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Power from F time v

  1. Nov 8, 2007 #1
    Suppose you have a frictionless and perfectly balanced conveyor belt going into space (and back). The belt is driven by a motor on earth. Suppose you want to send a person who weighs 100kg into space and you boost them onto the conveyor belt at 1m/s with the conveyor belt running at 1m/s (the boost is independant of the conveyor belt so it does not have to accelerate them). To maintain that velocity of 1m/s requires a force opposite to that of gravity which is approx. 1000N. Because Power = Fv does that mean that the motor must only generate a power of 1KW (1000N x 1m/s) ? i.e. with this configuration you can get into space with very little power if you are prepared to go slowly enough ?

    E.
     
  2. jcsd
  3. Nov 8, 2007 #2

    mgb_phys

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    Yes the power doesn't depend on the height only the speed.
    You are lifiting 100Kg ( 1000N) through 1m every second so it takes just 1kW.
    Exactly the same as it would take you to run up stairs at 1 (vertical)m/s.

    Space elevators are a very fuel efficent way of getting into space. If you have more mass coming down than going up they atually generate power!
     
  4. Nov 9, 2007 #3
    Yep. The mass in quesiton only takes 1KW.

    Space elevators are one of the more efficient and "green" methods for getting cargo to and from orbit.

    The real trick is in getting the elevator to work in the first place vis-a-vis accelerating the mass of the entire ribbon up to the 1m/s travelling speed. The amount of energy it takes to lift cargo is trivial compared to the amount of kinetic energy tied up in the ribbon itself.
     
  5. Nov 9, 2007 #4

    russ_watters

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    You could get into space this way pretty easily, but getting into orbit would be a lot tougher. Just riding it to low earth orbit altitude (about 200 km) doesn't accomplish you anything. Getting to geostationary orbit, iirc, is about 40,000 km up, so you'd have to ride this conveyor for 1.3 years to get there. So pack a hearty lunch.
     
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