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Assume that the moon it orbiting the earth in a circular trajectory.

It will experience an acceleration directed towards the centre of the earth given by $\frac{GM}{R^2}$ where $G$ is the universal gravitational constant and $M$ is the mass of the earth and $R$ is the radius of the moon’s orbit from the center of the earth. I guess I can equate this to $\frac{v^2}{R}$ from which I get

$$v=\sqrt{\frac{GM}{R}}$$

If the velocity for some reason becomes less than this, then will the moon proceed towards earth?

And did the moon at some point in the past naturally attain this velocity?

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  • $\begingroup$ I read somewhere that the inwards spiral due to tides will eventually stop and reverse. I think as the moon gets closer it will speed up and when it matches the earth's rotation this will happen. $\endgroup$ Commented Aug 5, 2023 at 21:03
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    $\begingroup$ The Moon's orbit decays outwards, not inwards. Related: physics.stackexchange.com/a/659775/44126 $\endgroup$
    – rob
    Commented Aug 5, 2023 at 21:47
  • $\begingroup$ If the question was in the context of general relativity then it would merge in $\rm t=5 c^5 r^4/256/G^3/(M_1 M_2)/(M_1+M_2)$ due to loss of kinetic energy to gravitational waves, but under Newton it wouldn't. $\endgroup$
    – Yukterez
    Commented Aug 5, 2023 at 22:45
  • $\begingroup$ Unless the reason for the decrease in the Earth's orbital velocity is given this question cannot be answered and here is an article about the effect of tides on the distance between the Earth and the Moon but this relates to the angular speed of rotation of the Earth. $\endgroup$
    – Farcher
    Commented Aug 5, 2023 at 23:18

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Small variations of velocity and/or mass happen all the time when meteorites collide with the moon. The effect is only a very small change in its elliptical orbit.

Considering the random probability of collisions at every directions, and the very small mass of the debris, compared with the mass of the moon, I would say that this changes are completely negligible.

If the orbit was a perfect circle, it would become slightly elliptic after a given change of velocity.

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For a circular orbit.

As you said, the Equations used in your question derive from $$ \begin{align*} F_z&=F_G\\ m_M\dfrac{v^2}{R}&=G_N\dfrac{m_Em_M}{R^2}\\ v&=\sqrt{\dfrac{G_Nm_E}{R}} .\end{align*} $$ Kepler's third law states $$ T^2\propto R^3 $$ where $$ T=\dfrac{\omega}{2\pi}=\dfrac{v}{2\pi R} .$$ If the velocity decreases (due to an external force), $T$ will decrease and due to Kepler's third law, $R^3$ will also decrease. This results in a spiral towards the earth, if the velocity decreases until it is zero. Otherwise, the orbit will be closer to earth (just because it drops to a value above zero, the moon won't spiral into earth).

A common theory for the origin of the moon is, that a large body collided with earth resulting in a debris ring, which formed the moon. Because Kepler's third law states, that $\tfrac{T^2}{R^3}=\text{const.}$, all satellites orbit at a distance from their planet corresponding to their velocity. So, to answer the question, the moon's velocity is the velocity of the debris (conversion of momentum). It not really attained it, rather it attains its orbit due to its velocity.

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