1
$\begingroup$

If we place an object in the earth's orbit around the sun, it should have the same orbital period right?

Lagrange points and another point in the orbit.

For example, what would be the behavior of an object in orbit at point A1? It is at the same distance from the sun as earth and is sufficiently far away from the earth to prevent earth from gravitationally influence it(is it?) . Shouldn't the object at A1 remain stationary with respect to the earth?

$\endgroup$
  • 3
    $\begingroup$ We can put it there, if we want, it's just going to need a little bit of fuel for the occasional orbital corrections. In the stable Lagrange point that is not the case... at least in theory. In practice there are perturbations even in the stable Lagrange points that one may have to correct for. $\endgroup$ – CuriousOne Mar 14 '16 at 3:47
  • $\begingroup$ @CuriousOne If we talk about the relative stability of the point L3 and A1, how less stable would the point A1 be? Please see the photo that i have included in the question. $\endgroup$ – Aniansh Mar 14 '16 at 10:35
5
$\begingroup$

Here's the key point: although they are often drawn this way, none of the Lagrange points lie exactly on the orbit of the less massive body. This is obvious for L1 and L2, but it is also true for L3, L4, and L5 as well.

First off, it's important to realize that the Earth does not technically revolve about the Sun. Rather, the Earth and the Sun both revolve around their common center of mass, a point called the barycenter of the system. This imaginary point is technically inside the Sun for the Sun-Earth system; but it's not exactly at the center of the Sun.

L3 lies slightly farther from the barycenter of Earth-Sun system than the Earth does. An object moving in a circular orbit of the same period as the Earth but at this slightly increased distance would need some additional force to keep it moving in this circle (as opposed to if it was just orbiting the Sun on its own). But it does in fact receive a little more force than it would in the absence of the Earth: first, the Sun is a little closer to it (since it's opposite on the opposite side of the Sun from the barycenter); second, the Earth is also pulling on it a little bit. The L3 point is the point where all of these forces come together perfectly, so that the net force on an object is of exactly the right magnitude to keep an object rotating about the barycenter with a period of one year, and such that the net force is directly towards the barycenter.

L4 and L5, meanwhile, form an equilateral triangle with the two massive bodies—not the barycenter. This means that neither the Sun nor the Earth is pulling an object at that point directly towards the barycenter. This would be true of your A1 point as well. But what's special about the L4 and L5 points is that the net force from the Earth and the Sun is, once again, of the correct magnitude and direction to keep an object in a circular path about the barycenter of the system. An object at the point A1, on the other hand, will experience a force that does not point towards the barycenter of the Earth-Sun system, and so eventually an object placed at that point will drift away.

These points are often obscured by the fact that the Sun-Earth system and the Earth-Moon system have rather lopsided mass ratios: 330,000 for the Sun-Earth system, and a "mere" 81 for the Earth-Moon system. This means that in these systems, the barycenter is actually inside the more massive body, and so it's difficult to distinguish between "both bodies revolve about the barycenter" and "the less massive body revolves about the more massive one." This is why, as I noted above, the Lagrange points L3, L4, and L5 are usually draw as though they lie on the orbit of the less massive body, and the more massive body is treated as stationary. But if you look at a system like Pluto-Charon, with a mass ratio of a little over 8, the distinction becomes much more important, and the deviations from the "idealized" locations become much more noticeable. I whipped up a diagram of the Lagrange points in the Pluto-Charon system below; Pluto and Charon themselves (the pink and grey circles) are not to scale, but everything else is. The yellow dot is the barycenter of the system, and the white crosses are the Lagrange points.

enter image description here

If you imagine increasing the mass of the more massive body relative to the less massive body, while leaving the distance between them fixed, three things would happen to this diagram: the more massive body would move closer to the barycenter, the less massive body would move further out, and all five Lagrange points would move closer to the orbit of the less massive object. In fact, if you treat the less massive object as having zero mass, your satellite can orbit directly on top of the orbit of the less massive object — at any location along it. This is probably the case you're thinking of, where you don't have to take the attraction of the less massive object into consideration. But since the Earth has mass (and we should all be grateful that it does), you have to take it into consideration when you calculate the Lagrange points.

$\endgroup$
  • $\begingroup$ Thank you very much for the answer! That helped a lot! I still have one little question though. What if two satellites of very less masses were placed in the same orbit? Will such a configuration be stable? $\endgroup$ – Aniansh Mar 14 '16 at 17:42
  • $\begingroup$ @Aniansh: That's probably worth asking another question about, to be honest. The derivation of the Lagrange points explicitly ignores any force of gravity exerted by the satellite/asteroid occupying the Lagrange point; but if you put two satellites in the same orbit, it's a different problem. You might look into the case of Janus & Epimetheus, two of Saturn's Moons that essentially "share an orbit". $\endgroup$ – Michael Seifert Mar 14 '16 at 17:47
  • $\begingroup$ Okay thanks i will look it up on the internet before posting another question! $\endgroup$ – Aniansh Mar 14 '16 at 17:48
1
$\begingroup$

Well, what generally happens when you put two massive objects floating in space with zero relative velocity is that they fall together. This is no different for the Earth and some other satellite in the same orbit around the sun: they may start out orbiting the sun at the same distance from the sun and some distance apart, but they still attract each other gravitationally. The presence of the sun and the angular momentum complicate how exactly they fall together, but it's easy to show that they won't maintain the original orbit indefinitely.

$\endgroup$
  • $\begingroup$ How can L3 function then? Wouldn't earth and the object be accelerated towards each other? P.S. I have edited the question. $\endgroup$ – Aniansh Mar 14 '16 at 10:34
  • $\begingroup$ You're right, orbits at the L3 point are also unstable. The only "stable" co-orbits are at the L4 and L5 points, and even they aren't perfectly stable - the satellite would orbit "around" the point, in a sort of kidney-shaped path - but the satellite wouldn't fall "out of" the L4/5 because the perturbations are counteracted by the effects they have on the orbit of the satellite in such a way that it stays in roughly the same place relative to Earth. $\endgroup$ – Asher Mar 14 '16 at 14:33
  • $\begingroup$ If that is the case then why is L3 considered a Lagrange point at all and why aren't all the points in the orbit considered as Lagrange points? Also thanks for explaining about L4 and L5. $\endgroup$ – Aniansh Mar 14 '16 at 15:13
  • $\begingroup$ The modelling in which the Lagrange points arise is highly simplified. It assumes that the planet keeps a constant circular orbit around the sun, and that besides the planet, sun and satellite the universe is empty, among other things. The L1,2,3 points are stable only in the three-body case; even the slightest perturbation causes the satellite to fall out of the circular orbit. On the other hand, satellites in the L4 and L5 points require a gravitational influence on par with the co-orbiting planet to be permanently de-orbited, so they are stable in the same sense the planet's orbit is stable $\endgroup$ – Asher Mar 14 '16 at 19:37

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.