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Projectile and Uniform Circular Motion

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    An object is launched horizontally from the highest peak on the Moon's surface. Calculate (a) the launch speed required so that the object travels completely around the Moon and returns to its original position, and (b) the time required to return to the launch point. Assume the free-fall acceleration at the launch point is 1/6 of the value at the surface of the Earth and that the radius of the object's circular path is 1.74 x 10^6 meters. BEGIN your solution by noting that this object simultaneously executes TWO types of motion, both of which you have studied. What are they?

    2. Relevant equations
    [tex] R = \frac{V_0} {g} sin(2 \theta ) [/tex]
    [tex] \vec a_r = \frac{v^2}{r} [/tex]

    3. The attempt at a solution
    I first tried solving it as a symmetric projectile motion problem which happened to be a circle which is the circumference of the moon but that ended when I realized the problem specified the launch angle to be horizontal or 0. That ruined the equation I used because Sin(0) is zero along with everything that Sin(theta) is multiplied by.

    After I gave up on projectile motion I thought about trying to think of it as uniform circular motion but it is not uniform as there is gravity pulling the object down? As I am writing this I think I might be able to think of the gravity as the radial acceleration and solve based off that if the gravity does not pull the object down.
     
  2. jcsd
  3. Feb 6, 2016 #2

    Nathanael

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    Homework Helper

    Sounds like a promising approach, let us know how it goes.
    By this you mean if it does not lose altitude, right? If you throw it too slow, it will lose altitude. If you throw it to fast, it will gain altitude. If you throw it at just the right speed, it will stay at the same altitude (i.e. it will move in a circle about the moon).
     
  4. Feb 7, 2016 #3
    It seems to have worked. Hopefully didn't make any mistakes
    [tex] \sqrt{r * a_r} = v [/tex]
    [tex] v = 1685.82 m/s [/tex] only accurate to 3 SF

    [tex] s = 2 \pi r [/tex]
    [tex] s/v = \Delta t [/tex]
    [tex] \Delta t = 6485.10 [/tex] only accurate to 3 SF
     
  5. Feb 7, 2016 #4

    Nathanael

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    Homework Helper

    Good job :smile:
     
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