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Archimedes Buoyancy Formula and washing astronauts

  1. Dec 15, 2008 #1


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    1. The problem statement, all variables and given/known data

    In the future, astronauts living on the moon will have baths in their accommodation. When using them, will they float higher in the water than on earth? Explain fully the reasoning behind your answer.

    2. Relevant equations

    Achimedes Buoyancy

    3. The attempt at a solution

    See, I know roughly what I need to do, but I can't find a formula that expresses Archimedes Buoyancy principle. I know it is upthrust force is equal to something like volume of water displaced, thus:

    [tex] F = mg = volume [/tex]

    but this is not right, since one side is Newtons, the other m^3

    I know that I need this formula, and that the force will change because the difference in gravity (from f = mg), but I just can't seem to see a formula. Any Ideas?


  2. jcsd
  3. Dec 15, 2008 #2


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    Homework Helper

    I know it is upthrust force is equal to something like volume of water displaced

    Upthrust force = weight of the displaced liquid.
  4. Dec 15, 2008 #3


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    Excellent. Thanks. So

    Upthrust force = weight of water displaced

    for this question, the weight of the water displaced is mg. This means that we need to deal with the mass in terms of its density and volume. Thus:

    [tex] F_{upthrust} = volume*density * g [/tex]

    since the astronaut won't change volume, this value is constant, so we can use any volume for the astronaut. Keeping it simple, I will say he has a volume of 1m^3

    density of water: 1000 kg/m^3


    [tex] F_{upthrust} = 1*1000 * g [/tex]

    [tex] F_{upthrust} = 1000*g [/tex]

    this means on Earth, the upthrust force will be 9800 N

    on the moon, g = 1.7 m/s^2

    thus the upthrust force will be 1700 N

    the Upthrust force is greater on Earth then it is on the moon, thus the astronaut will float lowere on the water then he does on Earth.

    Does this look right?


  5. Dec 15, 2008 #4
    Hey the astronaut is on the moon too.
    he has a moon weight
  6. Dec 15, 2008 #5


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    I din't think about that, so basically, its a case of the upthrust is acting upwards, and the weight is acting downwards?

    So on Earth, upwards we have:

    [tex] 1000*g [/tex]

    downwards we have mg

    On the moon we have upwards:

    [tex] 1000g_2 [/tex]

    where g2 is the moons g,

    and downwards we have

    [tex] mg_2 [/tex]

    Should I assume that the total force is 0, so that

    [tex] 1000g - mg = 0 [/tex]

    [tex] 1000g_2 - mg_2 = 0 [/tex]


  7. Dec 15, 2008 #6
    Surely the point is that the ratio between the astronaut's density and that of water will remain unchanged on the moon. Upthrust will decrease but so will the weight to be buoyed up. Both forces depend on the local gravity. So I guess he'll take his bath as usual.
  8. Dec 15, 2008 #7


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    I have to say, that makes a lot of sense, and since both force will decrease by the same amount, there won't be any change.

    From the question. See, I interpreted this as higher in the water, or lower. my first thought was that it wouldn't change, but my interpretation of the question made me think it couldn't be.


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