<p>"Perfectly" is a funny word.</p>

<p>Perfect circles are a mathematical abstraction. Real objects are not "perfect".  So supposing a "perfectly spherical planet" is to suppose something that does not and <em>could not</em> exist. All real planets are made of atoms and anything made of little clumps of matter cannot be perfectly spherical.  Even if you built a planet that was as spherical as possible, it would be distorted by its rotation and the tides. So there are no perfectly spherical planets.</p>

<p>Now you say "put in a perfectly circular orbit". This is like drawing a line that is exactly $\pi$cm long. Again you are supposing something that does not, and <em>could not</em> exist.</p>

<p>What we can do is consider a mathematical model of gravity. If you model the sun and the planet as "particles" (ie point masses) and you model gravity with Newton's law of universal gravity, and if you give the model the system with the exact amount of energy to give a perfect circle, then the system will remain in a perfect circle, it will never become elliptical.</p>

<p>If you use general relativity to model gravity, then the release of gravitational radiation will mean that no circular orbits are possible, all orbits will spiral inwards, however it would not become elliptical. Something similar will happen with quantum models of gravity. </p>

<p>So your question can only be answered in the context of a mathematical model of gravity. </p>

CC BY-SA 3.0
2017-09-16T21:45:55.977Z
This may seem a bit pedantic, but I can choose a number that is exactly Pi: Pi.  Unless you're saying that Pi doesn't exist as a number.  The rest of your point is well taken (that perfect mathematical abstractions don't generally show up in the real world.)
Beska
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2017-09-16T22:16:56.297Z
This could get very Platonic, I don't care to discuss if mathematical entities "exist", so I'll rephrase.
James K
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2017-09-16T22:49:10.677Z
For what it's worth, I think the rewording supports your point much more clearly!
Beska
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2017-09-17T00:48:52.440Z
Wow! Saying that "perfect" isn't possible is a bit **pedantic**, since real scientists make *Ceteris paribus* assumptions all the time.... ("Everything else being equal*-one in Bonehead English). So, stop **spanking** him and please, try again.
GwenKillerby
CC BY-SA 3.0
2017-09-17T07:22:25.197Z
Real scientists know exactly what is meant by "model", since the answer depends on the model of gravity used. Understanding that "perfectly" relates to a model and not to real objects is an important fact, that is often misunderstood. So this is not mere pedantry. Please re-read, since I have already addressed your point in the second half of the answer where I give the answer for both a Newtonian and a GR model, with the bodies modelled as particles.
James K
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2017-09-18T06:27:19.443Z
@Beska "Pi exactly" is not easy to do in a real world.
Thorbjørn Ravn Andersen
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2017-09-18T17:58:17.613Z
@ThorbjørnRavnAndersen Exactly.  That was my point.  Originally he said you couldn't pick a number exactly equal to Pi...and that's obviously possible; now that he has clarified that point to be related to the real world, it is correct and supports the rest of the answer more directly.
Beska
CC BY-SA 3.0
2017-09-16T18:50:03.100Z
2017-09-16T22:17:48.390Z
2017-09-16T22:17:48.390Z
James K
<p>No. Tidal friction will perturb your orbit out of spherical. Because your planet and moon aren't optimally shaped this will happen faster than if they were allowed to take on the liquid drop shape they would naturally have. Once having achieved the balanced shape and balanced orbit around the barycenter, your system is still not quite circular due to general relativistic effects.</p>

<p>This is the nature of the beast; circular orbits are inherently unstable and want to fall into precessing ellipses.</p>

CC BY-SA 3.0
2017-09-17T20:24:25.467Z
What if the moon is in tidal lock, and we ignore the (unnoticeable, anyway) gravitational wave energy loss? Are there tidal forces (generating heat) with tidal lock?
Peter - Reinstate Monica
CC BY-SA 3.0
2017-09-17T20:26:05.613Z
Unrelated to tidal forces is your statement "circular orbits are inherently unstable and want to fall into precessing ellipses." Is that indeed so?
Peter - Reinstate Monica
CC BY-SA 3.0
2017-09-17T21:16:15.473Z
@PeterA.Schneider: They tell me its true. Nice relativistic effect that one.
Joshua
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2017-09-17T04:17:14.810Z
2017-09-17T04:17:14.810Z
Joshua
<p>Short answer:  </p>

<p>Yes.  If you ignore the tidal effect and relativity and any change in mass (planets radiate light and lose atmosphere and add space dust and meteors all the time, so mass isn't constant), then in a two body system with no outside effects, the orbit would remain perfectly circular.   There would be no outside force to affect the circular orbit.    A circular orbit is impossible because nothing can be that exact, but on a computer simulation you could set it up and it would remain circular.</p>

<p>Long answer:</p>

<p>For your scenario to work you'd need to give both the planet and moon infinite hardness, so they wouldn't bend at all and fixed mass and space would need to be completely empty of anything else.  Needless to say that's impossible.  But only in Newtonian gravity.  </p>

<p>Relativity creates a very very tiny decay in orbits, in your system of a planet/moon that would be close to negligible but there would be a very tiny spiraling inward.    The relativistic effect on an orbit was first noticed with Mercury's orbit around the sun (and Mercury isn't falling into the sun, it was noticed by other effects - but lets not get into that here).   </p>

<p>Similarly, any loss in mass, gain in mass or orbital drag (because space is full of tiny particles, fast moving particles, photons and neutrinos, all of which cause a tiny but at least in simulation, calculatable drag), then the two body system would have an imperceptibly small spiral and not be a perfect circle.  You could say in a sense that it becomes elliptical but it would be more like a constant very tiny force where, once it was elliptical, it could wonder back to more circular.   Not all perturbations or drag on an orbit make that orbit more elliptical, it can work in either direction.</p>

<p>It's worth noting that "falling" or decaying towards the planet wouldn't "create" an elliptical orbit.   A circle is an ellipse.   You asked specifically about 2 body systems, where, ignoring tides, falling in or out would be more of a slow spiral.   An ellipse isn't the result of a decaying or perturbed orbit.  An ellipse is the baseline orbit.  Perturbations and orbital decay happen on top of the ellipse (if that makes sense), they don't cause the ellipse. </p>

<p>In a 3 or more body system you get orbital perturbations on the orbits.   Those often remain stable, they're just variations that mostly move back and forth.  See <a href="https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29" rel="noreferrer">Eccentricity variation</a> and <a href="https://en.wikipedia.org/wiki/Milankovitch_cycles#Apsidal_precession" rel="noreferrer">Apsidal Precession</a>. </p>

CC BY-SA 3.0
2017-09-18T05:02:07.850Z
What if the moon is in tidal lock (as it indeed happens eventually, as we know)? And if it is in a geo- (or rather planeto-) stationary orbit, so that there is no rotational acceleration from the gravitational field of the planet either?
Peter - Reinstate Monica
CC BY-SA 3.0
2017-09-18T06:53:27.520Z
@PeterA.Schneider  Maybe I'll update the answer with more details, but a perfect circle is impossible.  A perfect tidal lock is also impossible, there will always be some wobble and some transfer of energy into heat.   But you're right that a tidal lock like that is about as circular as it gets.   Pluto-Charon is like that and very close to a circle.
userLTK
CC BY-SA 3.0
2017-09-16T23:26:00.513Z
2017-09-17T02:24:13.337Z
2017-09-17T02:24:13.337Z
userLTK
<p>Let's assume that there is a perfectly spherical planet and there is a moon which is also perfectly spherical. Let's assume that there is no atmospheric drag and no other gravitational pull. If the moon is put in a perfectly circular orbit around the planet somehow, would the moon eventually "fall" towards the planet and form an elliptical orbit or would it continue to follow the perfectly circular orbit?</p>

<p>Edit: What i actually wanted to ask is that would the <em>gravity</em> of the planet cause the moon to "fall" towards the planet or would the gravity allow the moon to continue its orbit without bending its path any further towards the planet. I know that no planet can be a true sphere or a cube due to the shapes of the particles.</p>

CC BY-SA 3.0
2017-09-16T18:50:42.467Z
What model of gravity do you want to assume, Newtonian, or General relativity?
James K
CC BY-SA 3.0
2017-09-18T00:26:46.577Z
Yes, "the gravity of the planet would cause the moon to "fall" towards the planet"; that's exactly what an orbit is!
Keith
CC BY-SA 3.0
2017-09-18T13:01:42.583Z
One should note that the moon is not, strictly spoken, in an orbit around the planet; rather, both of them are orbiting the common center of mass. For small moons and large planets the difference is small, but for Pluto/Charon (and Charon's orbit is, userLTK says, fairly circular) it is significant: The center of the circle -- the *barycenter* of the Pluto/Charon system -- is outside Pluto.
Peter - Reinstate Monica
<orbit><gravity>
Is a perfectly circular orbit possible?
Wasat
# Is a perfectly circular orbit possible?

(Asked by: username_0 on 2017-09-16)

Let's assume that there is a perfectly spherical planet and there is a moon which is also perfectly spherical. Let's assume that there is no atmospheric drag and no other gravitational pull. If the moon is put in a perfectly circular orbit around the planet somehow, would the moon eventually "fall" towards the planet and form an elliptical orbit or would it continue to follow the perfectly circular orbit?

Edit: What i actually wanted to ask is that would the *gravity* of the planet cause the moon to "fall" towards the planet or would the gravity allow the moon to continue its orbit without bending its path any further towards the planet. I know that no planet can be a true sphere or a cube due to the shapes of the particles.

**Tags:** orbit, gravity

> **Comments:**
> **username_1**: What model of gravity do you want to assume, Newtonian, or General relativity?  
> **username_2**: Yes, "the gravity of the planet would cause the moon to "fall" towards the planet"; that's exactly what an orbit is!  
> **username_3**: One should note that the moon is not, strictly spoken, in an orbit around the planet; rather, both of them are orbiting the common center of mass. For small moons and large planets the difference is small, but for Pluto/Charon (and Charon's orbit is, username_8 says, fairly circular) it is significant: The center of the circle -- the *barycenter* of the Pluto/Charon system -- is outside Pluto.  

---

## 3 answers

### Answer by username_8 on 2017-09-16. Score: 7 | Accepted answer

Short answer:

Yes. If you ignore the tidal effect and relativity and any change in mass (planets radiate light and lose atmosphere and add space dust and meteors all the time, so mass isn't constant), then in a two body system with no outside effects, the orbit would remain perfectly circular. There would be no outside force to affect the circular orbit. A circular orbit is impossible because nothing can be that exact, but on a computer simulation you could set it up and it would remain circular.

Long answer:

For your scenario to work you'd need to give both the planet and moon infinite hardness, so they wouldn't bend at all and fixed mass and space would need to be completely empty of anything else. Needless to say that's impossible. But only in Newtonian gravity.

Relativity creates a very very tiny decay in orbits, in your system of a planet/moon that would be close to negligible but there would be a very tiny spiraling inward. The relativistic effect on an orbit was first noticed with Mercury's orbit around the sun (and Mercury isn't falling into the sun, it was noticed by other effects - but lets not get into that here).

Similarly, any loss in mass, gain in mass or orbital drag (because space is full of tiny particles, fast moving particles, photons and neutrinos, all of which cause a tiny but at least in simulation, calculatable drag), then the two body system would have an imperceptibly small spiral and not be a perfect circle. You could say in a sense that it becomes elliptical but it would be more like a constant very tiny force where, once it was elliptical, it could wonder back to more circular. Not all perturbations or drag on an orbit make that orbit more elliptical, it can work in either direction.

It's worth noting that "falling" or decaying towards the planet wouldn't "create" an elliptical orbit. A circle is an ellipse. You asked specifically about 2 body systems, where, ignoring tides, falling in or out would be more of a slow spiral. An ellipse isn't the result of a decaying or perturbed orbit. An ellipse is the baseline orbit. Perturbations and orbital decay happen on top of the ellipse (if that makes sense), they don't cause the ellipse.

In a 3 or more body system you get orbital perturbations on the orbits. Those often remain stable, they're just variations that mostly move back and forth. See [Eccentricity variation](https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_.28eccentricity.29) and [Apsidal Precession](https://en.wikipedia.org/wiki/Milankovitch_cycles#Apsidal_precession).

> **Comments:**
> **username_3**: What if the moon is in tidal lock (as it indeed happens eventually, as we know)? And if it is in a geo- (or rather planeto-) stationary orbit, so that there is no rotational acceleration from the gravitational field of the planet either?  
> **username_8**: @PeterA.Schneider  Maybe I'll update the answer with more details, but a perfect circle is impossible.  A perfect tidal lock is also impossible, there will always be some wobble and some transfer of energy into heat.   But you're right that a tidal lock like that is about as circular as it gets.   Pluto-Charon is like that and very close to a circle.  

---

### Answer by username_1 on 2017-09-16. Score: 27

"Perfectly" is a funny word.

Perfect circles are a mathematical abstraction. Real objects are not "perfect". So supposing a "perfectly spherical planet" is to suppose something that does not and *could not* exist. All real planets are made of atoms and anything made of little clumps of matter cannot be perfectly spherical. Even if you built a planet that was as spherical as possible, it would be distorted by its rotation and the tides. So there are no perfectly spherical planets.

Now you say "put in a perfectly circular orbit". This is like drawing a line that is exactly $\pi$cm long. Again you are supposing something that does not, and *could not* exist.

What we can do is consider a mathematical model of gravity. If you model the sun and the planet as "particles" (ie point masses) and you model gravity with Newton's law of universal gravity, and if you give the model the system with the exact amount of energy to give a perfect circle, then the system will remain in a perfect circle, it will never become elliptical.

If you use general relativity to model gravity, then the release of gravitational radiation will mean that no circular orbits are possible, all orbits will spiral inwards, however it would not become elliptical. Something similar will happen with quantum models of gravity.

So your question can only be answered in the context of a mathematical model of gravity.

> **Comments:**
> **username_4**: This may seem a bit pedantic, but I can choose a number that is exactly Pi: Pi.  Unless you're saying that Pi doesn't exist as a number.  The rest of your point is well taken (that perfect mathematical abstractions don't generally show up in the real world.)  
> **username_1**: This could get very Platonic, I don't care to discuss if mathematical entities "exist", so I'll rephrase.  
> **username_4**: For what it's worth, I think the rewording supports your point much more clearly!  
> **username_5**: Wow! Saying that "perfect" isn't possible is a bit **pedantic**, since real scientists make *Ceteris paribus* assumptions all the time.... ("Everything else being equal*-one in Bonehead English). So, stop **spanking** him and please, try again.  
> **username_1**: Real scientists know exactly what is meant by "model", since the answer depends on the model of gravity used. Understanding that "perfectly" relates to a model and not to real objects is an important fact, that is often misunderstood. So this is not mere pedantry. Please re-read, since I have already addressed your point in the second half of the answer where I give the answer for both a Newtonian and a GR model, with the bodies modelled as particles.  
> **username_6**: @username_4 "Pi exactly" is not easy to do in a real world.  
> **username_4**: @ThorbjørnRavnAndersen Exactly.  That was my point.  Originally he said you couldn't pick a number exactly equal to Pi...and that's obviously possible; now that he has clarified that point to be related to the real world, it is correct and supports the rest of the answer more directly.  

---

### Answer by username_7 on 2017-09-17. Score: 2

No. Tidal friction will perturb your orbit out of spherical. Because your planet and moon aren't optimally shaped this will happen faster than if they were allowed to take on the liquid drop shape they would naturally have. Once having achieved the balanced shape and balanced orbit around the barycenter, your system is still not quite circular due to general relativistic effects.

This is the nature of the beast; circular orbits are inherently unstable and want to fall into precessing ellipses.

> **Comments:**
> **username_3**: What if the moon is in tidal lock, and we ignore the (unnoticeable, anyway) gravitational wave energy loss? Are there tidal forces (generating heat) with tidal lock?  
> **username_3**: Unrelated to tidal forces is your statement "circular orbits are inherently unstable and want to fall into precessing ellipses." Is that indeed so?  
> **username_7**: @PeterA.Schneider: They tell me its true. Nice relativistic effect that one.  

---

