): $$v=\sqrt{gr} = 28,455\,\mathrm{kph}$$ but yes, assuming you were to get the pod moving at that speed before it enters the evacuated tube, then once it enters it will be in orbit just above the surface of the earth for the entire duration of its flight in the tube, in this case that duration would be roughly 8 minutes and 45 seconds.
I suspect the problem is more fundamental than money. Rather, high quality vacuums are hard. Call it 5 meters in diameter...I make it almost $8 \times 10^7\text{ m}^3$ to evacuate, and more than $6 \times 10^8\text{ m}^2$ of wall available to out-gas. Having fun trying to pump that down. Even neglecting the possibility of leaks (which you will have if only around the pumps). – dmckee Feb 8 '13 at 0:24
Well the drag power is proportional to $\rho v^3$ where $\rho$ is the air density, so a 10% increase in velocity requires you to reduce the density by 30%. It's not a fair game. :) – Michael Brown Feb 8 '13 at 1:02