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A GRAVITATIONAL POTENTIAL ENERGY questions

  1. Apr 22, 2007 #1
    1. The problem statement, all variables and given/known data
    determine the amount of work required to move a satellite from orbital radius of
    2r_e (twice the radius of the earth) to 3r_e, if it has a mass of 500 kg?


    2. Relevant equations
    Find change in Gravitation potential energy
    and also the change in the kinetic energy

    3. The attempt at a solution

    The solution is the book does not consider the change in the kinetic energy. Shouldn't it be considered as velocity is decreased when it moves to a higher orbit
     
  2. jcsd
  3. Apr 22, 2007 #2
    What is the relationship between either Work or Potentential energy and Kinetic? You didn't show your relavant equations. Show your attempt also
     
  4. Apr 22, 2007 #3
    it's simple though
    W = change in U + change in E_k
    <I solved that question using this method>
    and, the book solved it using
    W= change in U

    yea, i didn't show my work cuz I have the solution provided in the book, so I am just confused about how the book solved it
     
  5. Apr 22, 2007 #4
    what is the relationship between potential energy and kinetic? Would it make a difference if you write the change in potential or change in kinetic? It applies to what kind of question you are solving. You did not give us the exact problem you are asking, you did not give us all the relavant equations, you did not show us any of your work. Do you want to solve the problem?
     
  6. Apr 22, 2007 #5
    hmm.. certainly

    so, here's my exact question:
    determine the amount of work required to move a satellite from orbital radius of
    2r_e (twice the radius of the earth) to 3r_e, if it has a mass of 500 kg?

    as for relevant equations <they are just too messy>
    E_k = 0.5mv^2
    E_g = - (GMm)/r

    and here's the attempt
    Work = - (GMm)/r_f + - (GMm)/r_i + 0.5mv_f^2 - 0.5mv_i^2

    v^2 = GM/r
    0.5mv^2 = 0.5xGMm/r

    so
    Work = - (GMm)/r_f + (GMm)/r_i + 0.5(GMm)/r_f - 0.5(GMm)/r_i

    Work = -0.5(GMm)/r_f + 0.5(GMm)/r_i

    so, it is equal to 0.5(GMm)[1/r_i - 1/r_f]

    so this is how i did it
    but the book did it this way
    Work = - (GMm)/r_f + (GMm)/r_i
     
  7. Apr 22, 2007 #6
    Looks like the book did it correctly, Work is the difference between two energy states. Is that clearer? Work is transfer of energy. So you can say it is the difference of two inergy states. initial and final. How much work does it take to lift a object vertically up? Did the object change in potential energy? Also, when you look at a problem, see what they give you, and that is all you know. Given what they gave you, is it most logical to solve it the book way?
     
    Last edited: Apr 22, 2007
  8. Apr 22, 2007 #7
    yea,i understand the concepts of work

    How much work does it take to lift a object vertically up?
    but isn't this situation different, because initially the object is moving perpendicular to the applied force, and in its final state, it moving with a lower velocity, thus this decreased the kinetic energy. This happens along with the change in the gravitational potential energy.
     
  9. Apr 22, 2007 #8
    Read the question carefully, it says to move the satillite from one position to another. Did the problem give you velocity? If want to solve it using work-kinetic, be my guest. If they didn't give you the right pieces of the puzzle, how can you solve it?
     
  10. Apr 22, 2007 #9
    I solved for v using that equation

    Centripetal force = gravitational force

    (mv^2)/r = (MmG)/r^2
    <cuz it's orbiting>
    anyhow, now I got the question --> "from one position to another."

    now, this is my second question
    Prove that “No satellite can orbit Earth in less than about 80 min”?
    yes, i can do this numerically, and I have done that, and it gives minimum time of 85 min

    But, since they didn't mention anything about forces, so anyone can supply more downward force<towards the center of the earth>, so as to lower the time period.
     
  11. Apr 22, 2007 #10
    Yeah, but that is velocity in its orbit. That is not radial movement. Radial positions. Distance from earth. Lets say you are running with velocity and holding and object on top of your hand, you lift it up. is it the same amount of work if you were not in motion? Work on the object.
     
    Last edited: Apr 22, 2007
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