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The Fastest Speed You Can Walk

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

    Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface. In planning a future mission to Europa, what is the fastest that an astronaut with legs of length 1.1m could walk on the surface of Europa? Europa is 3100 km in diameter and has a mass of 4.8*10^22 kg.

    2. Relevant equations

    g planet = (G*Mplanet)/Rplanet^2
    vmax= square root of g*r, where r is the length of the leg.

    3. The attempt at a solution

    So I plugged in 1.1 for r, and found that g equals (6.6*4.4*10^22)/1550^2.
    This gave me an answer of 3.828672157·10^8 which according to Mastering Physics is wrong. Any ideas where I messed up?
     
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  3. Nov 1, 2009 #2

    rl.bhat

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    Calculate the escape velocity of the planet.
     
  4. Nov 1, 2009 #3
    Hmm I don't think I've learned how to do that yet. Is there a formula I can use?
     
  5. Nov 1, 2009 #4

    D H

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    Where did this second equation come from, Katlyn? I take it that this is what you were told to use to solve the problem. If so, you need to calculate g correctly and then apply this second equation.

    The problem is that you did not calculate g correctly. What is the value G? It is not just a number; it has units associated with it. You must make sure all of the units are commensurate. If you don't you will come up with nonsense.
     
  6. Nov 1, 2009 #5
    I know that G, according to my textbook, is 6.67 *10^-11 N*m^2/kg^2

    My textbook also said that gplanet = (G*Mplanet)/Rplanet^2

    So I would get gplanet= (6.67*10^-11N*m^2/kg^2)*(4.8*10^22 kg)/(3100/2)^2

    So that would give me 1.332611862·10^6 N*m^2/kg*km right?

    Is there a way to simplify the units? Or do I just use this in the second equation?

    vmax= square root of g*r, where r is the length of the leg.
    vmax= sqrt(1.332611862·10^6 N*m^2/kg*km) * 1.1 m
    vmax= 1.465873048·10^6 N*m/kg*km

    Is this the correct answer?
     
  7. Nov 1, 2009 #6

    D H

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    One newton is 1 kg*m/s2. Use that to come up with a more useful (in this case) set of units for G. G involves a unit of length. What is that unit of length? Is this commensurate with the value you are using for the diameter of the planet?

    Hint: I wouldn't be asking these questions if you didn't have a problem with your units.
     
  8. Nov 2, 2009 #7

    Borek

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    Is N*m/kg*km unit of speed?

    Note: in case you have any doubts - this is a rhetorical questions. It is not.

    --
    methods
     
  9. Nov 2, 2009 #8
    Oh ok, so N*m^2/kg^2 multiplied by kg and divided by m^2 gives me units of m/s^2

    I also re-did my calculations and came up with 1210 m/s as my answer.
     
    Last edited: Nov 2, 2009
  10. Nov 2, 2009 #9

    D H

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    You are missing the point, Katlyn. Let me illustrate with a simpler problem: How long does it take for a car is moving at 30.0 miles per hour to move 30.0 meters? Using t=d/v naively yields 1. One what, however? That result has goofy units: 30.0 meters / 30.0 mph = 1.00 hour*meter/mile. To use the equation t=d/v one must make the units commensurate with one another. Converting 30 mph to metric yields a speed of 13.41 m/s. Now using t=d/v yields 2.24 seconds.

    In the problem at hand, you have diameter in kilometers, leg length in meters, and G in units of N*m^2/kg^2 (which is the same as m^3/s^2/kg; you should verify this). These are not commensurate units. You either need to express leg length and G in kilometers rather than meters or express the diameter in units of meters.

    Another way to look at it: Does you computed value of g make any sense? Europa is smaller and less dense than the Earth, so acceleration due to gravity should be considerably smaller on Europa than 9.80665 m/s2, the acceleration due to gravity on the surface of the Earth.
     
  11. Nov 2, 2009 #10

    D H

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    As an aside, the escape velocity from Europa is 2 km/s, which obviously has nothing to do with the fastest speed an astronaut with 1.1 meter long legs can walk. The maximum walking speed formula cited in the OP, [tex]v_{\text{max}} = \sqrt{gr}[/tex] results from assuming that the astronaut's legs are simple pendulums.

    The small angle approximation for a simple pendulum of length [itex]r[/itex] yields a mean angular velocity of [tex]\omega = \sqrt{g/r}[/tex], which, combined with the relation [itex]v = r\omega[/itex] yields [tex]v=\sqrt{gr}[/tex].


    To Katlyn: Your incorrect result arises solely from an incorrect value for g for Europa.
     
  12. Nov 2, 2009 #11
    Ohhhh, I see what you're talking about. No wonder I was getting such weird answers. Thanks!!
     
  13. Nov 2, 2009 #12
    The value that you use for the gravitational constant(G) is for the planet Earth,therefore you need to calculate the constant for Europa..
     
  14. Nov 2, 2009 #13

    D H

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    Capital G is the universal gravitational constant. It has nothing to do per se with the Earth. Lower case g is every Earth-centric. It is the mean gravitational acceleration on the surface of the Earth. By extension, the gravitational acceleration on the surface of some other planet or moon is also called g. The gravitational acceleration on the surface of Europa is the quantity Katryn needed to determine, and she apparently finally did arrive at the right answer; see post #11.
     
  15. Nov 2, 2009 #14
    umm yea right (G) is also known as universal gravitational constant... ignore my post/

    my apologies
     
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