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B Moving mass in space

  1. Oct 16, 2015 #1
    This is not home work right off OK
    If there was a million pound rock in space how much pull (not a rocket pull) in pounds would it take to get it to 1 gee inlets say in a hour or less?
  2. jcsd
  3. Oct 16, 2015 #2
    I'm not sure you really mean 1g here since this is an acceleration.
  4. Oct 16, 2015 #3


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    What do you mean by 'a million pound rock'? Pounds is an (archaic) measure of weight (or, more generally, force), and the weight of an object is defined as its mass multiplied by strength of the local gravitational field. You haven't specified what the local gravitational field is.
  5. Oct 16, 2015 #4


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    Clearly he meant mass.

    If metric is your preference, let's assume, for round numbers, the rock masses 500 metric tons.

    He wants 1g of acceleration. There is no local gravitational field.

    marmstring, the issue here is that you've specified two criteria:
    - you want 1g acceleration, you could have that from the get-go with a given applied force.
    - but to have it reach 1g acceleration after an hour, means you're talking about an increasing acceleration, from 0 up to 1g over an hour.

    Normally, you'd specify either an acceleration of 1g, or you'd specify a speed you want to reach after one hour.

    Can you clarify?
  6. Oct 16, 2015 #5
    Ok sorry a Million pounds on earth and one gee it would have to stay at one gee
  7. Oct 16, 2015 #6


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    OK, that a 454 metric ton rock. Can we round to 500?
    You want 1g continuous acceleration.

    Simply F=ma

    a = 10m/s2.
  8. Oct 16, 2015 #7


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    it takes a mass of ~31250 slugs to weigh 1,000,000 lbs on the Earth, And 1,000,000 lbf to accelerate it at 1g
  9. Oct 16, 2015 #8
    Ok thats great. It's a idea I have for a electric drive. If it works 1 gee travel will be the road to the stars. Best part is most of the things have been made for other things.
  10. Oct 16, 2015 #9


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    What will you use for fuel? It's a loooooooooong way to those stars.
  11. Oct 16, 2015 #10
    Well if you can't get up to the speed of light you can never try and go faster then it?
  12. Oct 16, 2015 #11
    fuel around sol H2 and O the stars well nuclear
  13. Oct 16, 2015 #12


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    It's very difficult to understand you when you type in sentence fragments and use no punctuation.
  14. Oct 16, 2015 #13
    One other point it is not a rocket there is no exhaust as long as there is power you have fuel. so neer by stars are years away
  15. Oct 16, 2015 #14


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    You need fuel to provide power, not the other way around.
  16. Oct 16, 2015 #15
    Sorry better at making things in my mind and hands then putting them into words. My math is so so as well it does not make me wrong or a fool.
  17. Oct 16, 2015 #16


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    No, I'm just trying to help.
  18. Oct 16, 2015 #17
    H2 and O are the fuel and with no exhaust it just has to coll down to water and be split bac to H2 and O
  19. Oct 16, 2015 #18
    Ok the main drive part has been made and as funny it may soundit will need a boiler.
  20. Oct 19, 2015 #19


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    Generally, it's the exhaust that provides the propulsion. Conservation of momentum. The change in mass*velocity of the rocket ship in a forward direction equals the mass*velocity of the stuff you throw out the back.

    Doesn't mean that electric power isn't feasible. You can already use electricity to power an electromagnetic accelerator that sends charged particles out the back incredibly fast (30 km/sec, for example). That means you don't have to throw nearly as much stuff out the back as with conventional thrusters, but you still have to throw stuff out the back. Plus, the further away you get from the Sun, the less electrical power will be available (due to that annoying inverse square law).
    Last edited: Oct 19, 2015
  21. Oct 19, 2015 #20
    Ok this drive will not push it like a rocket it pulls it. There is no exhaust it is big and will need a big power supply 2 to 3 time the mass you want to move 5 to 10 times if you want to lift off of earth. Thats in watts The drive will shake a lot so the mass is at the back and the output is about 1/2 of 906480.59046179 joules or 668585.77257981
    foot-pounds for one unit
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