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Finding # of g's experienced on Jupiter

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Jupiter is about 320 times as massive as the Earth. Thus, it has been claimed that a person would be crushed by the force of gravity on a planet the size of Jupiter since people can’t survive more than a few g’s. Calculate the number of g’s a person would experience at the equator of such a planet. Use the following data for Jupiter, and take centripetal acceleration into account.

    Mass = 1.9 x 1027kg
    Equatorial radius = 7.1 x 102km (7.1 x 102m)
    Rotation period = 9 hours, 55 minutes (35700 seconds/rotation)

    2. Relevant equations

    F = GMm/r2 = ma (or mar)
    mar = mv2 / r

    3. The attempt at a solution

    F = (6.67 x 10-11)(1.97 x 1027)m / (7.1 x 107)2 = mar

    m's cancel, I get a centripetal acceleration of 26.06m/s2

    I also found velocity; 26.06m = mv2 / r
    m's cancel once again, V = [tex]\sqrt{}26.06(7.1 x 10^7[/tex] = 43014.65m/s. Not sure if that's important.

    And from there I have no clue. I don't know how to use the rotation period (I don't think they'd give me it if it wasn't important) and I have no idea how to use the centripetal acceleration to find g.
     
    Last edited: Oct 25, 2009
  2. jcsd
  3. Oct 25, 2009 #2

    ehild

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    You have to find the normal force Fn on the person at the surface of Jupiter. The gravity of Jupiter acts towards the centre, the normal force away from the centre, and as a result, the person moves on a circle with radius of Jupiter and period equal to the rotation period of Jupiter, so with speed equal to that of the equator. (The speed you calculated would be the speed of a space probe orbiting just above the surface of Jupiter. )
    So you have to calculate the speed of the equator first, and then Fcp belonging to that speed. This is equal to the difference between the force of gravity and Fn. From that, you get Fn, and Fn/m is the "G" in question.

    By the way, the Equatorial radius = 7.1 x 104km.
     
  4. Oct 25, 2009 #3
    Okay that was very confusing, but I thought of something else.

    Given the period time, I figured I could use the V = 2(pi)(r) / T equation.

    V= 2(pi)(7.1 x 10^7m) / 35700s = 12495.97m/s

    Fcp = mv^2 / r

    Fcp = (1.97 x 10^27kg)(12495.97m/s) / (7.1 x 10^7) = 4.33 x 10^27N

    I don't know where to go from here. This is force in the horizontal direction, is it not?
     
  5. Oct 25, 2009 #4
    Anyone?
     
  6. Oct 25, 2009 #5

    ehild

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    Well, again. The question was: "Calculate the number of g’s a person would experience". What does it mean? How many g-s you feel here on the Earth?

    ehild
     
  7. Oct 25, 2009 #6
    Ah. For some reasoning I was thinking that centripetal acceleration was horizontal.

    26.06 / 9.8 = 2.66 g's felt, correct?

    But then where does the rotational period come into play?
     
  8. Oct 26, 2009 #7

    ehild

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    The 2.66 g would be the case if the Jupiter had not revolved. But it does and together with its equator, the person moves on a circle. For that, a certain centripetal force is needed, and the magnitude of this force can be calculated from the speed of the man (here comes in the period) and the radius of Jupiter. You have calculated that speed, V= 2(pi)(7.1 x 10^7m) / 35700s = 12495.97m/s. Determine the centripetal acceleration which corresponds to this speed.

    The person moves with this centripetal acceleration round, and this motion is the result of two forces: one is the gravitational pull of Jupiter, FG = GMm/r2,
    the other force is the normal force Fn from the ground, it points upward. If m is the mass of the person
    macp =mv2/r= GMm/r2-Fn.

    Determine Fn/m and divide it with g=9.8.

    ehild
     
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