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Gravtational Potential energy

  1. Feb 18, 2014 #1
    When a rocket is launched into space, its GPE increases and its KE decreases.

    If we equate the change in both energies we can find out the velocity of the rocket at a certain height.
    Δ GPE = Δ KE

    Does the equation already take the opposing gravitational force of earth into account?
     
  2. jcsd
  3. Feb 18, 2014 #2

    Nugatory

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    Yes. The GPE is represents the work done against the gravitational force of the earth. If it weren't for that force, the potential energy wouldn't increase and the kinetic energy wouldn't decrease.
     
  4. Feb 18, 2014 #3
    Isnt GPE the work done in bringing a mass from infinity to a point inside a particular gravitational field?
     
  5. Feb 18, 2014 #4

    Nugatory

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    No, it's defined as the work done in bringing a mass from one point to another in a particular gravitational field, plus the PE at the initial point. The value of the PE that we assign to that initial point is completely arbitrary.

    You are thinking of the very useful and widely used convention in which we choose to assign zero to the potential energy at infinity. Then the PE at any other point is, as you say, the (negative of) the work done on bringing a mass in from infinity. However, we could just as reasonably choose to assign zero potential to a point at the surface of the earth, and this choice makes the calculations easier if you're working with a rocket launched from the surface of the earth.
     
  6. Feb 18, 2014 #5
    Does changing the position of zero position change the formula of GPE?
     
  7. Feb 18, 2014 #6

    jtbell

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    Yes. If we define GPE = 0 at r = ∞, then
    $$GPE = -\frac{GMm}{r}$$
    where M is the mass of the earth, m is the mass of the object in question, and r is the distance from the center of the earth. (This applies only above the surface of the earth)

    If we define GPE = 0 at r = R, where R is the radius of the earth, then
    $$GPE = GMm \left( \frac{1}{R} - \frac{1}{r} \right)$$

    You should be able to check for yourself that both formulas give the same result for ΔGPE between two different values of r.
     
  8. Feb 19, 2014 #7
    In this case of rocket launch what force is doing work?
     
  9. Feb 19, 2014 #8

    Nugatory

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    Gravity is doing work on the rocket, and the thrust of the rocket motor is doing work on the rocket.
     
  10. Feb 19, 2014 #9
    So can the potential energy of the rocket can be calculated by the formula

    (Thrust - Gravitational force) * distance

    In my book the gain in GPE of the rocket is calculated by -GMm/r which basically came from integrating Gravitational force* distance

    Is there any difference?

    I am really confused in the case of rocket launching
     
  11. Feb 19, 2014 #10

    jbriggs444

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    Potential energy applies when the work it takes to reach a certain position against a particular force does not depend on the path you take to get there. Gravity has this property. Gravitational force is a "conservative field". So it makes sense to talk about the potential energy of gravity. Thrust does not have this property. So it does not make sense to talk about the potential energy of thrust.
     
  12. Feb 19, 2014 #11

    jtbell

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    No, because the rocket thrust is not a conservative force. You can define potential energy only for conservative forces such as gravity.
     
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