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Moon orbit and conservation of energy

  1. May 6, 2010 #1
    I have seen this topic discussed a bit on the web, but I don't see any clear answers.

    I takes a lot of energy to make the moon go in a curve, if you had to do it with rockets you would need a lot of fuel.

    Therefore it takes a lot of energy to make the moon go round the earth for 4 billion years. As the moon and earth are more or less in the same place as they were 4 billion years ago, where does this vast amount of energy come from?

    Most of the answers I have seen seem to follow the line of: it doesn't actually require energy, because the velocity/potential energy is the same, or because it is like a frictionless pendulum...

    It seems to me any explantion has to explain why it does take a lot of energy to move the moon if you do it with a rocket, but the exact same "flight path" can be done with no energy spent using gravity. This seems a contridiction.

    I'm guessing nature has some conservation of energy "gotcha' somewhere, but I can't see where it is.
  2. jcsd
  3. May 6, 2010 #2


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    This is a misunderstanding of what an orbit is and what energy is. Energy (work) is force times distance. Since the distance between the earth and moon never changes from orbit to orbit (minus some losses), there is no energy required to maintain an orbit.
  4. May 6, 2010 #3
    I can see how that makes sense for me sitting in this chair, Gravity applies a force to me, but I do not gain kinetic energy and lose potential energy, as I am not moving. Therfore there is force, but not distance, so no energy.

    However my understanding of newtonian physics is that something changes direction, this is a type of acceleration. In the case of me sitting on a chair there is no acceleration. Is it possible to accelerate an object but not spend any energy doing it?
  5. May 6, 2010 #4


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    Yes: when the acceleration vector is perpendicular to the velocity vector, there is no speed - and therefore no distance - associated with the change in velocity.

    Consider that many of the same forces exist in any spinning solid object, such as the earth or a bike wheel. If you flip over a bike and spin the wheel, it'll spin for a long time (slowed by friction) while forcess inside the wheel and between the wheel and spokes counter the force tending to want to pull the wheel apart.
    Last edited: May 6, 2010
  6. May 6, 2010 #5

    D H

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    Welcome to PhysicsForums, matt_nz!

    A lot of energy is needed for an object such as the Moon to be in orbit about the Earth. The Moon has a lot of kinetic energy, about 3.8×1028 joules, relative to the Earth. It does not have nearly enough energy to escape the Earth's gravitational pull. The potential energy of the Earth-Moon system is about -7.6×1028 joules. The total energy of the Earth-Moon system (kinetic+potential) is about -3.8×1028 joules.

    What you are missing, matt_nz, is that the total energy does not change1 month-to-month. Ignoring eccentricity, zero work is performed keeping the Moon along its orbit. The Moon's orbit is not circular. This means that sometimes the Moon's kinetic energy is a bit less than the average value, sometimes a bit more. However, the total energy remains constant. The work performed over the period of one orbit is zero.


    1 The above is not quite true. The Moon is slowly receding from the Earth due to transfer of angular momentum from the Earth's rotation about its axis to the Moon's orbit. This is a small effect. Conservation of energy still applies of course. You just need to draw the system boundaries a bit broader to encompass the Earth's rotational kinetic energy.
  7. May 6, 2010 #6
    I am still a bit unclear about this.

    The core of my question is more:

    "Why do I have to burn 1 million litres of fuel,to power the rockets that move the moon 1 degree off it's trajectory, but the earth can do it for free."

    The rockets could not have an acceleration vector perpendicular to the velocity vector, and burn no fuel for the trajectory change, why is the earth different?
  8. May 6, 2010 #7


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    Your question centers on: where does the leverage come from in each case?

    The Earth-Moon system has a common center of mass, and both the Earth and the Moon are moving around that common center of mass.

    Take a strong rope, tie a brick on one end and start swinging that around, in a circle. The rope is used as a tether. Gradually you release more rope, but you keep the rotation so fast that the brick doesn't touch the ground. By the time that tether is several meters long you really have to brace yourself. You will be leaning backward, and your torso will not be on the axis of rotation; your torso is rotating around the axis of rotation.

    In fact, the further your torso is from the axis of rotation, the more leverage you have, enabling you to keep that brick swinging. In all of that, the common center of mass of you and the brick is not moved.

    Now, what if you would not have that rope?
    As you write, rocket propulsion can maintain that circular trajectory too, and your qeustion is, where does the leverage come from.

    Here's why a rocket is able to propel mass:
    A rocket shoots out exhaust at extremely high velocity. You have rocket exhaust going in one direction, and the rocket plus payload is going in the other direction. In rocket propulsion, as in any dynamics, the common center of mass is not moved.
  9. May 6, 2010 #8


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    Welcome to PF!

    Hi matt_nz! Welcome to PF! :smile:
    The work-energy theorem says that change in energy = work done (ie = force "dot" distance).

    As russ_watters :smile: says, if the acceleration (or force) is perpendicular to the velocity, then the work done is zero, and the energy stays the same.

    This happens all the time in magnetic fields (eg all those pictures from CERN) … magnetic force is always perpendicular to velocity, and so magnetic fields never change the energy. :wink:
    If the rocket force is radially outward or inward, then it will do no work, but the Earth's gravitation will do work, and that causes the change in energy.

    In every other direction, there is a magnitude for the rocket force that will do work that cancels out the work done by gravity, and the energy will stay the same.
  10. May 6, 2010 #9
    Does this mean that in some sense the moon and earth can be treated as one object that is spinning.

    If it was the case that there was no gravity (say both the earth and moon where an empty shell 1mm thick), but there was a steel bar connecting the 2 of them, and the whole object was rotating around a centerpoint a little outside the earths center, it would not make sense to ask what energy is needed to make the moon change direction, as it is just the spinning objects momentum (angular momemtum I guess?).

    Given the gravity is a force that holds the objects together, just like the steel bar, and given that there earth and moon do rotate around a point close to the earths center, is it accurate to say the steel bar case and gravity case are more or less the same thing?
  11. May 7, 2010 #10


    Staff: Mentor

    Because rocket engines are very inefficient for this purpose, and 100% of their energy is wasted in the exhaust. The KE of the moon is not changed in its orbit, so with an efficient mechanism it can be done without any energy.
    Last edited: May 7, 2010
  12. May 8, 2010 #11
    How many rockets of the power of the Saturn V series, would it take to accelerate or retard the moons orbital speed by 1 second? If the rockets were positioned with their exhaust vertical and they were placed at the point on the moon that is 90 degrees to the moon to earth centerline in the direction of the orbital plane. Like a chinese wheel driven by fireworks. Would this not accelerate the moon to a higher or lower orbit and thus change its orbital period?
  13. May 8, 2010 #12

    Andy Resnick

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    I'm not sure you understand that the moon is in free fall- there is no energy required to *keep* the moon in free fall. To deviate the moon from free fall requires energy. That is, your million litres of fuel is needed to *prevent* an object from continuing to fall freely.
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