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Law of Gravitation Problem

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data
    At the Earth's surface a projectile is launched straight up at a speed of 8.1 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.

    2. Relevant equations
    F = GMm/r2
    Ug = -GM/r2
    G = 6.67 e -11 whatever units it happens to be
    That's all I'm 'given' for this problem.

    3. The attempt at a solution
    Well, I figured that the kinetic energy that the rocket/thing starts off with converts into gravitational energy.
    1/2 mv2 = GM/x2
    where x is the distance between the centers of the rocket and the earth.
    So that would mean x = r + h?
    So if we rearrange terms to solve for x, we get:
    x = ±2GM/v2 - r
    However, that doesn't seem to work, either sign for the potential energy.
    Am I missing something?
     
  2. jcsd
  3. Dec 11, 2012 #2

    rock.freak667

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    Your initial equation should be

    ½v2= GM/x


    But in any event, you did the rearranging correctly. Why doesn't it work?
     
  4. Dec 11, 2012 #3

    haruspex

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    That's not right, but you seem to have (almost) used the correct version below.
    Looks like you originally had x2, but when you corrected it you forgot to get rid of the ±. And you've left out the PE it started with.
     
  5. Dec 11, 2012 #4
    Terribly sorry about that. I used the formula U = -GMm/x for that, where x still is what I defined it to be previously.

    So you're saying that the equation looks like:
    1/2 mv2 + GMm/r = GMm/x ...? where r is the radius of the earth?
    Let me try this out... looks promising.
    But here's another question.

    A long long time ago I learned that there was kinetic energy and potential energy, mgΔh. This type of potential energy is gravitational and so is GMm/r, so would the problem also be solved by replacing GMm/r with mgΔh?
     
  6. Dec 11, 2012 #5

    haruspex

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    No, that's only valid for constant g.
     
  7. Dec 11, 2012 #6
    aahhh... makes sense. It's similar to the reason why we don't use kinematics for equations with velocities accelerating at increasing/decreasing rates.
     
  8. Dec 11, 2012 #7
    Got it. Thanks!
     
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