# Why are planets in unstable equilibrium in a stable orbit? [closed]

It is well known that a planet in stable orbit is in unstable equilibrium. If e.g. the moon was just a few m/s slower in velocity, or a few m closer to the earth than it is, the gravity would constantly be more than the centripetal force and over millions of years, it would be pulled towards the earth. As it moved towards the earth, the gravity would increase still further, causing it to crash to earth even faster. THERE WOULD BE NO POINT WHERE THE GRAVITY WOULD BECOME EQUAL TO THE CENTRIPETAL FORCE AGAIN, ALLOWING IT TO ENTER A STABLE EQUILIBRIUM. Similarly, if it was a little faster or a little further away, it would fly off in space, and never return to equilibrium.

It is also well-known that systems not in equilibrium cannot exist in practice, e.g. for a pencil standing on its point, where it would fall if tipped slightly, we would not expect millions of such pencils standing on their point routinely existing in nature. Similarly, if evaporation of water in a closed vessel were to cause more evaporation, then all the water would be evaporated and the vapor would never be in equilibrium with the liquid.

Yet, we see millions of planets and moons in stable orbits yet in unstable equilibrium in nature. How is this possible?

When each planet (with respect to its star) and each moon was first created, or first passed by one another, how did they come to be at the very exact velocity and very exact distance from each other so that they could get trapped in a stable orbit, so that if the velocity or distance was just a little bit more or less, the orbit would have decayed and would never be stable? How did this very rare coincidence happen billions of times so that there are billions of planets and moons in stable orbits?

## closed as off-topic by Emilio Pisanty, Jon Custer, David Hammen, Bill N, ZeroTheHeroJun 29 '17 at 21:28

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• What is the basis for your first sentence? Give some good references. NASA/NOAA satellites prove that statement wrong all the time. – Bill N Jun 28 '17 at 16:29

Firstly, as mentioned in the comments, in a two body system orbits are not unstable. The types of orbit possible are described in the Wikipedia article on Kepler orbits. The bound orbits can be circular or elliptical, and all the bound orbits are stable i.e. the orbital parameters do not change with time.

However it's not obvious why the orbits have to be stable, and in fact it's only in three spatial dimensions that we get stable orbits. If the universe had less than three or more than three spatial dimensions there would be no stable orbits.

The stability of orbits in three spatial dimensions is because in three dimensions the gravitational potential decreases with distance $r$ as $r^{-1}$. Bertrand's theorem tells us that for a central potential with an $r^{-1}$ dependance on distance the orbits are all stable.

The trouble is that understanding Bertrand's theorem requires some mathematical sophistication, and I don't know of any intuitive way to explain this to the budding physicist. The nearest I've seen is the answer to this question.

• You could say that the derivation is the fairly standard stability analysis of "fixed points" of DEs: one derives the the linear constant co-efficient DEs that describe first order perturbations from circular orbits. Linear cc DEs have solutions that are sums of terms of the form $\exp(\alpha_j\,t)$. You check the sign of the $\alpha_j$: if any are positive, you have a solution which is driven away the equilibrium points. Maybe even: the Banach fixed point theorem shows that there is a stable neighborhood of the equilibrium point if and only if the linearized equations have all $\alpha_j<0$ – WetSavannaAnimal Mar 29 '15 at 8:49
• Isn't it no stable closed orbits in other dimensions? I thought that in fewer dimensions we could have stable orbits except that they would not be closed. – badjohn Jun 27 '17 at 10:40

Your question seems to be based on a misunderstanding of what remains constant when varying orbital parameters.

For a simple, circular orbit: $$\frac{GMm}{r} = \frac{mv^2}{r}$$

and thus: $$v = \sqrt{\frac{GM}{r}}$$

In the comments to your question, you seem to imply that $v$ should remain constant as $r$ changes (which, you're right, would result in a non-equilibrium orbit); however, $v$ is a function of $r$ and varies accordingly for the required orbital radius.

• This seems to be not the correct answer as far as I can tell - may be I am wrong, but would appreciate some more answers. True that for a body revolving in orbit all possible (v,r) pairs are possible. For every v, there is an r at which it is stable. But for it to first come into that stable orbit, the first body must be travelling at a v, pass by the second body at a very precise distance r corresponding to that v, and be travelling at a tangential path at that moment. Otherwise it will fall into the other body or fly off in space. It will never go into a stable orbit. – Khushro Shahookar Jan 22 '15 at 11:44
• That only applies to exactly circular orbits - there are a wide range of acceptable $v$s for a given $r$ that permit capture into an elliptical orbit. – Kieran Hunt Jan 22 '15 at 11:45
• Do not confuse circular orbits with stable orbits. Changing the orbital velocity of a satellite with a circular orbit will make the orbit more elliptical, not less stable. – Kieran Hunt Jan 22 '15 at 11:49
• @KhushroShahookar maybe this would clear things up for you. – fibonatic Jan 22 '15 at 14:48