I Will an elliptical orbit self-correct into a circular one?

1. Jan 28, 2017

Edward Barrow

I know that moons tend to orbit their planets in a slightly elliptical orbit rather than a perfectly circular orbit. But for the purpose of this thread, lets assume that moons effectively orbit their planet in a circular orbit.

So here is the question...

If our moon was struck by an object and sent into an distinctly elliptical orbit around earth, would the moon orbit the earth for a couple of months in an elliptical orbit before slowly self-correcting into a more circular one?

And if something like this happened in earths past, do you think this could account for the earths wobble on its axis (Precession)? That is, would this cause the earth to wobble as the moon got near the earth and far from the earth as it orbited on its elliptical orbit?

2. Jan 28, 2017

Bandersnatch

Hi Edward Barrow, welcome to PF.

Tidal interactions in any 2-body system do circularise elliptical orbits over time. The timescales are geological, though. Definitely not months - more in the vein of hundreds of millions of years.

Precession of the Earth's axis is also caused by tidal interactions with the Moon and the Sun, both applying torque on the equatorial bulge, which is offset from the plane of the ecliptic (for solar tides) and plane of lunar orbit (for lunar tides). Both bodies act on the bulge trying to bring it to the level with their respective orbital planes. This torque is what changes the angular momentum of the spinning Earth by shifting its axis, which in their absence would have to be conserved - i.e. no precession.

The dynamics of precession are well understood and their descriptions easy to find (typically a first year undergrad exercise, so no higher maths necessary).

High ellipticity of the lunar orbit would cause the precession to increase its fluctuations with a period of one lunar month. But it wouldn't be the underlying cause of precession.
The main point, though, is that you need forces acting NOW for precession to occur, not something acting in the past.

3. Jan 29, 2017

Edward Barrow

What's your basis for assuming an elliptical orbit would take hundreds of millions of years to self-correct? I don't see how it would take this long.

Could it happen in the space of 1000 years?

4. Jan 29, 2017

Janus

Staff Emeritus
Mainly because there is nothing more "correct" about a circular orbit over an elliptical one. Mercury and Pluto both have noticeably elliptical orbits and show and have so for a long time with out circularizing. If you put something in an elliptical orbit, it will be perfectly happy and remain in it unless some outside influence acts to change this. In other words, Unless something actively acts on it to cause it, a elliptical orbit will not naturally circularize.
No. Consider this: Tidal interaction between the Earth and Moon is at present increasing the Moon-Earth distance by ~4cm per year. This same tidal interaction is what would work to circularize the orbit. Now if we just assumed that this varied directly by tidal force, and since tidal force varies by the cube of the distance, then if the Moon was put in an orbit so with a perigee 1/2 its present distance, then at perigee, the Moon would recede at 8x4cm = 32 cm per year while at apogee it would still recede at 4cm/ year. At this rate it would take ~ 621 million years for the perigee to catch up with the apogee and the orbit to become circular.

However, this assumes that the rate at which the perigee and apogee increase will stay the same,and they won't. As they move away, the tidal influence weakens and the recession rate for each will decrease. This decrease effects the perigee the most and thus the difference in recession speed will decrease as time goes on, and this will increase the time needed for circularization to an even longer time.

And even taking this into account you will get an answer that is too small, as the tidal circularization effect is not even as strong as we assumed above. I was heavily fudging in favor of the effect and even then got an answer in 100's of millions of millions of years.

5. Jan 29, 2017

Staff: Mentor

The force between two objects can be approximated by the force between two point-masses. In this case, all ellipses are stable - they won't change their shape at all. The same is true for perfectly spherical masses
Changes in the shape have to come from deviations from those assumptions (and from General Relativity). Earth and Moon are nearly spherical, however. The tiny oblateness of Earth will influence the orbit a bit, the tides on the surface of Earth will influence the orbit a bit, the sun and the other planets will influence the orbit a bit. All those effects are well-studied and their size is known. They influence the orbit notably over a billion years. Within 1000 years, those things are completely negligible.

6. Jan 29, 2017

Edward Barrow

So, who's telling the truth?

Bandersnatch says:

And Janus says:

7. Jan 29, 2017

Staff: Mentor

Tides are an "outside influence" in that context. They are not the center of mass of either object.
Janus explains how tides circularize the orbit in his post.

8. Jan 29, 2017

Edward Barrow

If the moon was knocked into a distinctly elliptical orbit around the earth (by an outside force such as an asteroid), would the gravitational pull of the sun (on the earth and moon) help correct the elliptical orbit of the moon around the earth?

Could the gravitational force of the sun help correct this more quickly (say in 1000 years) than hundreds of millions of years?

9. Jan 29, 2017

Janus

Staff Emeritus
No. It would tend to stretch the orbit along the Sun-Earth line. When the long axis of the orbit is along this line, it will increase its eccentricity and when it perpendicular to it, it will decrease the eccentricity, but this will cyclic (one cycle per synodic month) and not lead to any circularization of the orbit over the long run. (The Moon in its present orbit undergoes this same variation.)
No. See above.

10. Jan 29, 2017

Staff: Mentor

Do you expect different answers if you ask the same question multiple times?

The laws of the universe didn't change since yesterday.

11. Jan 30, 2017

sophiecentaur

Is this the plot of an intended SF story?

12. Jul 19, 2017

Apoart

It seems that it is not well publicised that the circular equivalent of an orbital ellipse, while sharing the same orbital period (T=2pia/[GM/a]^½]), does NOT share the same orbital angular momentum.

Let the mass of the satellite be m, the circular orbital speed be v = [GM/a]^½], and as is usual let a and b represent the semi-major & semi-minor axes respectively. Additionally, Lc and Le are the angular momenta of the circular and elliptical orbits respectively: Lc = mva and Le = mvb.

Physically, this means that there cannot be a circular equivalent of an elliptical orbit unless the former gains angular momentum from somewhere (a>b), which necessarily negates the assumed equivalence by modifying period and energy. Perhaps, this supports the idea that lower energy elliptical orbits are more stable that their more energetic circular ‘equivalents’.

13. Jul 19, 2017

snorkack

Moon IS in a distinctly elliptical orbit. Eccentricity 0,055.
Meaning that the tidal forces on Moon at perigee are 30 % stronger than at apogee. Moon deforms every month, and dissipates a lot of energy in so doing.
How fast is the orbit of Moon circularizing? When was the eccentricity 0,060 rather than the present 0,055? And what were then the lengths of month and day?

14. Jul 19, 2017

Apoart

According to:
https://eclipse.gsfc.nasa.gov/SEhelp/moonorbit.html
MEAN moon orbit eccentricity 0.0549. The ratio of the inverse laws is 1.25 i.e. 25% increase in gravity at perigee over apogee. Also, apparently the moon’s eccentricity varies from ~0.035 to ~0.075 between 2008 -2017 cyclically.

In accordance to my posting and
https://astronomy.stackexchange.com/questions/9937/shouldnt-all-moon-orbits-be-inherently-unstable
“You can have a perfectly circular orbit in a mathematical thought experiment.”

Additionally, you know that the moon keeps it ‘face’ towards Earth because it rotates synchronously with its orbital period, which was established due to the ‘brake blocks’ of tidal friction, I believe fairly soon after the moon was created. The lunar recession from Earth is due to its tidal bulges being carried forward marginally due to its rotation so that they can produce a recession rate from Earth of 3.8 cm/yr.

Briefly, because I have just learned the cyclic magnitudes of the Earth, Sun and Moon interactions, I cannot give specific answers to your interesting questions. My non-specific answers: I would not expect the moon to be dissipating much energy due to largely stable tidal forces and I believe that elliptical orbits need extra angular momentum and energy to become circular. It is not obvious to me why the moon would circularise its orbit.

15. Jul 19, 2017

snorkack

And tidal forces are actually increased by 40 %, because they increase with inverse cube of distance.
But Moon has to librate because of orbital eccentricity. The angular velocity of rotation has to be constant through orbit, the angular velocity of orbit is not.
Moon could only keep a face towards Earth if the orbit were circular.
But the tidal forces are not stable. Over the elliptical orbit, the size of tidal forces changes through a large range, and their direction also changes, though through a modest angle not full circle.

16. Jul 19, 2017

Janus

Staff Emeritus
Just because the angular momentum for a circular orbit with radius A is not equal to the angular momentum of an elliptical orbit of semi-major axis A, does not mean that the circular orbit has a higher energy.

The total energy of an object in orbit is
$$E = \frac{mv^2}{2}- \frac{GMm}{r}$$
If we assume a circular orbit r is constant and always equal to a.
If we substitute the orbital speed from the equation above for v, this reduces to
$$E= -\frac{GMm}{2a}$$

If we now consider an elliptical orbit, we need to consider that both The orbital speed and radius changes over the course of the orbit. But since the total energy remains constant throughout the orbit, we only need to consider one point, so we'll use the periapis.
In respect to a, periapis is found by

$$r_p = a(1-e)$$

The orbital speed at any point of an elliptical orbit is
$$v = \sqrt {Gm \left ( \frac{2}{r}- \frac{1}{a} \right ) }$$

Substituting rp for r gives us

$$v= \sqrt{ \frac{GM(1+e)}{a(1-e)}}$$

Substituting this for v in our total energy equation and rp for r

we get

$$E = \frac{GMm(e-1)}{2a(1-e)}$$

Which is the same as

$$E = \frac{-GMm(-(e-1))}{2a(1-e)}$$
$$E = \frac{-GMm(1-e)}{2a(1-e)}$$
$$E= -\frac{GMm}{2a}$$

Which shows that regardless of the eccentricity of the orbit(circle or ellipse), As long as a is the same, the orbital energy is the same.

17. Jul 20, 2017

Apoart

Further to my earlier post, I thought it would be interesting to plot elliptical orbit eccentricity (e) versus relative angular momentum (Le) for constant a. A line applies when e = 1 and a circular orbit (radius a=b) for e = 0. Since Le = m.v.b, the plot is really just e vs b. Clearly, for a natural satellite to convert from an elliptical orbit into a circular one requires a variable amount an angular momentum boost dependent on initial e.

Responding to SNORKACK: I agree that: “tidal forces are actually increased by 40 %” not 30% (Incidentally, it was Newton who first proposed the inverse cubed law); “Moon could only keep a face towards Earth if the orbit were circular” and since I have not done the work I trust that you have, so I assume that “the size of tidal forces changes through a large range”. Thank you for drawing my attention to these extra facts.

Responding to JANUS: I accept your algebra that is found in textbooks and web articles. The energy that you cite is of course total system energy summing potential gravitational with kinetic energies. However, orbital angular momentum is a function of the varying radius and varying kinetic energy and is constant for a given eccentricity. My errors were not to specify ‘kinetic energy’ for the orbits and also to overlook that the period IS constant for all eccentricities as indicated in the statement “...orbital period (T=2πa/√[GM/a])...” Thank you for drawing my attention to my errors and accepting that angular momentum is variable with eccentricity for constant a.

“It's only those who do nothing that make no mistakes, I suppose.” Joseph Conrad.

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18. Jul 20, 2017

Staff: Mentor

It is not a secret that you can fully determine the shape of an orbit (e. g. determine semi-major axis and eccentricity) by giving energy and angular momentum. To change something, you have to change energy or angular momentum. We know mechanisms to do so - tides, for example. The Moon is gaining energy and angular momentum from Earth.

19. Jul 21, 2017

Apoart

In keeping with PF’s principle of problem solving, I would like to say that the debate stemming from Edward Barrows post of 28 Jan. 2017 has been illuminating for me for my quest to understand elliptical orbits a bit more. With thanks to those who responded to my first post, I have learned that the moon’s orbit is expanding but not necessarily into a circular orbit. I believe that my first assertion that it is not possible to add angular momentum to an orbit without changing other parameters (as per moon) is correct.

Just because all orbits with a constant semi-major axes (a) share the same total energy and period there is no simple physical mechanism that allows one to migrate to another and ultimately to a circle. No doubt NASA can manipulate artificial satellites to perform such feats but most probably with several manoeuvres and expenditure of energy. An excellent qualitative description of such manoeuvres may be found at:
http://www.rcptv.com/spacetec-ita/changing_orbit.htm under The Effect of a Rocket "Burn"

A misconception seems to be that the equivalent circular orbit of an elliptical orbit is centered at the middle of the ellipse presumably because it still has a radius of a. Without elaborating further, I present a figure below that shows how a family of constant a ellipses morph into a their equivalent circular orbit all having a common (attractor) focus. I leave it to the reader to deduce the interesting geometrical features.

Recall that total energy comprises kinetic and potential components. Therefore as b increases so the angular momentum increases and when b = a then the kinetic and potential energies equate to a maximum for the former.