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gravitational field Questions

  1. Jul 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume you are agile enough to run across a horizontal surface at 8.50 m/s, independently of the value of the gravitational field. What would be (a) the radius and (b) the mass of an airless spherical asteroid of uniform density 1.10 310 3kg/m3 on which you could launch yourself into orbit by running? (c) What would be your period? (d) Would your running significantly affect the rotation of
    the asteroid? Explain.


    2. Relevant equations



    3. The attempt at a solution
    I have a few conceptual questions that pertain to this problem. I have a belief that the force with which the person running exerts on the surface is equal and opposite to the gravitational force that the asteroid provides, and is the reason why this person is running at a constant speed. However, I am not sure of why I have this suspicion. Also, I know this is a very elementary question, but how is it possible for a person to run if the ground provides and equal and opposite force to the force that the person exerts on the surface with there foot?
     
  2. jcsd
  3. Jul 10, 2013 #2

    tiny-tim

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    Hi Bashyboy! :smile:
    Why? One is horizontal and the other is vertical, so why should they cancel? :confused:
    The only (horizontal) force on the person is the force from the ground.

    Start again. For parts (a) (b) and (c), you can pretend the speed is supplied by a rocket.​
     
  4. Jul 10, 2013 #3

    haruspex

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    If they are equal and opposite, how are you able to keep running around the curve of the surface (instead of traveling in a straight line)?
    Since speed is constant, both forces are vertical.
     
  5. Jul 13, 2013 #4
    I think I am experiencing a little confusion myself. How is the force with which the runner pushes against the ground vertical?
     
  6. Jul 13, 2013 #5

    tiny-tim

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    the push force (using friction) is horizontal

    i think haruspex is thinking of the normal reaction force
     
  7. Jul 13, 2013 #6
    Okay, so the person pushes against the ground, and the ground provides friction that propels the person forward because there is no other force that keeps the person from not moving, right?
     
  8. Jul 13, 2013 #7

    tiny-tim

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    no other horizontal force, right :smile:
     
  9. Jul 13, 2013 #8
    Hmm, well what does the speed being constant have to do with their being a normal force every instant that the person plants their foot on the ground and the gravitational force, the two vertical forces haruspex is speaking of.
     
  10. Jul 13, 2013 #9

    haruspex

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    tiny-tim, I think you are misreading the question.
    A steady speed is attained by some means. There is now no horizontal force; the tangential speed is constant. The only acceleration (while staying on the surface) is vertical (centripetal). There is no friction, no drag. The question is, what is the size of the sphere if 8.5m/s is just enough to start losing contact with the surface?
    Bashyboy, it will only complicate matters unnecessarily if you consider actual bipedal motion. Simpler to treat it as bicycling.
     
  11. Jul 14, 2013 #10

    tiny-tim

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    uhh??

    the question was …
    if you're not exerting a horizontal force, you're not running
     
  12. Jul 14, 2013 #11

    haruspex

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    That is not true. If there is no drag and you're not gaining or losing speed or changing direction then there cannot be a horizontal force. Your feet are hitting the ground merely to provide a vertical force. In practice, there will be small variations to the horizontal velocity of your mass centre, but they will average out to zero.
     
  13. Jul 14, 2013 #12

    tiny-tim

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    exactly … not running! :smile:

    (more sort of rolling)
     
  14. Jul 14, 2013 #13

    haruspex

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    No, doesn't have to be rolling, though that would be a less confusing model with which to discuss the intended question.
    Running/walking means placing feet in front of each other alternately in a manner that avoids falling over. The difference between them is that running swings the legs faster than their natural period. It normally entails a propulsive force because of resistance, but if running down a slight incline it is obvious that a propulsive force provided by friction on the ground need not be present. If it weren't for the inherent instability of our upright posture we could run some distance across ice, somewhat like a bouncing ball.
    But let's get back to the OP. It is quite clear that the question is really asking what the radius of the asteroid is if 8.5m/s is just fast enough for a surface-level orbit.
     
  15. Jul 16, 2013 #14
    I still don't quite understand how the person is able to move forward--or is it the planet that is rolling backwards?
     
  16. Jul 16, 2013 #15

    tiny-tim

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    if a stationary person wants to start moving forward, he uses his leg muscles to rotate at the knee, which makes the foot want to move backward relative to the body

    since gravity results in a normal force between him and the planet, this movement produces a horizontal friction force

    so instead of the body rotating slightly (but not moving forward), the whole body moves forard and the planet very slightly moves backward :smile:
     
  17. Jul 16, 2013 #16

    haruspex

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    Agreed. But for the purposes of the question we need not be concerned with how the speed of 8.5m/s was achieved. All that matters is that if it were just slightly faster you would take off.
     
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