# How is it possible for orbits to maintain stability?

According to $a = v^2/R$, the circular velocity and radial distance between two attracting objects (such as planets), must remain in perfect proportion in order for orbital motion to take place. How is it possible for objects in nature to achieve this proportion perfectly?

Not only that, to maintain an orbit seems to be impossible. For example, assuming that the moon is orbiting the earth 'perfectly'. Let us say the moon is then hit by a series of meteorites. This would shift the balance slightly, and cause the moon's orbit to decay? Apparently not... How is it possible for the moon to remain in orbit for so long?

-

## 3 Answers

If the velocity of a satellite differs from the right velocity of a circular orbit, Newton's equations imply that the object will simply move along a non-circular orbit, an ellipse. This fact as well as the detailed parameters of this ellipse were already known to Johannes Kepler.

All planets and moons in the real world orbit around their stars or planets along ellipses and there is no fine-tuning here whatever. The deviation from a circular orbit is known as "eccentricity" of the ellipse and it is nonzero for all real celestial objects: none of them has a fine-tuned velocity. For any initial position or velocity, one finds an ellipse (which may be a circle if someone, e.g. NASA, fine-tunes the parameters) or a hyperbola or a parabola (if the speed exceeds the escape speed or is equal to it) and the object will move along it, in agreement with Newton's laws of motion.

All the elliptical trajectories of the 2-body system are stable (and the elliptical ones are periodic): a small perturbation of the initial state only leads to equally small perturbations of the final state. This proposition has to be modified for 3 bodies and larger numbers (chaotic behavior) and for nearby orbits around very heavy objects in general relativity that may be unstable. But in Newton's theory for 2 bodies, everything is easy.

-
"If the velocity of a satellite differs from the right velocity of a circular orbit, Newton's equations imply that the object will simply move along a non-circular orbit, an ellipse" aren't other conic sections like parabola or hyperbola permitted? – user1355 Feb 22 '11 at 16:32

3
@sb1: not without seriously modifying the energy of the orbit --- parabola and hyperbola solutions require you to be able to escape to infinity. – genneth Feb 22 '11 at 17:02
Thanks, @genneth, that's what I meant, indeed. The other conic sections are also discussed in my answer. – Luboš Motl Feb 23 '11 at 6:51

Let's treat the case of one single particle going around a 1/r potential (i.e. Moon around Earth). In the rotating frame of the Moon's orbit, there is an effective potential given by: $$V(r) = -\frac{GMm}{r} + \frac{J^2}{2mr^2},$$ where $J$ is the angular momentum of the orbit (and is conserved). The problem then reduces to that of a single particle moving in a one-dimensional potential, which has a well-defined minimum:

Thus any small displacement is stable, and simply results in an oscillation of the orbital radius.

-

Of course, this completely ignores the effect of other objects. Some of the trajectories which one assumes to be independent in the Solar system could get close (e.g. in early Solar system, or for comets) and that would change the picture drastically. – Marek Feb 22 '11 at 12:04
@Marek: I read the OP as asking for the 2 body case --- it is a mis-understanding of how basic orbits work as opposed to the much more sophisticated question of how many-body orbits are stable (do we even have a complete answer to that? My knowledge stops at KAM theorems...) – genneth Feb 22 '11 at 14:13
I guess you're right, I misinterpreted the question. And yeah, my knowledge also stops at KAM (or rather, even before; basically I just know that the theory exists and that's it). But I'd like to learn more one day. – Marek Feb 22 '11 at 16:40

That is the conditions for circular orbits.

Orbits have no trouble existing in non-circular (elliptical) varieties. In this case the velocity and radius vary in such a way as to keep the angular momentum constant.

You can find the angular momentum at a particular point in time as $L = m * v_t$ where $v_t$ is the velocity transverse to the line connecting the two bodies. From the considerations you stated for circular orbits you should be able to deduce when the orbit is heading outward and when inward.

-

– Marek Feb 22 '11 at 10:30
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9249294400215149, "perplexity_flag": "middle"}