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Need help finding energy for escape velocity

  1. Mar 29, 2017 #1
    1. The problem statement, all variables and given/known data
    The gravitational potential energy of a certain rocket at the surface of the Earth is -1.9x10^12 J. The gravitational potential energy of the same rocket 300km above the Earth's surface is -1.8x10^12 J. Assume the mass of the rocket is constant for this problem.
    A) How much work is required to launch the rocket from the surface of the earth so it coasts to a height of 300km? (starting and ending at rest, no orbit, just straight up and down): Found to be deltaU= -1.8x10^12- -1.9x10^12= 1x10^11 J. (this mas be wrong but teacher said to just make corrections).

    B) What additional Kinetic energy is required to put the rocket into a circular orbit? Found to be KE= 1/2(1x10^11)= 5x10^10 J

    Here is where I have trouble.
    C) How much extra energy is required for the rocket to reach escape velocity from this orbit?

    2. Relevant equations
    V_esc=(2GM/R)^1/2
    I'm sure I am missing something. Also sure it's really easy just blanking on it.


    3. The attempt at a solution
    I get V_esc= 10927.99m/s but then I go to use the equation for KE=1/2MV^2 but I dont know how to find the mass of the rocket because we were told it was constant. So I'm just not sure if I should be using a different equation or what.
     
  2. jcsd
  3. Mar 29, 2017 #2

    haruspex

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    By what reasoning?
    If an object just reaches escape velocity, where can it go, and what PE and KE will it have when it gets there?
     
  4. Mar 29, 2017 #3
    To be honest I'm not actually sure. It seems to be correct on my quiz, but I just did KE=1/2U_g (gravitational potential energy) which was found in A.

    As to your second part, if it reaches escape velocity doesn't it just leave the Earth's orbit and go into space?
     
  5. Mar 29, 2017 #4

    haruspex

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    Well, -1/2U_g, but 1x10^11J is not its PE; that was the change in PE.
    Yes, but to what altitude, in principle?
     
  6. Mar 29, 2017 #5

    Well it asks for the additional energy.

    I dont know. How would I find that?
     
  7. Mar 29, 2017 #6

    haruspex

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    It asks for the additional energy to go from hovering at a height of 300km to orbiting at a height of 300km. That can have nothing to do with how it got to 300km. Your 1x10^11J was the energy to lift it from Earth's surface to 300km. If it had started at 299km it would have needed far less energy to reach 300km. Would you then have taken that much smaller amount of energy and halved it to find the extra energy to make it orbit at 300km?
    What does escape velocity mean? If it were enough velocity to get 1000000km from Earth, but no further, would it have escaped Earth's gravity? Where does Earth's gravity end?
     
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