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Let's assume there is a small natural satellite oribiting a massive planet. The satellite rotates synchronously (revolution period = orbital period) on a circular orbit, both bodies are homogenous and rigid. The question is, how to find the Roche limit with regard to centrifugal forces?

With use of Wikipedia I understand that the Roche limit is basically the orbit where difference of gravitational forces between the planet and two points in the satellite body (its centre and a surface facing to the planet) cancels the gravitational pull of the surface of the satellite to its centre. Wikipedia also states that more accurate is to include also the centrifugal force on the surface of the satellite due to its rotation.

Why is not included the difference of the centrifugal forces acting on satellite centre and surface facing to the planet, caused by satellite's orbital movement?

Are there any other forces I need to take into account, for example orbit of the whole sytem around a star?

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The roche limit is a calculation of when pieces of a satellite will be pulled off the satellite. Any force that affects the whole satellite equally can be ignored, as this will affect the motion of the satellite as a whole but not (directly) whether pieces are pulled off.

If you consider an object which maintains a constant orientation, but rotates about another body, the paths each part of the object take differ only by a translation in space, and so each part will experience the same centrifugal force. So, the only rotation that needs to be considered is the rotation of the body around its own axis.

In the special case of a tidally locked satellite in a circular orbit, the satellite's rotation around the planet, and the satellite's rotation around its own axis, are the same rotation, just differing in what axis we are considering the rotation to be around.

In the wikipedia article's calculations, the centrifugal force for the rigid body case is considered around the satellite's own axis, but for the fluid body case it is considered as a rotation about the primary. This is an equally valid calculation for the special case of a tidally locked satellite in a circular orbit.

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  • $\begingroup$ Thanks, bu I still have problems to imagine it right. If the rotation is "locked", there is only one centrifugal force, while if the rotation is not locked, there would be two ? $\endgroup$
    – David Komanek
    Mar 19, 2018 at 10:15
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    $\begingroup$ The locked rotation about the main body can be viewed as a superposition of a rotation of the satellite as a whole around the main body and the rotation of the satellite around its own axis. But only the rotation about its own axis differently affects different parts of the satellite, and thus affects the Roche limit. $\endgroup$
    – simon
    Mar 19, 2018 at 15:08
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    $\begingroup$ Another way to explain what DK is correctly saying here is that we can imagine a frame that is locally translating in a circle so as to fix the location of the planet. That is not a "rotating" reference frame, it has a constant fictitious force per gram everywhere because the circle has the same radius everywhere, so it's not a "centrifugal force" in the usual sense of a force that increases in proportion to radius. In fact there isn't a good word for this type of fictitious force, as it is used so rarely! But it's good for this problem whenever you can't assume the spin is locked. $\endgroup$
    – Ken G
    Mar 19, 2018 at 19:32
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    $\begingroup$ To clarify, we can say that the Roche limit has nothing to do with orbits at all, but in the case of a locked rotation, it is easy to mistake the effect as being due to the orbit because in that special case you get the right answer if you mistakenly think the orbit is doing it. All that actually matters is the tidal effects of gravity, and the spin of the satellite relative to the distant stars. $\endgroup$
    – Ken G
    Mar 19, 2018 at 19:40
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When a satellite is in a circular orbit around a planet you may consider its motion in a rotating frame of reference; the frame rotates at the same rate as the satellite orbits. In this frame the satellite doesn't move. But because this is a rotating frame the formula $F=ma$ is doesn't work, you need to add extra terms for the centrifugal force and the corilis force.

In the rotating frame there are two forces acting on the satellite: gravity and the centrifugal force, and since the satellite isn't moving (in this frame) these two forces must equal each other.

If the satellite is modelled as a particle, all is well. But if the satellite is a rigid body the part of the satellite that is closer to the planet will experience more gravity than centrifugal force. The part of the satellite that is furthest from the planet will have the opposite: the centrifugal force will be greater than gravity. The result is that part of the satellite is pulled towards the planet, and part of the satellite is pushed away. This is called the "tidal force", as it is the primary reason for tides in the oceans.

The satellite has its own gravity, and normally that is stronger than the tidal force. But if the satellite is sufficiently close to the planet, the tidal force may become greater than the self-gravitation of the satellite. The point at which this occurs is the Roche limit.

The basic calculation of the Roche limit is based on finding when the tidal force (caused by the different gravitational and centrifugal forces over the satellite) exceeds the gravitational force.

A more subtle calculation can take into account other factors: The tidal force can distort the satellite, the satellite may have significant rotation, there may be significant strength in the materials that form the satellite. These factors can cause a satellite to break up earlier or later than a simple calculation suggests.

However, The "centrifugal forces acting on satellite centre and surface facing to the planet, caused by satellite's orbital movement?" are the tidal forces. So these are already accounted for in the simple calculation.

The big uncertainty in calculating the break-up of a satellite is the strength of the chemical bonds that hold the satellite together. Most artificial satellites orbit will inside their Roche limit, but they don't break up because they are held together by strong metallic bonds.

If a satellite is rigid, then it probably has some tensile strength and will hold together even if tidal forces are tending to break it apart. If a satellite is a "rubble pile" and doesn't have any significant strength, the assumption of it being "rigid" must be questioned.

These factors introduce uncertainties that are much greater than any other forces, such as solar tides.

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  • $\begingroup$ Thank you for the explanation, but I still don't get it: centrifugal force acting on an unit mass (small part of the satellite) is m * r * omega^2, right ? In the orbital movement, the "r" is different for the centre and for the surface, so the force is also different and pushes parts of the satellite apart. As the "orbital omega" equals the "rotational omega" both centrifugal forces have the same value, but there are still two forces of that type, so why should I include only one of them in the calculation? I definitely have problem to imagine the frames of reference in a right way. $\endgroup$
    – David Komanek
    Mar 19, 2018 at 10:13
  • $\begingroup$ My point is that you have already included the "orbitial motion centrifugal force" You don't need to include it again. $\endgroup$
    – James K
    Mar 19, 2018 at 11:03
  • $\begingroup$ I've extended this, to explain what tidal forces are, how the relate to the centrifugal force and what I man by "you have already included" it. $\endgroup$
    – James K
    Mar 19, 2018 at 12:55
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    $\begingroup$ A correction must be made to what you mean by "the centrifugal force." In order for the difference between gravity and the centrifugal force to be the "tidal force," it is necessary to use a nonstandard meaning of centrifugal force. Normally, centrifugal force is for a rotating frame, but your statement about the tidal force is only correct if you use the fictitious force of a frame going everywhere in a circle of constant radius, like taking a piece of paper and keeping it at fixed orientation as you make its center go in a circle. There really isn't a good word for a frame like that! $\endgroup$
    – Ken G
    Mar 19, 2018 at 19:35
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    $\begingroup$ In particular, to clarify what I'm saying, a satellite that is tidally locked actually experiences two separate effects that control the Roche limit. One is the tidal effect, which is purely due to how gravity works and has nothing to do with either orbits or rotations, and the second is the effect of the spin of the satellite, which is a type of centrifugal force but has nothing to do with the orbit. In short, the Roche limit has nothing to do with an orbit and also applies to a freefalling satellite, it just wouldn't be interesting in that case because freefall is too short. $\endgroup$
    – Ken G
    Mar 19, 2018 at 19:38
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I think if the satellite doesn't rotate about its own axis the centrifugal force will be of the same strength for each point of the satellite because the radius of each trajectory is the same. But if the satellite rotates synchronously there should be different centrifugal forces causing the Roche limit to decrease by a factor of $\sqrt[3]{2}$
Edit: Okay, I think the factor was not correct. If $r$ denotes the distance between the centers of masses, $M$ is the big mass $m$ is the satellite mass and $R$ denotes the radius of the satellite the accelaration on a point at $r+\Delta r$ will be $$\omega^2(r+\Delta r)-\frac{GM}{(r+\Delta r)^2} \approx \omega^2r-\frac{GM}{r^2}+\omega^2\Delta r +2\frac{GM}{r^3}\Delta r$$ for small $\Delta r$. Note that $\omega$ is independant of $\Delta r$ if and only if the satellite rotates synchronously. Now since $\frac{GM}{r^2} = \omega^2 r$ it follows that the total accelaration at $r+\Delta r$ ist just $\frac{3GM}{r^3}\Delta r$. Setting $\Delta r = R$ and comparing it to the gravitational accelaration $\frac{Gm}{R^2}$ one obtains $$3MR^3 = mr^3 \Rightarrow r = R\sqrt[3]{\frac{3M}{m}}$$ (instead of $r = R \sqrt[3]{\frac{2M}{m}}$ for the case without rotation of the satellite about itself.)

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    $\begingroup$ Do you have any scientific evidence to support your answer? $\endgroup$
    – WarpPrime
    Sep 29, 2021 at 17:32

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