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Gravitional Energy

  1. Jan 28, 2007 #1

    fsm

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    1. The problem statement, all variables and given/known data
    A projectile is shot straight up from the earth's surface at a speed of 1.40×10^4 km/hr.

    How high does it go?

    2. Relevant equations
    [tex](1/2)m(v_0)^2-((G*M*m)/(R+y_0))=(1/2)mv^2-((G*M*m)/(R+y))[/tex]


    3. The attempt at a solution
    I assumed this is a gravitational energy problem instead of a flat Earth energy problem because of the speed for which it launched. Converting to m/s I get 3889 m/s. Using the equation above, the mass of the projectile cancels and is not needed. That provides the equation:

    [tex](1/2)(v_0)^2-((G*M)/(R+y_0))=(1/2)v^2-((G*M)/(R+y))[/tex]

    Since y_0 is 0 and assuming the max height is being asked for, v=0 m/s.

    [tex](1/2)(v_0)^2-((G*M)/(R))=-((G*M)/(R+y))[/tex]

    Solving for y:
    [tex]x=-((4*G^2*M^2)/((R*(v_0)^2-2*(G*M)))*(v_0)^2)-R-((2*G*M)/((v_0)^2))[/tex]

    The answer I get is 12,700km which is wrong. When I do this problem like an ordinary energy problem I get 770km for the height. Any help?
     
  2. jcsd
  3. Jan 28, 2007 #2
    I would work this problem by integrating the force (due to Earth's gravitational field) function with respect to displacement. That will yield a change in potential energy, which should be equal but opposite to the original energy (kinetic) of the object in question.

    Someone please feel free to correct me if I'm wrong.
     
  4. Jan 28, 2007 #3

    fsm

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    Huh? I don't understand. How does that get me the height?
     
    Last edited: Jan 28, 2007
  5. Jan 28, 2007 #4
    The upper limit of the integral is height, the lower the earths radius. The integral = Vi^2/2. Sort of like a kinematics soln where sqrt(2a*y)=v^2 but you cannot assume costant a as that varies with inverse square.
     
  6. Jan 28, 2007 #5

    fsm

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    Where did you get Vi^2/2? I don't know what you mean about the kinematics soln either. This seems WAY more complicated than the book and the professor's lectures.
     
  7. Jan 28, 2007 #6
    let me try again; potential energy gained=kinetic lost
    for linear accelaration and a vertical shot like the one you descibe,h is height

    m(9.80)h=1/2mv^2

    But this is like a mach 10 shot w/o drag--gonna go real high so we need to aaccount for the change in g, so one ends up with an integral from the surface of the earth to the peak altitude with the force of gravity varying as a function of height. At least I believe that was what Capt Zappo was referring to. I believe the soln is in natural logs.
     
  8. Jan 28, 2007 #7

    fsm

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    Ok I understand the explanation but they way that is being suggested to solve the problem is foreign to me. It seems like this method is a new method to me by comparing my notes and textbook. While I don't doubt that this way is correct and valid I just don't want to learn another method when the methods presented to me are troublesome enough.
     
  9. Jan 29, 2007 #8
    actually the methodology is the same, just expressed a bit differently. The eqn you cited lumps all the energy both potential and kinetic on each side of the eqn instead of the difference in kinetic energies being equal to the gain of potential.

    I believe if you let Yinit=0, and collect all the terms,
    Y(distance from the center of the earth) and consider the rocket at rest at apex:

    2GM/R-2GM/h=V^2, solve for h, and then subtract R from it as again, is referenced to the earths center.
     
  10. Jan 29, 2007 #9

    fsm

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    My equation is perfectly valid. Energy is being conserved throughout the problem because gravity is the only force acting on the projectile sans drag. No external forces are acting on the system. This allows the starting and ending energies to be equal. All I did was to forget to subtract the Earth's radius from my answer. Thanks for the help.
     
  11. Jan 29, 2007 #10
    never said it wasn't valid. just diffeent, glad I was of assistance.
     
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