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Circular motion of mars

  1. Mar 9, 2008 #1
    The curvature in Mars is such that its surface drops a vertical distance of 2.0 meters for every 3600 meters tangent to the surface. In addition, the gravitational acceleration near the surface is 0.4 times near the surface of Earth. What is the speed a golf ball would need to orbit Mars near the surface, ignoring the effects of air resistance?


    First let me tell you that I am teaching myself physics, forgive me if I sound completely clueless. I am not quite sure how I should start. Should I use Newtons law of Universal gravitation F=Gm1m2/r^2. Circular motion equations or Hooke's law? Any other clues to get me started would be appreciated.

    Thanks.
     
  2. jcsd
  3. Mar 9, 2008 #2

    tiny-tim

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    Yes, F=Gm1m2/r^2, and also the formula for the acceleration of an object in circular motion, a = v^2/r, and of course F = ma. :smile:

    [size=-2](Hooke's law is about the force in springs, and has nothing to do with hooking golf balls!)[/size]​
     
  4. Mar 9, 2008 #3
    if the distance is too small

    you can use mv^2/r = mg
    then u find v= sqrt gr for circular around the planet you want...
    Suppose at this speed, the golf ball will move aroung the planet with a circular, low-attitude(the ground) orbit.
     
  5. Mar 9, 2008 #4
    Ok, sorry guys I know which equations I should use now, but please give me a small hint as to the first step to this problem.
     
  6. Mar 9, 2008 #5

    Doc Al

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    No need for all those formulas.
    Hint: How long does it take for the golf ball to fall 2.0 meters on Mars?

    (You could do this problem using force laws and centripetal acceleration. But that's the hard way and I don't think that's what they want you to do.)
     
  7. Mar 9, 2008 #6
    Ahhh, thank you DocAl :)
     
  8. Mar 9, 2008 #7

    tiny-tim

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    Hint: the acceleration from gravity must exactly equal the acceleration for circular motion.

    oops! … :redface: I hadn't noticed :redface: … they've told you that mars-gravity = .4 x earth-gravity, so you don't need F=Gm1m2/r^2 at all. Just use .4 x g (look the value of g up in a book if you don't know it).

    So work out what r is, and then put .4g on the left, and v^2/r on the right … :smile:
     
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