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Thread: How do rockets work in space w/ nothing to thrust against

  1. #1 How do rockets work in space w/ nothing to thrust against 
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    How do rockets turn in space when there is nothing to thrust against being that its a vacuum. feel free to contact me at penguinhomocide@hotmail.com


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    Forum Isotope Zelos's Avatar
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    they use thrust, according to newtons law, when they thrust somethign out they experience a equal amount of thrust in the opposite direction


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  4. #3 Re: How do rockets work in space w/ nothing to thrust agains 
    Forum Professor wallaby's Avatar
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    Quote Originally Posted by PenguinHomocide
    How do rockets turn in space when there is nothing to thrust against being that its a vacuum. feel free to contact me at penguinhomocide@hotmail.com
    its not that there thrusting against anything its that the mass of the fuel being combusted pushes on the mass of the rocket when it is expelled from the combustion chamber.
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    Forum Sophomore Elbethil's Avatar
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    To my basic knowledge of the topic, the propulsion makes use of the Law of Conservation of Momentum.

    Basically, if you have two objects, the total momentum (equal to mass times velocity) of the two objects before and after their interaction (collision, explosion, etc) is the same as the total momentum of them afterwards.

    So:
    m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub> = m<sub>1</sub>v<sub>1</sub>' + m<sub>2</sub>v<sub>2</sub>'

    Where m = mass and v = velocity, and the <sub>1</sub> indicates the first object and the <sub>2</sub> indicates the second. The prime symbols indicate the velocities after their interaction; their masses are constant here, so they lack the prime symbols.

    Now, let's say that we have an explosion: one object seperates into two objects. As the object before the seperation has the mass of the two objects totaled together, our formula becomes:
    (m<sub>1</sub> + m<sub>2</sub>)v<sub>0</sub> = m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub>

    Which just means:
    (momentum of the big object) = (momentum of its first part) + (momentum of its second part)

    For a rocket, this means:
    (momentum of the rocket and its fuel) = (momentum of the rocket) + (momentum of its fuel)

    If a rocket is stationary in space, it's velocity is 0, and thus it's momentum. Leaving us with:
    0 = (momentum of the rocket) + (momentum of its fuel)
    (momentum of the rocket) = -(momentum of its fuel)

    Mass doesn't have direction, but velocity does. Hence, the velocity of one side is negative and the other positive. So, logically, if you expel the fuel with velocity in one direction, the rocket will have velocity in the opposite direction.
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    Forum Isotope Zelos's Avatar
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    you should notice this only worx in none-relativistic situations
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  7. #6  
    Forum Sophomore Elbethil's Avatar
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    True. But its generally a good idea to cover the basics (ie. classical mechanics) before going onto the more advanced stuff.
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  8. #7  
    Forum Isotope Zelos's Avatar
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    yepp, but its also a good idea making it clear its classical so new people dont run of thinking that covers anything at anytime
    I am zelos. Destroyer of planets, exterminator of life, conquerer of worlds. I have come to rule this uiniverse. And there is nothing u pathetic biengs can do to stop me

    On the eighth day Zelos said: 'Let there be darkness,' and the light was never again seen.

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  9. #8  
    Forum Freshman starry_eyed_guy's Avatar
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    Quote Originally Posted by Elbethil
    To my basic knowledge of the topic, the propulsion makes use of the Law of Conservation of Momentum.

    Basically, if you have two objects, the total momentum (equal to mass times velocity) of the two objects before and after their interaction (collision, explosion, etc) is the same as the total momentum of them afterwards.

    So:
    m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub> = m<sub>1</sub>v<sub>1</sub>' + m<sub>2</sub>v<sub>2</sub>'

    Where m = mass and v = velocity, and the <sub>1</sub> indicates the first object and the <sub>2</sub> indicates the second. The prime symbols indicate the velocities after their interaction; their masses are constant here, so they lack the prime symbols.

    Now, let's say that we have an explosion: one object seperates into two objects. As the object before the seperation has the mass of the two objects totaled together, our formula becomes:
    (m<sub>1</sub> + m<sub>2</sub>)v<sub>0</sub> = m<sub>1</sub>v<sub>1</sub> + m<sub>2</sub>v<sub>2</sub>

    Which just means:
    (momentum of the big object) = (momentum of its first part) + (momentum of its second part)

    For a rocket, this means:
    (momentum of the rocket and its fuel) = (momentum of the rocket) + (momentum of its fuel)

    If a rocket is stationary in space, it's velocity is 0, and thus it's momentum. Leaving us with:
    0 = (momentum of the rocket) + (momentum of its fuel)
    (momentum of the rocket) = -(momentum of its fuel)

    Mass doesn't have direction, but velocity does. Hence, the velocity of one side is negative and the other positive. So, logically, if you expel the fuel with velocity in one direction, the rocket will have velocity in the opposite direction.


    How did you type m1v1 and m2v2 . See I typed them and they don't look like yours!!
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