This is my first question. I am a space enthusiast but I don't have a formal understanding of celestial mechanics, so I would like to ask how stable against perturbation an orbit is, given the following assumptions:

  • orbiting mass negligible vs massive center of orbit
  • no atmosphere, no friction, no space dust in the way
  • completely rigid bodies so no tidal concerns
  • no solar radiation pressure, no Yarkovsky effect, etc
  • no obscure funky disturbances not listed here
  • object is in orbit at a given altitude with the proper speed for equilibrium

My perturbation would be a nudge either inwards or outwards, e.g. turning on a propeller aimed at the orbit's center (or away from it) for a small amount of time.

What are the mechanics of this: Will the object settle in another stable orbit or will it slowly drift away / crash into the center?

My naive understanding is that, if I push inwards (thus not changing the tangential speed), the object would move to a closer potential orbit but its speed would be too low and it would spiral down to a crash, while it I push outwards the object would have a too fast tangential speed so it would drift away.

According to my (potentially flawed) reasoning, all orbits are unstable, meaning that you always tend to "diverge away" from that equilibrium, like a marble on top of a hill.

  • 1
    $\begingroup$ Welcome to Space! You've done well for your first question. I am going to do some edits so readers don't mistakenly believe there is more than one question here. $\endgroup$ – DrSheldon Apr 4 '20 at 19:32
  • 1
    $\begingroup$ With all of those conditions, pretty darn stable. For instance, if you give it a nudge downward, the orbit will become a little elliptical, but it wouldn't fall. If it did, it would gain speed as it falls and zoom back out of it. Consider that if you were on the space station with a rifle, you literally could not shoot the earth because the bullet doesn't go fast enough to deorbit. But add asymmetrical mass, moons, etc., and it gets tricky. $\endgroup$ – Greg Apr 4 '20 at 20:10
  • 1
    $\begingroup$ It's not a complete answer to your question, but this answer contains a "cheat sheet" showing the effect of "nudges" to a low Earth orbit space.stackexchange.com/a/12014/6944 Your intuition isn't exactly right, a radial "in" burn shifts the perigee towards the burn point. $\endgroup$ – Organic Marble Apr 4 '20 at 21:28

Your question indicates to me that although you aren't familiar with celestial mechanics, you do have some knowledge of physics and astronomy. You're a space enthusiast? Good for you!!

The situation you describe is very much aligned with what we call Keplerian orbits, named for their pioneer, Johannes Kepler, publishing in the early 1600's.

In the absence of perturbations by other gravitating bodies, which I assume comes under your prohibition on "obscure funky disturbances", Keplerian orbits are indeed stable.

Where did your reasoning lead you astray? Assume the orbiting object is in a circular orbit, as shown by the dark line in the figure, orbiting in the direction shown. At the large red arrow you gave the orbiting object an instantaneous gentle nudge inward, indeed not changing the tangential velocity at that point.

A circular Keplerian orbit (dark green) around a spherically symmetrical primary (blue), and the orbit after a vertically downward perturbation at the large red arrow

But you did change, even if only a little, the flight path angle, to that indicated by the small red arrow on the red orbit. The object is no longer traveling horizontally, it's headed a bit downhill. When the object was traveling strictly horizontally (circular orbit) the local gravity vector was perpendicular to the velocity vector. When those vectors are perpendicular there is no change in the object's speed. (Speed is a scalar quantity, the magnitude of velocity, which is a vector quantity with both speed and direction) In the downhill case, post-nudge, just as in the case of something rolling down a hill on Earth, there's a (small) component of the gravity vector parallel to the velocity vector. The object accelerates, i.e. its speed increases with time, so its speed doesn't remain constant after the perturbation. That change in speed corresponds to the change in gravitational potential energy due to the changing radius from the primary's center: the farther downward it goes, the faster it goes.

This increase in speed with decreasing altitude causes the orbit's radius of curvature to be larger than that of a circular orbit at that altitude, so the orbit eventually bottoms out at the periapsis, 90° away (as measured from the center of the primary) from the perturbation point. It then rises back to the original altitude, 180° away from the perturbation, with the same tangential speed and the same vertical speed, just upward instead of downward, as you see at the bottom of the diagram.

That vertical speed carries the object higher, and that decelerates it. The reduced speed decreases the orbit's radius of curvature, so it peaks out at apoapsis 270° from the perturbation and begins descending. At 360° from the perturbation — one orbit — it is back to exactly where it started, at the instant of the perturbation, with the same velocity, same flight path angle, same everything, and this repeats ad infinitum.

This orbit, like all bound (i.e., not escaping) Keplerian orbits, is perfectly stable. Given the constraints you listed, it would remain exactly as shown forever, without any kind of control.

If you made the perturbation not small, say a significant fraction of the orbit speed, then you could make the object collide with the primary. "If you push something hard enough, it will fall over."

Once you start complicating the picture — the planet isn't spherically symmetrical, the planet rotates, there are other gravitating bodies involved, part of the orbit is in full sunlight and part is eclipsed, etc. — then those perturbations make the orbit evolve (change with time), in a few cases to the point of colliding with the primary or even being ejected from the system. Orbit evolution happens to everything orbiting Earth, even the moon.

  • 2
    $\begingroup$ Your answer is exactly what I needed to disentangle a cognitive knot that prevented me from getting the full picture: even if aware of elliptical orbits, my mind was stubbornly thinking exclusively of circular orbits, and assuming that a nudge could either move us to another circular orbit or escape/crash. As I now understand it, if we start with a circular orbit, a nudge in whatever direction in the orbit plane will always result in an elliptical orbit. This just fueled in me a sudden urge to dig deeper into the wonders of orbital mechanics... So fascinating! Thank you $\endgroup$ – Unai Vivi Apr 4 '20 at 23:44
  • 1
    $\begingroup$ Thank you! Indeed I find orbital mechanics fascinating. After a couple of courses from the late John Breakwell, so many things that had been mysteries to me were illuminated! $\endgroup$ – Tom Spilker Apr 4 '20 at 23:47
  • $\begingroup$ What still feels "magic" and keeps puzzling me is the fact that the point where the gravity and velocity vectors become perpendicular (and speed starts decreasing) happens exactly at 90° and not later or sooner. Is that always true or is it just for the sake of this example? $\endgroup$ – Unai Vivi Apr 5 '20 at 0:16
  • 2
    $\begingroup$ @UnaiVivi it happens to be 90° in this case because the initial perturbing impulse was purely radial; it didn't change the tangential component of the velocity. If there's a tangential component to that perturbation this would no longer be the case. For example, if the perturbation was purely tangential (only speed up or slow down, no direction change) then the apses will be 0° and 180° from the point at which the impulse was applied. $\endgroup$ – Kyle Apr 5 '20 at 5:56
  • 1
    $\begingroup$ @Kyle is right on the money! $\endgroup$ – Tom Spilker Apr 5 '20 at 6:25

So the expression is this, if you are thinking of a small object in orbit around a sun or a planet that contains infinitely more mass:


Assuming $\bar{v}$ is the velocity vector of the orbiting object and $\hat{r}$ is a unit vector pointing from the orbiting object to the center of mass of the planet/sun.

This is Newton's classical law of gravitation.

Somehow you can analytically solve the equation above and find that the orbiting body must always move in an ellipse unless it has enough speed to escape from the planet/sun.

Basically, if you do not push the object hard enough to escape the gravity of the sun/planet it will always follow an ellipsoid motion and come back to exactly the same point where you pushed it from.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.