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I About gravitational binding energy and kinetic energy

  1. Jul 2, 2018 #1
    If a celestial body's kinetic energy (say, the Moon's) surpassed the energy necessary to exceed the gravitational binding energy (GBE), would said body break apart because of it?

    For example, the Moon is currently orbiting the Earth at a speed of 1020 meters per second, giving it a kinetic energy of roughly 3.8e28 joules. Its gravitational binding energy is approximately 1.24e29 joules. If it suddenly (and somehow) accelerated to 100,000 meters per second, grossly exceeding its gravitational binding energy, would it break apart or anything from the strain?

    Thanks in advance.
     
  2. jcsd
  3. Jul 2, 2018 #2

    phinds

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    Uh ... I think you need to go back to the basic definition of kinetic energy, since you seem to think it has an absolute value. Hint: It's frame dependent.
     
  4. Jul 2, 2018 #3
    Correct me if I'm wrong, but isn't the kinetic energy that is conserved non-reference frame dependent?
     
  5. Jul 2, 2018 #4

    phinds

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    If you pick a frame of reference and stick with it the energy in a closed system is conserved. (1) That doesn't mean that kinetic energy is conserved. Think of an apple falling to the ground. It has kinetic energy on the way down but if your reference frame is the surface of the Earth, it has zero after it hits the ground. (2) Your original post seems to imply that you think the moon has some specific, absolute, kinetic energy. it doesn't. (3) "If it suddenly (and somehow) accelerated to 100,000 meters per second" is just handwaving magic unless you explain HOW the external force is applied to make that happen. The result of such a force application might well cause the moon to fragment.

    You need to be very specific when asking scientific questions, and hand-waving is not allowed.
     
  6. Jul 2, 2018 #5

    mathman

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    Wouldn't speeding it up make the moon escape from the earth?
     
  7. Jul 2, 2018 #6

    CWatters

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    Look up how fast the solar system is moving around the galaxy.
     
  8. Jul 2, 2018 #7

    CWatters

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    Depends how much you speed it up.
     
  9. Jul 2, 2018 #8

    Bandersnatch

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    When talking about binding energy of astronomical objects, you're talking about supplying enough kinetic energy to some system to bring all of its components outside the gravitational potential well produced by those same components.
    This means that you have to be careful to identify which gravitational potential you're overcoming, because only that potential matters.

    When you're talking about the Moon in its orbit, the system bound by gravity is the Earth-Moon system.
    You're measuring both kinetic and potential energy (i.e. binding energy) of the components of this system w/r to a stationary reference frame of the Earth-Moon barycentre (or Earth's centre, if you're lazy).

    Increasing KE of the Moon sufficiently for it to exceed the escape velocity would unbind the system in the sense that the Moon would stop orbiting Earth and fly away.

    When you're talking about the binding energy of the Moon itself, the system bound by gravity is all the matter that the Moon is made of - and nothing else.
    You're measuring both KE and PE of elements of this system w/r to a stationary reference frame of the centre of mass of the Moon.

    The components of this system (all the lunar rocks etc.) don't have any KE energy w/r to the centre of the Moon, apart from whatever comes from its monthly rotation and residual temperature.
    In this reference frame, the Moon has zero energy associated with its orbit around Earth. This is, of course, because all components of the Moon orbit the Earth at the same speed, so they're all stationary w/r to each other.

    If you wanted to unbind the Moon, you'd have to increase the KE of the particles making up the Moon in the frame of reference associated with its centre of mass.
    Spinning it up is one way of doing it. Heating it up is another.
    Making the entire Moon go faster w/r to some external body won't achieve anything (unless they then collide).
     
  10. Jul 3, 2018 #9

    mathman

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    I presume it is speeding up enough relevant to the initial question.
     
  11. Jul 3, 2018 #10

    phinds

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    Uh ... what do you reckon would happen if you applied enough force to accelerate the moon out of the solar system in a period of, say, 5 seconds? Unless you could apply the force uniformly to every molecule in the moon, it would break apart. That's how I read what the OP is asking about.
     
  12. Jul 3, 2018 #11

    Bandersnatch

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    1. Why not 58 quintillion seconds? Why non-uniform acceleration instead of uniform? Let's not add content to the question that isn't there.
    2. If you did manage to accelerate the entire Moon to the speed specified by the OP, then the application was uniform and/or gradual enough to prevent breaking. If you didn't manage to keep the Moon intact, then it was functionally equivalent to a collision, and you only have bits of the satellite moving at the specified orbital speed. And that's a different setup than what is asked for.

    While the mention of 'strain' in the OP might suggest that they're concerned with the method of acceleration breaking apart the Moon, I think it's a red herring. The rest of the post suggests it's just a matter of confusing unrelated quantities (lunar orbital KE and its binding energy), and that they think all that extra energy would cause some strain (because where did it go, eh?).
    As you said in your first response, the whole thing smells of a fundamental misunderstanding of what it means for KE to be frame dependent, or not knowing it altogether.
     
  13. Jul 3, 2018 #12

    phinds

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    Agreed
     
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