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A math question about acceleration over a distance
"Dr J R Stockton" wrote in message nvalid... In sci.math message oglegroups.com, Sat, 31 Dec 2011 22:49:23, STJensen posted: Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? Since the elevator cable cannot be infinitely rigid, and will in practice be quite flexible, one will need to consider the effect of the sideways forces on its shape - unless the mass of a shortish length of the cable is large in comparison with that of your human and his accessories. Really good point. And the Earth is rotating so the cable must rotate with the same angular speed as the Earth. So, to do this calculation properly requires consideration of tangent speeds, tangent acceleration (tangent to the cable for those) and so forth. _ |
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A math question about acceleration over a distance
In sci.math message ,
Sun, 8 Jan 2012 02:11:06, K_h posted: "Dr J R Stockton" wrote in message . invalid... In sci.math message oglegroups.com, Sat, 31 Dec 2011 22:49:23, STJensen posted: Let us say that you had a 62,000-mile-long Earth-anchored space elevator and let us say it has an electro-magnetic repulsion accelerator along its entire length. If you were to accelerate a human so that he experienced only 2g (twice the force of gravity) during the entire length of the space elevator, what velocity would that human be expelled from the space elevator and how long would it take the human to travel the entire length? Since the elevator cable cannot be infinitely rigid, and will in practice be quite flexible, one will need to consider the effect of the sideways forces on its shape - unless the mass of a shortish length of the cable is large in comparison with that of your human and his accessories. Really good point. And the Earth is rotating so the cable must rotate with the same angular speed as the Earth. So, to do this calculation properly requires consideration of tangent speeds, tangent acceleration (tangent to the cable for those) and so forth. It should not be difficult for the rigid-cable case, where the cable is kept under sufficient tension by having an adequate mass at its tip.. Perhaps the rigid case is best done in natural units, in which GSO is at unit radius and the sidereal period at GSO is also unity. The variables are then, in GSO units, the starting and finishing heights, and the constant thrust, which ISTR was going to be twice of what is needed to be felt to remain stationary at the starting point, a.k.a. ground level. The acceleration outwards results from the combination of the thrust, the inverse square gravity, and the linear centripetal pseudo-force. Integrate once with respect to the time to get the outwards speed as a function of time, and again to get the outwards distance as a function of time. Then determine the time to release from the second integral, and substitute in the first integral to get the release speed. Unchecked. -- (c) John Stockton, near London. Web http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, and links. Correct = 4-line sig. separator as above, a line precisely "-- " (RFC5536/7) Do not Mail News to me. Before a reply, quote with "" or " " (RFC5536/7) |
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