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Gravitation question

  1. Jan 2, 2012 #1
    while in downward motion or in an orbit astronomers experience weightlessness
    their apparent weight (the mass they feel is m = (g - a)
    where g is acceleration due to gravity and a is acceleration with which the person is going downwards.
    now my question is what will the person feel if a>g???
     
  2. jcsd
  3. Jan 2, 2012 #2
    If he is in a falling lift and a is less than g then you are on the right track. His apparent weight would be m(g-a) this is what would be indicated on bathroom scales on the FLOOR of the lift.
    If a is greater than g then the bathroom scales would have to be on the ceiling of the lift.........this would now become a "floor"
     
  4. Jan 2, 2012 #3
    Your equation doesn't make much sense since it says that [itex]a = g - m[/itex] and the units are wrong since mass is a scalar and acceleration is a vector, rather than the "mass" that they feel it is, as you said previously, their apparent weight (I'm not sure if this was a slip of the tongue but it is a distinction you will need to be comfortable with).
    Now if [itex]a>g[/itex] is an interesting question, imagine you're in an elevator going up, what do you think the answer is?

    Edit: beaten to it :)
     
  5. Jan 2, 2012 #4
    mg is the force holding you in place....what you feel near earth is relative to that.

    ma is the force from some form of acceleration...maybe an elevator accelerating, maybe a
    rocket launching.
     
  6. Jan 2, 2012 #5
    Mg is the force on an object of mass m due to gravity.
    Ma is the value of the resultant force on an object of mass m as in
    F = ma
     
  7. Jan 2, 2012 #6
    Let us suppose that you are in free fall in orbit and you are wearing a suit made of ferrous materials. A bad guy flies near, hovers below you and turns on a powerful magnet which exerts a force in the same direction as the force due to gravity. What you will experience is a decay in your orbit because your velocity is no longer adequate to keep you at the same distance from the body you were orbiting. If you can turn on your thrusters and increase your tangential velocity to compensate for the additional radial force you'll be back to the same "weightless" feeling but at a different orbital speed.

    Regarding what you feel - according to those who do the "weightless" thing which is really free fall without ever hitting the ground, the sensation makes you literally lose your lunch. Nausea is a significant problem in free fall. But otherwise, when you think about it, you can free fall orbit any large body. These bodies do not have the same force due to gravity and you can free fall orbit at different altitudes where the force due to gravity will be larger or smaller, so the sensation will be the same - you'll feel like you are going to puke no matter what g is at that point.

    IF you should hit the body of course the value of g and the height from which you fall will determine how hard you hit but free fall with different values of g will just give you a slightly different feel when it first begins but after that it's just floating around in that space station doing your job or hanging out.
     
  8. Jan 3, 2012 #7
    @JHamm dude plz see its written m(g-a) up there and not "a (g - m) as u thought it was



    hey guys thanx a lot for ur help :)
    Είστε το καλύτερο
     
  9. Jan 3, 2012 #8
    [tex]m = g - a[/tex]
    [tex]m + a = g - a + a [/tex]
    [tex] m + a - m = g +(-1 + 1)a - m = g - m [/tex]
    [tex]a + (-1+1)m = a = g - m [/tex]
     
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