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I'm a physic student of Italy. I'm moving my first steps in this beautiful world, and as every new entry, my simple and stupid questions are sometimes not so simply resolved. I'm trying to know the why planets aren't falling on each other, like the moon on the earth.

I thought it could be explained with the Lennard Jones potential, which says us there is a $r_0$ that before or after the potential is repulsive and attractive. But I read that this potential is designed for the atomic model. Could I extend this concept also for planets?

I tried to think to the uniform circular motion, where the centripetal force constantly modify the velocity direction: so the body would escape (also for the planet) but is redirected every time. Through this example I imagined that the moon is trying to escape as the body in the circular motion but is constantly redirected. This could work, but I think this is not correct because, according to Einstein theory, the spacetime near Earth should be curved and constantly pull the moon toward Earth. This means it should be a collision. But collision doesn't happen, so according to my last theory, it should exists a force opposite to the gravitational one that should make all radiant forces null. I think the best reason is about Lennard Jones potential.


marked as duplicate by ACuriousMind, Gert, John Rennie newtonian-mechanics Mar 3 '16 at 6:42

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    $\begingroup$ Hi! I think you should worry about studying Newtonian Mechanics, which requires calculus to learn. Right now you're thinking a lot of things that just aren't physics, and are misunderstanding a lot of concepts. Maybe study a bit of calculus (to where you can differentiate/integrate simple functions), and then get the book Newtonian Mechanics by A P French. $\endgroup$ – user12029 Mar 2 '16 at 18:27
  • $\begingroup$ Which concepts am I misunderstanding? Can you explain me what am I thinking wrong and why it should be wrong? Thanks $\endgroup$ – Andrea Martinelli Mar 2 '16 at 18:29
  • $\begingroup$ Is dued to the centrifugal force which compensate the centripetal one? But how can this accord to the spacetime? $\endgroup$ – Andrea Martinelli Mar 2 '16 at 18:43
  • $\begingroup$ Well, basically Lennard Jones and general relativity have nothing to do with it! Think of swinging a string with a ball on the end of it around over your head. the only force on the ball is inwards, towards your hand, and yet it's stable! You don't need an outwards force. The same is true for the moon/earth or earth/sun, but the force isn't from string, it's from Newtonian gravity. khanacademy.org/computer-programming/… $\endgroup$ – user12029 Mar 2 '16 at 18:44
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/9049 and links therein. $\endgroup$ – Qmechanic Mar 2 '16 at 20:54

The Lennard-Jones potential takes it form because charged particles can both repel and attract. In contrast, gravity is always attractive. So forget about the Lennard-Jones potential.

This thought of yours:

the moon is trying to escape as the body in the circual motion but is costantly redirected.

is what comes closest to the right answer. Gravity ensures that the moon is accelerated in the direction of the planet while, at any given point in time, its velocity is perpendicular hereto$^1$. If by magic you suddenly removed the planet, the acceleration would disappear and the moon would continue in a straight path.

This is true whether you describe gravity in a Newtonian or Einsteinian$^2$ framework. In both frameworks, the result is an acceleration $a$. In the case of circular motion around a central object, this acceleration is perpendicular to the velocity $v$, and just "happens" to be exactly enough to cause $a$ to keep being perpendicular to $v$.

Perhaps your confusion rises from wondering "how can it be that this acceleration deflects the moon just exactly enough from its path so a to keep it from either falling into the planet or escape?" Well, it doesn't have to be "just exactly". If it's a little too small or too large, it simply won't deflect the path of the moon by the right amount, so $a$ and $v$ won't be perpendicular, and hence the orbit won't be circular, but instead elliptical (and if $a$ is too small, the moon will crash, while if its too large, the moon will escape).

$^1$You have a similar scenario if you swing a rock tied to a line around your head; here the centripetal acceleration is just provided by the string's tension rather than gravity.

$^2$Although in general relativity, this won't happen instantaneously, as any perturbation of space only propagate at the speed of light.

  • $\begingroup$ Thank you, I understand. But, geometrically, I can't explain how the moon can escape if the spacetime near the massive planet is curved like a hole. $\endgroup$ – Andrea Martinelli Mar 2 '16 at 20:14
  • $\begingroup$ Maybe the velocity of the moon is enough to "climb" a bit the spacetime hole (for inertial) and refall after? $\endgroup$ – Andrea Martinelli Mar 2 '16 at 20:22
  • $\begingroup$ @AndreaMartinelli: See my update. As for your ball-on-a-curved-surface analogy, yes, you could put it this way, but the same counts as for the "real" moon: It doesn't have to be exactly enough to "climb"; the orbit will just be elliptic rather than circular. $\endgroup$ – pela Mar 2 '16 at 20:55
  • $\begingroup$ Thats really clear. Thank you very much. I'm trying extend the simple newtonian concept to the deformed spacetime. Am I thinking right that the plane where the moon orbit lay isn't parallel with the plane tangent to the point at the bottom of the hole? $\endgroup$ – Andrea Martinelli Mar 3 '16 at 19:12
  • $\begingroup$ @AndreaMartinelli: Yes, if the orbit isn't circular, the moon alternates between being deep in the potential and being higher up, so in the ball-on-a-curved-surface analogy, its plane has some inclination wrt. "horizontal". $\endgroup$ – pela Mar 4 '16 at 11:06

As stated in one of comments, the L-J Potential does not apply to your query. The reason is that it is for modeling a situation of competing attractive and repulsive forces. In your model you need deal only with the attractive gravitational force.

It might help your mental visualization to break it down into velocity vectors. If you have a moon orbiting a planet, then at any instance in time, the moon has a velocity vector at or near 90 degrees to the line running through the 2 bodies. The gravitational force is inducing a separate velocity vector towards the planet. The resultant vector keeps it on the circular or elliptical path. Note that this is a seamless process, not step out and then fall back in.

You started down this path in your initial post, but you might have gotten sidelined by thinking of the planets unaffected path as circular instead of straight once you added in the curved space idea. This is what we call in English, comparing apples to oranges. Look at the problem in either Newtonian or Einsteinian terms, not both at the same time.

If you end up trying to calculate a moon's orbit, I would caution you not to choose the Earth and its moon. With the rather unique combination of sizes and distances of the sun/earth/moon triad, the sun actually exerts a little over twice as much pull on the moon than the earth does. So we ought to say that the moon orbits the sun and the earth warps its orbit significantly.


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