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The moon is rotating around the earth in elliptic or approximately circular orbit with speed $v$. Its acceleration have two components tangential and radial one. It is the radial acceleration of the moon towards earth or more precisely towards a common center which is the core of their mechanical equilibrium through the equality $$\frac{G~m_1~m_2}{R^2}= m_1~ \frac{v^2}{R}$$ ($m_1$,$m_2$,$G$ mass of moon , earth's mass and universal gravitational constant respectively $v$ is the velocity of moon, $R$ is the radius of circular motion of moon ) . The system's total energy i.e. Kinetic + Potential is negative and corresponds to its binding energy. However, this is only a partial answer and not complete since this equilibrium is not stable and the orbit radius $R$ is not stationary i.e. it can can increase or decrease. It means that the moon could loose heat or mechanical energy hence approaches the earth or gain more energy (heat from the sun or mechanical from earth tides) and hence going far towards the sun. Recalling the analogous case of electron revolving around the nucleus and how Bohr coined his quantization formula $$2\pi mvR=nh$$ (quantization of orbital angular momentum $2\pi mvR$) just to solve this dilemma and stabilize the atom. When there is no such quantization of the moon orbit, the moon should go far and far if its net energy gain is positive (which is the case in the present days enlarging the moon orbit by 5 cm a year) . On the other hand if the energy balance formula is reversed,due to climate changes, the moon will continue approaching the earth in shorter and shorter radii again and again until it finally collides with the earth and destroys many sorts of life.Planet Mars did it before in the old ages. A serious question is which scenario for the moon future is more likely: going slowly far or coming faster towards the earth ?your answer is helpful and appreciated.

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    $\begingroup$ the moon is actually very slowly moving away from earth at the rate of around 4 cm per year, and the earths rotation slowing, due to tidal forces. $\endgroup$ – Robert DiGiovanni Nov 25 '19 at 9:27
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    $\begingroup$ The expression $$\frac{G~m_1~m_2}{R^2}= m_1~ \frac{v^2}{R}$$ is valid everywhere only for a circular orbit. As is the case here, starting with a bad assumption oftentimes results in a contradiction. The contradiction confirms that one or more of the starting assumptions was incorrect. $\endgroup$ – David Hammen Nov 25 '19 at 12:51
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First a nitpick: the mars-sized planet called Thea which is hypothesized to have collided with Earth long ago did not do so because it used to orbit earth and this orbit decayed. Instead it is assumed that the turbulent interactions of all the planets in the early solar system just put it onto a collision course by chance.

Your question has several layers to it that I'll attempt to untangle:

The system total energy Kinetic +Potential is negative and corresponds to its binding energy . However,this is only a partial answer and not complete since this equilibrium is not stable and the orbit radius R is not stationary i.e. either can increase or decrease.It means that the moon could loose heat or mechanical energy hence approaches the earth or gain more energy (heat from the sun or mechanical from earth tides ) and hence going far towards the sun.

It actually is a stable equilibrium within the mechanics you are describing here. Two perfectly rigid Newtonian bodies will orbit around each other for eternity once they start out this way. All that happens in an elliptical orbit is that energy is shuffled back and fourth between the kinetic and the potential energy term but there is no dissipative term there so it happens with perfect efficiency.

Recalling the analogous case of electron revolving around the nucleus and how Bohr coined his quantization formula mv2πR=nh (quantization of orbital angular momentum mv2πR) just to solve this dilemma and stabilize the atom.

You should have a bit of a closer look into why one expected an electron orbit to decay. Because again, from naive two-body mechanics the situation should be perfectly stable. The answer is in the electromagnetic field: An electron is a charge and an accelerating charge (such as one that is on a circular orbit) emits electromagnetic waves. These carry energy away from the system and therefore the orbit would have to decay.

The analogous situation when talking about a gravitationally bound system of orbiting bodies are gravitational waves. As you may have heard in the last few years it became possible to detect these waves as they are produced by the most gravitationally intense events in the cosmos: To black holes orbiting each other closely and just like the classical electrons the lose energy due to the waves they emit and finally fall into each other.

However, none of this matters to the Earth-Moon system because the masses of the Earth and the Moon are so low that the gravitational waves produced by this system carry a power in the milliwatt range. Much too small to make any noticeable difference.

When there is no such quantization of the moon orbit, the moon should go far and far if its net energy gain is positive (which is the case in the present days enlarging the moon orbit by 5 cm a year) .

So you have heard about this but think for a moment: This means the moon is gaining potential energy - which isn't explained by any mechanism we've talked about so far. Something else has to be going on. In particular this thing is called tidal forces. Neither the moon nor Earth are perfectly rigid and homogenous bodies. Instead the gravitational interaction of the two stretches both bodies along their connecting axis. Meanwhile the Earth rotates once per day but this "squishing" can't rotate with it. The result is that the planet gets a constant massage of stretching and pushing - slowing down our rotation and turning that kinetic energy into heat of our core. But that's not all!

Now it helps to remember Newtons third law though: every force goes both ways. Just as Earths rotation is slowed by the Moons tidal squishing so does the Earth end up pulling on the moon in the direction of its rotation. So while Earth is slowing down the Moon is speeding up, gaining kinetic energy and - per the equilibrium you wrote - also potential energy. So it slowly recedes from us.

On the other hand if the energy balance formula is reversed,due to climate changes, the moon will continue approaching the earth in shorter and shorter radii again and again until it finally collides with the earth and destroys many sorts of life.

Climate change has no effect whatsoever on this.

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  • $\begingroup$ S.Move Unfortunately you did,nt get enough comprehension of the question $\endgroup$ – user230437 Nov 25 '19 at 9:15
  • $\begingroup$ you did,nt get enough comprehension of the question..word mars-sized existed but lost by Ed machine ii-Collision by Chance ? you are only one say it.iii- stable equilibrium for non stable one you better read about atom then planetary systems where all radii are permissable.perfectly.iii- you slipped when telling that the moon is an isolated system loosing or gaining no mechanical energy or heat energy at all .better know something about thermodynamics ? iv-Tidal forces are mentioned in the question but you slipped it v-After arriving to your scientific term Earth Massage " better to stop $\endgroup$ – user230437 Nov 25 '19 at 9:42
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    $\begingroup$ @IsmailAbbas, what kind of answer are you looking for? Everything S. Move is writing is correct and according to the scientific concensus as far as I can tell. $\endgroup$ – Marius Ladegård Meyer Nov 25 '19 at 11:56
  • $\begingroup$ Marius:S.Move Answer is full of mistakes .In addition it does not comprehend the earth-moon system binding energy which is the core of the question .Please read the question again and my first comment. $\endgroup$ – user230437 Nov 25 '19 at 16:31
  • $\begingroup$ This Answer is full of mistakes .In addition it does not comprehend the earth-moon system binding energy which is the core of the question .Please read the question again and my first comment.Moreover he seems not knowing the meanings of stable equilibrium and stationary orbits. $\endgroup$ – user230437 Nov 25 '19 at 16:38

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