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If energy is force times distance and I use the Newtonian formula to calculate the force between the earth and the moon $$ F=\frac{G m_1 m_2}{r^2} , $$ then multiply it by the circumference of the orbit (distance), will this give me the amount of energy in joules spent by the moon in orbit for one revolution? As a curiosity, where did the moon get the energy in the first place if it is conserved? (theories welcome).

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    $\begingroup$ No it doesn't, and no it won't. The orbital energy of the moon is slowly decreasing due to tidal forces, but that effect is tiny and unrelated to your question. $\endgroup$ – J. Murray Jan 17 '18 at 2:09
  • $\begingroup$ One more point, the moon earth orbit is not perfect circle, there're little fluctuations between PE and KE. See another answer for your further interest. $\endgroup$ – Ng Chung Tak Jan 17 '18 at 13:11
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No.

The problem with your understanding is in the definition of work. While the work is given naively by "$W = Fd$" which looks like a scalar equation, it is in fact a vector equation $W = \mathbf{F} \cdot \mathbf{d}$ where this is a vector dot product (and more generally, for varying displacement and force, an integral of this dot product.). Thus in terms of scalars only this is $W = Fd \cos \theta$, where $\theta$ is the angle between the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$ of the point of application of that force. The other form applies only when the force and displacement are in the same direction.

In the case of a circular orbit, which the Moon's is approximately, $\mathbf{F}$ and $\mathbf{d}$ are at right angles with each other, so $\theta = 90^{\circ}$ and thus $\cos \theta = 0$. Thus $W = 0$, no work is done, and thus no energy need be expended. So the basic error is not taking into account the direction the forces and displacements have, and work depends crucially upon these.

Now of course, one may object that the Moon's orbit is not perfectly circular, thus $\theta$ will be something else and so it will do work. And this is true. But because gravity is a conservative force, the work done around the whole orbit is zero - during one part of the orbit, the gravitational force does positive work, and during the other, negative work, both of which exactly cancel, and so there is no net flow of energy required. The change in energy during each phase comes as a conversion between gravitational potential energy and kinetic energy, so that as the Moon approaches the Earth, the gravitational potential energy becomes kinetic energy and it speeds up, and as it gets farther away, the opposite occurs and it slows down again.

ADD: You ask another question and that's where the Moon gets its energy from in the first place. Now I'm not sure what energy you're asking about - namely is it the energy it has now in its orbit or this putative "sustaining energy"? Since we've already shown that the latter doesn't exist, it is not possible to answer that, but the other is definitely a suitably sensible substitute and we can ask about it.

The answer to this requires an answer to the reason for the formation of the Moon. This is still not an entirely settled question, but the most credible theory so far is the "giant impact hypothesis" which states the Moon was formed near the beginning of the history of the Earth by a collision between it and a planetoid known as "Theia" which, perhaps in some sense could be regarded as the last of a series of major collisions between planetary-scale bodies that formed the Earth in the first place.

And this gives two sources of energy. This impact broke off a large amount of debris, lofting it into orbit around the Earth. Physically, this corresponds to an increase in gravitational potential energy. The source of this energy was the kinetic energy of the impactor. This debris formed a small disc around the Earth similar to the one that was around the Sun and from which the Earth itself formed. The same process of coalescence, called accretion, that formed the Earth in orbit around the Sun also happened here, and formed the Moon at a distance much closer than now. Over time, however, the Moon would gradually recede as the tidal forces from it stole some rotational energy from the Earth, causing the day to lengthen to the present 86.4 kiloseconds (24 hours) length. This, then, is the other source of energy to the existing Earth-Moon orbital system. And it is still receding - in fact, in the far future, it will become far enough away that total solar eclipses will no longer occur. Even further, it is theoretically possible the Earth's rotation will be slowed to match the Moon's orbital period, at that point exceeding the current ~2.5 megasecond period. But in reality, this will not get a chance to happen, for the timescale required exceeds the remaining lifetime of the Sun, and it is expected the Earth-Moon system will be incinerated in the red giant phase at the end of the Sun's lifetime as it expands to reach Earth's orbit and the drag from the atmosphere causes it to fall inwards and be consumed.

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As other answers say there is no work done because of the fact that the force of gravity of the earth on moon and moon on earth are nearly orthogonal to their orbit around the barycenter. The tidal energy drain is not insignificant. it has to do to viscosity of earth and moon and the fact that earth has to be deformed by a huge pair of bumps( not just the oceans, the entire mass of earth) constantly runing along its surface with a huge speed of one rotation a day, approximately.

This protruding couple of masses tend to drag to a stand still and calmness, if it were not for moon's gravity pulling it around like a tail in its orbit.

This gradual bleeding of moon's angular momentum will cause it to go to higher and lower energy orbits and also increase the period of earth's day.

Another side effect of this tidal movement on earth is creating tectonic strains and stresses which as predicted has increased the odds of some big earthquakes around California next decade.

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The moon-earth system is in almost perfect gravitational equilibrium, which means it requires no work be expended to maintain the system in its current configuration. Friction, in the form of what are called "tidal effects," very slowly perturbs this system, causing the moon to very very very slowly spiral away from the earth.

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    $\begingroup$ Actually, the "tidal effects" that you mention are causing the moon to very very very slowly spiral outward, not inward. The tides are robbing the Earth of its rotational energy, and transferring that energy to the Moon's orbit. The Earth's rate of rotation slows down, the Moon's orbital distance increases, and both will continue until the Earth's rotational period becomes equal to the Moon's orbital period. $\endgroup$ – Solomon Slow Jan 17 '18 at 2:59

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