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Can a planetary orbit have a zero eccentricity?


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Hi everyone,

We see that Kerbin's orbital eccentricity equals zero.  In other words, its orbit is circular.

In our real solar system, every planet's orbit has a non-zero eccentricity -- that is, a non-circular orbit.  (For the most part, the eccentricity of every planetary orbit in our solar system is small.  Nevertheless, they all have a positive value.)

So my question, please, is this: In the real world, can a planet actually have a zero-eccentricity orbit around its star?  Is that physically possible, even if it is rare?

Thank you.

Stanley

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10 minutes ago, MetricKerbalist said:

Hi everyone,

We see that Kerbin's orbital eccentricity equals zero.  In other words, its orbit is circular.

In our real solar system, every planet's orbit has a non-zero eccentricity -- that is, a non-circular orbit.  (For the most part, the eccentricity of every planetary orbit in our solar system is small.  Nevertheless, they all have a positive value.)

So my question, please, is this: In the real world, can a planet actually have a zero-eccentricity orbit around its star?  Is that physically possible, even if it is rare?

Thank you.

Stanley

 

Would be rare if at all.

Because planetary formation is messy with several worlds forming.

 

Best shot is probably a single planet solar system.....which is either rare or unlikely.

 

 

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5 minutes ago, MetricKerbalist said:

Hi Spacescifi,

Thank you for your response.

But a zero-eccentricity orbit is possible.   Is that correct?

Stanley

Your welcome.

In theory yes....though unlikely.

 

Try it. Play spacewar online and try getting a perfect circular orbit.

https://www.masswerk.at/spacewar/

Remember that planets do not have engines....where they go is dependent on a lot of things besides gravity....and add gravity of other planets to the mix really complicates matters.

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Venus approaches zero eccentricity, and googling shows  0.007.  My guess for a "real" zero eccentricity would be a bi-star system, or otherwise have a large body that has more effect on the planet than any apparent eccentricity.  Unfortunately, if it moved that much it might have an observable eccentricity, at least until batted around again.

The real issue in trying to claim something is "zero" vs. "less than a specified epsilon" is that the real world simply doesn't allow perfect measurement.  Any solar system with 3 bodies not locked into Lagrange points will be unstable, so there simply won't be an absolute value of eccentricity.  Even with a two body problem, there are only so many zeros you can pile on before eventually you hit a non-zero number.

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2 hours ago, MetricKerbalist said:

Hi wumpus,

OK.  I will ponder what you say.  Thank you for weighing in.

Stanley

1.   Venus has a very, very small eccentricity.  I think it wasn't measured until the 20th century.

2.  Zero doesn't exist in physics, except as weird quantum effects (like superconducting).  Orbits are old fashioned physics, so literal zero isn't possible.

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Triton gets about as close to zero as you'll find in the solar system, with an orbital eccentricity around Neptune of 0.000016

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Posted (edited)

Eccentricity is really just a way to describe the shape of the orbit geometrically, assuming that it's a perfect ellipse/other conic.  In real life, the trajectories of planets would only approximate these shapes.  You can still assign an eccentricity based on the best-fitting shape that describes the trajectory though.  As a quantity, the eccentricity is more just useful for us humans to describe what the shape of the orbit is in a way that is both easy to understand and mathematically convenient.  Nothing particularly exciting happens if you have exactly 0 eccentricity vs 0.000001 eccentricity, though the boundary at eccentricity 1 is a bit more exciting since then the orbiting body would escape!

Physically, there is no reason why you couldn't have perfectly circular orbits, since in physics we can have a perfect universe with only two objects in it in exactly whatever configuration we want.  Real life isn't quite so perfect, but it doesn't really matter.  As others said, you can still get as close to zero eccentricity as you like within measurement error.

Edited by StopIteration
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All accurate answers so far, but since it hasn't been discussed yet....

During the formation of a solar system, there are two driving forces that tend to circularize a planetary orbit. The first is the clearing of the local neighborhood. Objects which cross the orbit of a growing protoplanet due to high eccentricity are more likely to have high-energy collisions, while objects which orbit more or less circularly are more likely to have low-energy collisions. Since low-energy collisions are more likely to cause the growth of an object, objects will tend to grow larger in circular orbits.

The second driving force is the solar wind. Early in the life of a star, the outward pressure of the solar wind is going to be much greater because the star is still in the process of collapse. Outward pressure does not have any net effect on a circular orbit but it tends to circularize eccentric orbits.

This is not true, however, for objects orbiting very close to a star, because tidal locking and deviations in the energy output of a star can increase eccentricity. That's why Mercury has the most eccentric orbit (because it is so close) and Venus has the least eccentric orbit (because it is far enough out that tidal forces aren't an issue but it is also close enough that the solar wind forcing reduced its eccentricity).

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Posted (edited)
1 hour ago, sevenperforce said:

This is not true, however, for objects orbiting very close to a star, because tidal locking and deviations in the energy output of a star can increase eccentricity.

Does this mean hot gas giants tend to have high eccentricities ?

One thing I'll say here is that eccentricity is only useful to describe the average ellipses that approximates the orbit. In reality planetary orbits and trajectory have perturbations that means they're not exactly elliptical. For long-term predictions (currently up to 1000 years for eclipses and 500 years for general solar system bodies) as well for probe navigation and deep-space observation we use common barycenter rather than the parent star itself, but you can only predict so much.

Edited by YNM
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5 hours ago, YNM said:

Does this mean hot gas giants tend to have high eccentricities ?

I'm not sure what the current state of prediction is regarding the eccentricities of gas giant exoplanets very close to their stars.

If I had to guess I would say that tidally-locked exoplanets are likely to be more eccentric, while exoplanets that are far enough out to not be tidally locked would be more circular.

5 hours ago, YNM said:

One thing I'll say here is that eccentricity is only useful to describe the average ellipses that approximates the orbit. In reality planetary orbits and trajectory have perturbations that means they're not exactly elliptical. 

Yes, eccentricity is only constant in a two-body system.

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