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How does the Moon stay in orbit? The Moon has not reached escape velocity from the Earth, then how does the Moon stays in orbit?


marked as duplicate by Emilio Pisanty, Gert, HDE 226868, user36790, John Rennie Dec 7 '15 at 6:27

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The answer is the same as the answer to the question "why do satellites stay in orbit": the gravitational pull of the earth is just strong enough to keep it in orbit at the altitude it is, given the angular momentum (velocity) that it has. In equations:


where $r$ is the distance from the center of the earth to the center of the satellite (moon), $v$ is the orbital velocity, $G$ is the gravitational constant, and $M_{earth}$ is the mass of the earth.

It follows that for any value of $r$ there is a velocity $v$ for which there is a stable orbit. The moon found such an orbit.


It's moving the orbital velocity for its particular altitude, which means it has the exactly the speed at which Earth's gravity supplies the centripetal force needed to accelerate it in a circle.

Now, I'm guessing what you mean by your question is that at the surface of the Earth, we need to boost a rocket to about $v_e(R_\oplus)=11{\rm km s^{-1}}$ (let $v_e(R)$ be the escape velocity from a point at distance $R$ from the Earth's centre) for it to escape Earth's gravitational field: that's if you're going to escape by a short burn, which is what a chemical rocket can do. In principle, there is nothing stopping something escaping Earth's gravity by moving steadily upwards at $1{\rm km\,h^{-1}}$ if there were a way to thrust continuously.

So imagine we built the moon here and wanted to send it off into its present orbit by a chemical rocket. Let's first boost it into low Earth orbit and watch it for a bit: actually we couldn't do this because the Moon is too big - so let's imagine a Moon sized black hole instead so that we can simplify things and talk about a point mass like we do a rocket. We should need to boost it to its orbital velocity $V_o(R_\oplus)$ which is $v_e(R_\oplus)/\sqrt{2}$. This is about $7.5 {\rm km\,s^{-1}}$ : a great deal faster than the orbital velocity of the Moon we ken here on Earth (whose orbital speed I work out to be about $1{\rm km\,s^{-1}}$. After we'd admired the LEO Moon for a bit, we'd then boost it to a higher speed such that its kinetic energy + gravitational potential in LEO were equal to the kinetic energy + gravitational potential for the Moon in its real orbit. This would be rather more than $7.5 {\rm km\,s^{-1}}$ but a little less than $v_e(R_\oplus)$. So the Moon would now coast off into space, converting its kinetic energy to potential as its altitude got higher and higher. If our burns were all correct, it would slow down as it coasted into the real Moon orbit, at the real, present orbital speed of the Moon. It's a little more complicated than this because we'd have to choose the appropriate transfer orbit, such as a Hohmann Transfer and undertake a series of burns to achieve the near circular orbit the real Moon has, but the principle of boosting it to a higher kinetic energy than its LEO kinetic energy and letting it coast off to convert this extra kinetic energy to the higher potential energy of the higher altitude orbit is the right idea.


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