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GPE, ME, and Binding Energy

  1. Jan 25, 2013 #1
    Hello,

    I am currently reading about the conception of a general expression for gravitational potential energy. I understand that we have to use methods of calculus to generate a general expression of gravitational potential energy, because when considering alterations to the configuration of an object in a gravitational field that are large, the force varies. The force is varying as it is being applied over a distance to counter the gravitational force it is experiencing from another object. What I am getting confused over is this part of the derivation:

    [itex]U_f-U_i = -GM_Em(\frac{1}{r_f}-\frac{1}{r_i})[/itex]

    The paragraph accompanying this fuction:

    "As always, the choice of a reference configuration for the potential energy is completely arbitrary. It is customary to choose the reference configuration for zero potential energy to be the same as that for which the force is zero. Taking [itex]U_i=0[/itex] at [itex]r_i= \infty[/itex], we obtain the important result,

    [itex]U(r)=-\frac{GM_Em}{r}[/itex]"

    Th portion in red I don't quite understand. Is that saying when the two bodies in question are separated by an infinite distance, the initial potential energy is zero? What is that suppose to mean, and why does it allow us to do away with the [itex]U_i[/itex] term?

    Also, in my textbook it says, "When two particles are at rest and separated by a distance r, an external agent has to supply an energy at least equal to [itex]+Gm1m2/r[/itex]to separate the particles to an infinite distance." I know that if I wanted to separate, say, the moon and earth, which as a system possess gravitational energy, that I would need to apply a force on the moon. Moreover, I would have to be applying this force over a distance, meaning I am putting energy into the moon. But how am I suppose to know that the equal that I provide by applying a force over a distance to free the moon from earths gravitational pull is exactly [itex]+Gm1m2/r[/itex]? Honestly, if someone was in the enterprise of moving the moon, they'd be more concerned with the force that earth pulls on the moon, so that they could counter it and move the moon, and not with the gravitational potential energy of the moon. Again, this infinite distance business comes up, what exactly do they mean by this?

    Finally, I am also reading about Energy Considerations in Planetary and Satellite Motion. The mechanical energy equation the derive, of which I understand, is [itex]E = 1/mv^2 - \frac{GMm}{r}[/itex], where we don't consider the kinetic energy of the object the satellite is rotating around. One thing the author says in my textbook is, "...[itex]E[/itex] may be positive, negative, or zero, depending on the value of [itex]v[/itex]." Could someone give scenarios that would correspond to those possibilities, that [itex]E[/itex] is positive, negative, zero?

    I would appreciate the help. Thank you in advance!
     
  2. jcsd
  3. Jan 25, 2013 #2

    Doc Al

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    Staff: Mentor

    As the first sentence in the quoted paragraph reminds you, the choice of U = 0 is arbitrary. But it's very convenient to choose such a reference (U = 0 when r = ∞) as it simplifies any calculations.
    The power of energy considerations is that you can answer the question of how much energy is needed to separate two masses without worrying about the details of the force involved.
    If the total energy is negative, the bodies are bound together. (The satellite cannot escape.) If the satellite has too much energy (it has a speed greater than the so-called escape velocity), the 'orbit' is free. The satellite will just keep going (assuming it's not on a collision course).

    Do a search on "orbits" and you'll find various classifications.
     
  4. Jan 25, 2013 #3

    mfb

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    2016 Award

    Staff: Mentor

    Right.

    The first formula does not fix the potential to a unique value - you could use any initial position you like to define the potential.

    In a lab, it is conventional to define the floor as "0 potential energy here". But if you do not have a floor, how do you define a unique potential? The easiest way is "0 potential energy is at infinite distance".

    This is constant - and in classical physics, you do not have to care about absolute energy. Only energy differences matter, and they remain the same if you remove a constant term.

    This is a direct result for the formula above. The motion of the moon is ignored here.

    But that is the same!

    Well, in our universe, objects cannot have an infinite distance. But you can get quite close - if you move moon to 1000 times its current distance from earth, you have to provide 99,9% of the energy needed to move it "to infinity" (or "arbitrarily far").

    This should be 1/2mv^2, I think you just forgot the 2.

    It is negative for everything orbiting another object, e.g. the moon. It is positive for objects so quick that they leave the other object permanently, for example the Voyager probes, leaving the solar system. It is zero for objects directly in between.
     
  5. Jan 26, 2013 #4
    Okay, I believe all of my questions have been sufficiently answer. Except, however, how do I know that [itex]+Gm_1m_2/r[/itex] is the minimum amount of energy I need to provide an object in orbit, in order that it may leave its orbit?
     
  6. Jan 26, 2013 #5
    The quote you posted actually says objects at rest, not in orbit. That changes the answer. The point is that the total energy needs to be at least 0 for the object to escape (again using the convention where the potential is zero at infinity). If it is at rest, the total energy is [itex]-\frac{Gm_1m_2}{r}[/itex] and therefore you need to supply at least [itex]\frac{Gm_1m_2}{r}[/itex] to get it to 0. If it is in orbit, it has some kinetic energy and the minimum energy will be different.
     
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