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No terminal velocity in a vacuum?

  1. Oct 28, 2004 #1
    if we made a vacuum tube ten thousand feet high, and dropped a million tonne weight into it, it would have no terminal velocity right? how fast would it hit the ground? armageddon? :P
     
  2. jcsd
  3. Oct 28, 2004 #2
    Impact velocity would be approx. [tex]v=\sqrt{2gh}[/tex] and it has nothing to do with mass. When you put numbers in you get that v is around 250 meters per second. The same speed would be if you have dropped the feather. Now, destructive effect depends on mass because kinetic energy which is spent in the moment of "braking" during the impact depends on mass [tex]T=\frac{1}{2}mv^2[/tex]. When you calculate this you get that T is approx 30TJoules which is one-hour energy output of 10 larger power plants.
     
  4. Oct 28, 2004 #3
    So basically, yes, armageddon. :smile:
     
  5. Oct 28, 2004 #4
    someone should try this out.. just for fun..
    lol
     
  6. Oct 28, 2004 #5

    Gokul43201

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    Correct, there would not be a terminal velocity - since there's no drag, and the weight doesn't matter. In the non-relativistic approximation, assuming 'g' is roughly constant, the final velocity will be [itex]v = \sqrt {2gh} =~about~800 m/s [/itex]
     
  7. Oct 28, 2004 #6
    lets set up a research lab deep underground, with a vacuum tube leading from the surface. We could accelerate particles that would normally be disturbed by other means.. ok now im talking out of my ass :)
     
  8. Oct 28, 2004 #7

    Integral

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    Actually not, a possible means of rapid transit would be to dig a parabolic tunnel between London and New York, Drop a train in in London and if falls out in New York. What an engineering feat that would be!
     
  9. Oct 28, 2004 #8
    I think you forgot to convert feet to meters.
     
  10. Oct 29, 2004 #9

    Mk

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    Shouldn't the parabolic curve be a little off to work?
     
  11. Oct 29, 2004 #10
    terminal velocity appears due a viscosity force like F = -kv ... if this force don't exists, then...
     
  12. Oct 29, 2004 #11

    Integral

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    Not sure what you mean by this?

    Actually any smooth curve would do the trick.
     
  13. Oct 29, 2004 #12

    arildno

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    If "g" is not constant, assuming we start at rest from infinity, the terminal velocity of an object when it hits Earth's surface equals the (minimum) escape velocity from Earth.
    The [tex]\sqrt{2gh}[/tex] approximation is only good when h is much less than the radius of the Earth.
     
  14. Oct 29, 2004 #13

    Gokul43201

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    I think not. I just wrote "ft" as "m"...my bad.
     
    Last edited: Oct 29, 2004
  15. Oct 29, 2004 #14

    pervect

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    I get 244 m/s using google calculator. You can see the calculation

    here

    (this turns out to be 800 ft/sec)
     
  16. Oct 29, 2004 #15

    Gokul43201

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    Ummm...yes, my mistake is not that I didn't convert, but I mistyped. :redface:

    I got 800 ft/s using my head, g=32 ft/s2 and sqrt(64) = 8 :smile:
     
    Last edited: Oct 29, 2004
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