# Hill spheres, Roche limits and Periods, OH MY!

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### #1 Janus

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Posted 06 June 2006 - 06:28 PM

Over in the Space Voyage thread I mentioned that it was not possible to place satellites in selenostationary orbits around the moon. I went on to say that this was true for any tidally locked body (any body that rotates with the same period as its revolution.) I thought that here I would show how this is determined to be true

$R \approx a \sqrt[3]{\frac{m}{3M}}$

Where a is the distance between our two main bodies.
m is the mass of the body that we are finding the Hill sphere for.
M is the mass of the body that the smaller body is orbiting.

Now we take the formula for the period of a satellite:

$T = 2 \pi \sqrt{\frac{r^3}{Gm}}$

If we substitute R from the first equation for r in in the second, we get an equation that gives the period for a satellite orbiting at the edge of the Hill sphere. Since no satellite can exist outside the Hill sphere, this is also the longest period a satellite can have around the body:

$T = 2 \pi \sqrt{\frac{ \left(a \sqrt[3]{\frac{m}{3M}} \right)^3}{Gm}}$

$T = 2 \pi \sqrt{\frac{ a^3\frac{m}{3M}}{Gm}}$

$T = 2 \pi \sqrt{\frac{ a^3}{3GM}}$

Now we compare this period with the rotation of our satellite, which is equal to the orbital period of that satellite in a tidal lock situation:

$T_2 = 2 \pi \sqrt{\frac{a^3}{GM}}$

Comparing the two, we get:

$ratio = \frac{ 2 \pi \sqrt{\frac{ a^3}{3GM}}}{2 \pi \sqrt{\frac{a^3}{GM}}}$

$ratio =\sqrt{ \frac{ \pi \frac{ a^3}{3GM}}{2 \pi \frac{a^3}{GM}}}$

$ratio =\sqrt{ \frac{ \frac{ a^3}{3GM}}{ \frac{a^3}{GM}}}$

$ratio = \sqrt{1/3} = 0.57735...$

Which shows that the longest period our satellite can have will always be shorter than the orbital period of the body it is orbiting, and by extension, shorter than that of the rotational period of that body if it is tidally locked.

Now let's look at the other extreme for satellite orbits; the Roche limit. The Roche limit is the point where tidal forces become great enough to tear the body apart. (one small caveat: The Roche limit only effects satellites large enough that the tidal forces can overcome the tensile strength of the satellite.)

Often you will see the Roche Limit expressed as

$r_{Roche} \approx 2.423 R$

Where R is the radius of the body our satellite is orbiting.

This expression is somewhat vague as it makes two assumptions. One is that the satellite and main body are of equal density, and the other is that the bdy is "fluid" in nature. (not rigid, which is more descriptive of most moons)

Correcting for these assumptions we get the following relation:

$r_{Roche} \approx R \sqrt[3]{2 \frac{\rho M}{\rho m}}$

Where $\rho M$ and $\rho m$ are the densities of the primary and satellite respectively.

Now we can determine the period of a satellite orbiting at the Roche limit by inserting this equation into the orbital period equation as we did above:

$T = 2 \pi \sqrt{\frac{\left( R \sqrt[3]{2 \frac{\rho M}{\rho m}}\right)^3}{GM}}$

$T = 2 \pi \sqrt{\frac{ R^3{2 \frac{\rho M}{\rho m}}}{GM}$

$T = 2 \pi \sqrt{\frac{2R^3 \rho M}{GM \rho m}}$

Now we're going to get a little sneaky.

The density of a body is equal to its mass divided by its volume, and since the volume of a sphere is expressed as:

$V = \frac{4 \pi R^3}{3}$

we can express the density of the primary as

$\rho M = \frac{3M}{4 \pi R^3}$

If we now insert this expression into our orbital period equation, we get:

$T = 2 \pi \sqrt{\frac{2R^3 \frac{3M}{4 \pi R^3} }{GM \rho m}}$

After canceling and simplifying, we get:

$T= \sqrt{ \frac{6\pi}{G \rho m}}$

Since $\pi$ and G are constants, the only variable left is $\rho m$.

This means that the minimum period for a moon that is subject to break up from passing within the Roche Limit is determined by the density of the moon alone and no other factor.

DISCLAIMER
The above works under the following conditions: That the moon is much less massive than the planet it revolves around and that the planet is nearly spherical and not too much of an oblate spheroid.

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### #2 Mercedes Benzene

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Posted 06 June 2006 - 07:51 PM

OH dear!

That is way too many equations in one post, for me.
haha.

### #3 CraigD

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Posted 06 June 2006 - 09:11 PM

Over in the Space Voyage thread I mentioned that it was not possible to place satellites in selenostationary orbits around the moon. I went on to say that this was true for any tidally locked body (any body that rotates with the same period as its revolution.) I thought that here I would show how this is determined to be true …

My confidence in Janus’s reputation for sound, rigorous mechanics, is enough for me to just enjoy reading his demonstration, without any attempt at criticism. I applaud you, Janus!

To play the devil’s advocate, though, the L1 and L2 Earth-Moon Lagrangian points, located by definition at points on the Moon’s Hill sphere, are effectively selenostationary. Objects placed in these points would appear stationary from the Moon’s surface.

### #4 coldcreation

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Posted 07 June 2006 - 12:41 PM

Over in the Space Voyage thread I mentioned that it was not possible to place satellites in selenostationary orbits around the moon. I went on to say that this was true for any tidally locked body (any body that rotates with the same period as its revolution.) I thought that here I would show how this is determined to be true

Wow, that was a tremendous post.
Question: why are you interested in showing the impossibility of placing satellites in selenostationary orbits around the moon?
CC

### #5 Janus

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Posted 07 June 2006 - 05:16 PM

To play the devil’s advocate, though, the L1 and L2 Earth-Moon Lagrangian points, located by definition at points on the Moon’s Hill sphere, are effectively selenostationary. Objects placed in these points would appear stationary from the Moon’s surface.

And to play the devil's advocate to your devil's advocate, actually, they are not selenostationary, but selenosychronous. They will trace out a 12 degree arc as seen from the Moon's surface. This is due the Moon's libration.

While the Moon's orbital and rotational periods are the same, they do not always share the same angular velocity. The Moon rotates at a steady rate, however, since the Moon's orbit is elliptical, is orbital velocity changes over the course of its orbit. As a result, the Moon's rotation alternately lags behind and and pulls ahead of its orbit. The Moon therefore "wobbles" with respect to its orbital radial (The line running through both the Moon's and Earth's center) Since the L1 and L2 point lay on that line, the Moon also "wobbles" with respect to these points and they will not be stationary with respect to the surface of the Moon.

### #6 Nootropic

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Posted 07 June 2006 - 09:17 PM

DISCLAIMER
The above works under the following conditions: That the moon is much less massive than the planet it revolves around and that the planet is nearly spherical and not too much of an oblate spheroid.

So what exactly happens then? Much more complicated? I would attempt to look it up, but I'm currently taking a break from my math project and getting even more distracted is not a good thing for me...

And what's the largest ratio of moon to planet discovered?

### #7 coldcreation

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Posted 08 June 2006 - 10:17 AM

And to play the devil's advocate to your devil's advocate, actually, they are not selenostationary, but selenosychronous. They will trace out a 12 degree arc as seen from the Moon's surface. This is due the Moon's libration.

While the Moon's orbital and rotational periods are the same, they do not always share the same angular velocity. The Moon rotates at a steady rate, however, since the Moon's orbit is elliptical, is orbital velocity changes over the course of its orbit. As a result, the Moon's rotation alternately lags behind and and pulls ahead of its orbit. The Moon therefore "wobbles" with respect to its orbital radial (The line running through both the Moon's and Earth's center) Since the L1 and L2 point lay on that line, the Moon also "wobbles" with respect to these points and they will not be stationary with respect to the surface of the Moon.

Question: why are you interested in showing the impossibility of placing satellites in selenostationary orbits around the moon?

Forgive me if it's a personal question, but I would like to know, before I jump in...
CC

### #8 Janus

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Posted 08 June 2006 - 04:39 PM

Question: why are you interested in showing the impossibility of placing satellites in selenostationary orbits around the moon?

Forgive me if it's a personal question, but I would like to know, before I jump in...
CC

The reason that I brought the subject up was that in the Space Voyage thread, mention was made of using such oribts for communication satellites.

As to extending the argument to the more general case: This was something I discovered some time back while indulging my own intellectual curousity and I found it interesting.

The reason I posted it here was that I thought that others would find it interesting.

### #9 Janus

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Posted 08 June 2006 - 05:00 PM

So what exactly happens then? Much more complicated? I would attempt to look it up, but I'm currently taking a break from my math project and getting even more distracted is not a good thing for me...

And what's the largest ratio of moon to planet discovered?

Well, for one thing, if the Moon is massive enough, then the equation for the orbital period is more accurately stated as:

$T= 2 \pi \sqrt{\frac{r^3}{G(M+m)}}$ and thus the mass of the Moon becomes a factor in determining the orbital period.

As for the oblate spheroid, the equation for volume becomes:

$V = \frac{4 \pi r^3 \epsilon}{3}$

Where $\epsilon$ is the oblateness of the planet.

If you use this equation for in our substituition for the planet density as we did above, the oblateness does not cancel out and also becomes a factor in determining the period of the orbit.

At present the moon that is closest to its planet in mass is Charon, the moon of Pluto.

### #10 coldcreation

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Posted 09 June 2006 - 12:24 PM

And to play the devil's advocate to your devil's advocate, actually, they are not selenostationary, but selenosychronous. They will trace out a 12 degree arc as seen from the Moon's surface. This is due the Moon's libration.

While the Moon's orbital and rotational periods are the same, they do not always share the same angular velocity. The Moon rotates at a steady rate, however, since the Moon's orbit is elliptical, is orbital velocity changes over the course of its orbit. As a result, the Moon's rotation alternately lags behind and and pulls ahead of its orbit. The Moon therefore "wobbles" with respect to its orbital radial (The line running through both the Moon's and Earth's center) Since the L1 and L2 point lay on that line, the Moon also "wobbles" with respect to these points and they will not be stationary with respect to the surface of the Moon.

And to play the devil's advocate to your devil's advocate to CraigD's devil's advocate, I would go even further with a fundamental generalization...

The necessary stability conditions for real systems to exist is attained through gravitational interactions leading to periodic recurrences of resonance patterns, which in turn lead to symmetries in the mean motion of objects and ensuring repetition of certain geometrical configurations. And, the existence of special areas and points in space positioned in conspicuous locations within and around the intrinsic and combined gravitational fields of massive bodies__providing a basic mechanism for long-term stability__is present in a wide variety of dynamical systems and on all scales.

It is difficult to overestimate the importance of the Lagrange discovery. Not only does it confine the number of possible structures that can exist, it demonstrates a regularity and pattern among those systems that do exist.

The Lagrange system reveals how interacting gravitational fields of massive bodies generate periodicity, just as the Pauli principle generates the periodicity among atomic elements in the case of electrons orbiting atomic nuclei__as in the Mendeleev periodic table of elements__and the systematic pattern among quark clusters that represent the subsistence of a profound layer of reality on the smallest scales. (The forces that cluster quarks together are not yet fully understood but some of the patterns and features have already been identified).

Our ability to recognize that nature forms regular patterns at all scales (from quarks to superclusters) and limits the number of available structures, rather than giving way to disorganized chaos, is essential if we are to make any progress at all in cosmology.

In Théorie des fonctions analytiques (1796), more than one hundred years before Albert Einstein and Hermann Minkowski, Joseph-Louis Lagrange referred to dynamics as a “four-dimensional geometry.” Without doubt, such illustrations aimed at their audiences, with physical intent, were certainly not geometric in any ordinary sense; but unquestionably they were geometric in their concern with planimetric space of gravitating systems, and in the fundamentals of their relationship with matter.

I can (and should elaborate) if the above is yet unclear.

From mechanics (for starters), it is thought that the existence of stable orbits depends on the distance, the mass, and the angular momentum of given bodies (stemming from some ad hoc initial condition); where centrifugal energy (or force) exactly counters the tendency of gravitational attraction. A stable state is possible only if the velocity of a massive body (e.g., a planet) is precisely adjusted. For example, if the velocity of the earth were any different it would either crash into the sun or break away from the solar system. In the absence of the existence of this dependence, conditions for the stability of motion are not satisfied.

Note, in passing, a similar problem occurs when contemplating the expansion of the universe (where a fine-tuning problem exists between the pull of gravity and the expansive tendency due to lambda, quintessence, dark energy or other; you choose).

CC

### #11 Janus

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Posted 09 June 2006 - 06:17 PM

From mechanics (for starters), it is thought that the existence of stable orbits depends on the distance, the mass, and the angular momentum of given bodies (stemming from some ad hoc initial condition); where centrifugal energy (or force) exactly counters the tendency of gravitational attraction. A stable state is possible only if the velocity of a massive body (e.g., a planet) is precisely adjusted. For example, if the velocity of the earth were any different it would either crash into the sun or break away from the solar system. In the absence of the existence of this dependence, conditions for the stability of motion are not satisfied.

CC

This is simply not true. There is no such fine balance for stable orbits. The Earth's velocity can be changed quite a bit before it would do either of the two things you mention. You would have to decrease the Earth's velocity to about 10% of its present value to cause it to crash into the Sun and increase it by 42% to cause to to break away from the solar system. Anything less than this will simply put it into a new, different stable orbit.

### #12 CraigD

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Posted 10 June 2006 - 11:53 AM

…For example, if the velocity of the earth were any different it would either crash into the sun or break away from the solar system. In the absence of the existence of this dependence, conditions for the stability of motion are not satisfied.

This is simply not true. There is no such fine balance for stable orbits. The Earth's velocity can be changed quite a bit before it would do either of the two things you mention. You would have to decrease the Earth's velocity to about 10% of its present value to cause it to crash into the Sun and increase it by 42% to cause to to break away from the solar system. Anything less than this will simply put it into a new, different stable orbit.

I think Janus is right, though I calculate 10% to much too small a change to cause the Earth’s orbit to intersect the surface of the Sun.

Making the unrealistic assumption that the earth’s orbital speed were reduced suddenly, the % change in velocity required for a transfer orbit from Earth’s current orbit to the surface of the sun is $1-\sqrt{\frac{2r_1}{r_1+r_2}}$, where $r_1$ is the radius of the sun, about 696000000 m, and $r_2$ is the orbit of the Earth, between about 147098074000 and 152097701000, which calculates to about 90%

From $\frac{v_1-v_0}{v_0} = Q = 1-\sqrt{\frac{2r_1}{r_1+r_2}}$

we can derive $\frac{r1}{r2} = (\frac{1 - Q}{1 + Q})^2$

A 10% reduction is speed, then, results in a transfer orbit intersecting an circular orbit about .6694 the radius of the initial circular orbit’s radius.

I’m always reassured of the validity of the use of “canned” formulae after confirming them via simulation. Here’re a couple using XGRAVSIM2:
XGRAVSIM2>KILL,TI=60
XGRAVSIM2>1M=1.9891e30,2M=5.9742e24,2X=147098074000,2VY=2938,2Y=0,2VX=0
XGRAVSIM2>G 10080,W T,1-2D,1-2V,/LF,R 8
T=604800,1-2D=145983891948.8553055,1-2V=4733.72001466362388
T=1209600,1-2D=142606970337.0407956,1-2V=8091.777959808350465
T=1814400,1-2D=136858050672.5705681,1-2V=11986.94572400672793
T=2419200,1-2D=128531731618.3976197,1-2V=16412.34076860132352
T=3024000,1-2D=117278727420.9593602,1-2V=21623.48572426255307
T=3628800,1-2D=102498451264.4597731,1-2V=28178.6711678731093
T=4233600,1-2D=83060161158.62824095,1-2V=37420.1054679312208
T=4838400,1-2D=56315348724.45282885,1-2V=54022.10595594028115
T=5443200,1-2D=8190676643.699498085,1-2V=174986.3000632678207
XGRAVSIM2>G 10,W T,1-2D,1-2V,/LF,R 10
T=5443800,1-2D=8090274426.783181915,1-2V=176132.0597225815228
T=5444400,1-2D=7989268959.2419484,1-2V=177306.0055398931496
T=5445000,1-2D=7887646536.09340199,1-2V=178509.3698476551481
T=5445600,1-2D=7785392898.89187068,1-2V=179743.4629243502723
T=5446200,1-2D=7682493203.55573021,1-2V=181009.6794915693444
T=5446800,1-2D=7578931985.75151138,1-2V=182309.505887389083
T=5447400,1-2D=7474693123.606006255,1-2V=183644.5280009369547
T=5448000,1-2D=7369759797.492215095,1-2V=185016.4400655650122
T=5448600,1-2D=7264114446.606375095,1-2V=186427.054422727972
T=5449200,1-2D=7157738722.02102846,1-2V=187878.3123858878082
T=5449800,1-2D=7050613435.862595845,1-2V=189372.2963540543664
XGRAVSIM2>G 1,W T,1-2D,1-2V,/LF,R 9
T=5449860,1-2D=7039858916.263961375,1-2V=189524.1293886055948
T=5449920,1-2D=7029096680.425772825,1-2V=189676.4142399705774
T=5449980,1-2D=7018326707.66104398,1-2V=189829.1532516772573
T=5450040,1-2D=7007548977.186235745,1-2V=189982.3487845253117
T=5450100,1-2D=6996763468.12062073,1-2V=190136.0032167509424
T=5450160,1-2D=6985970159.485642535,1-2V=190290.118944193596
T=5450220,1-2D=6975169030.20426958,1-2V=190444.6983804646381
T=5450280,1-2D=6964360059.10034354,1-2V=190599.7439571180114
T=5450340,1-2D=6953543224.89792224,1-2V=190755.2581238229035
T=5450400,1-2D=6942718506.22061698,1-2V=190911.2433485384531
T=5450460,1-2D=6931885881.590924255,1-2V=191067.7021176905233
Earth hits the Sun between T=5450280 (seconds) and T=5450340.

Interestingly, reducing the initial velocity by 1 m/s less results in the earth missing the sun by a mere 11000000 meters
XGRAVSIM2>KILL,TI=60
XGRAVSIM2>1M=1.9891e30,2M=5.9742e24,2X=147098074000,2VY=2939,2Y=0,2VX=0
XGRAVSIM2>G 10080,W T,1-2D,1-2V,/LF,R 8
T=604800,1-2D=145983899302.2356199,1-2V=4734.33105757472755
T=1209600,1-2D=142607000326.9633533,1-2V=8092.11690514685932
T=1814400,1-2D=136858120479.9825214,1-2V=11987.14958684849842
T=2419200,1-2D=128531862261.293635,1-2V=16412.45584399717345
T=3024000,1-2D=117278947241.9985589,1-2V=21623.52349726916746
T=3628800,1-2D=102498803527.2855686,1-2V=28178.61748146735761
T=4233600,1-2D=83060725116.5213141,1-2V=37419.89398640776504
T=4838400,1-2D=56316328888.43167725,1-2V=54021.400853655891
T=5443200,1-2D=8194757059.08165373,1-2V=174940.1895103219091
XGRAVSIM2>G 60,W T,1-2D,1-2V,/LF,R 8
T=5446800,1-2D=7583188931.760209665,1-2V=182255.5771699500134
T=5450400,1-2D=6947180244.95342135,1-2V=190846.9212307504088
T=5454000,1-2D=6282636665.09304711,1-2V=201159.4771898013579
T=5457600,1-2D=5584103159.29164916,1-2V=213896.6012512719975
T=5461200,1-2D=4844032889.283851505,1-2V=230252.3240616983176
T=5464800,1-2D=4051487511.967748804,1-2V=252464.7066601551794
T=5468400,1-2D=3189742112.04700061,1-2V=285382.712389301768
T=5472000,1-2D=2232613696.478998848,1-2V=342239.4903970319643
T=5475600,1-2D=1164821365.375232411,1-2V=475542.5363133236275
XGRAVSIM2>G 1,W T,1-2D,1-2V,/LF,R 44
T=5475660,1-2D=1147020446.67227793,1-2V=479247.293190331188
T=5475720,1-2D=1129302183.21669423,1-2V=483021.2703055113585
T=5475780,1-2D=1111676408.229058064,1-2V=486864.4889940764965
T=5475840,1-2D=1094153724.451833985,1-2V=490776.6871108019755
T=5475900,1-2D=1076745559.700428167,1-2V=494757.2726895303325
T=5475960,1-2D=1059464225.063372224,1-2V=498805.2719952039145
T=5476020,1-2D=1042322975.492971191,1-2V=502919.2715474033805
T=5476080,1-2D=1025336072.40341982,1-2V=507097.3537378542315
T=5476140,1-2D=1008518847.740764483,1-2V=511337.0257386368485
T=5476200,1-2D=991887768.804055779,1-2V=515635.141511825624
T=5476260,1-2D=975460502.87555315,1-2V=519987.8168954436995
T=5476320,1-2D=959255980.456248399,1-2V=524390.337966845121
T=5476380,1-2D=943294455.598464479,1-2V=528837.0631859683285
T=5476440,1-2D=927597561.478549078,1-2V=533321.3202109447085
T=5476500,1-2D=912188358.96075375,1-2V=537835.2987704444665
T=5476560,1-2D=897091375.4727253625,1-2V=542369.9415820402575
T=5476620,1-2D=882332631.052830448,1-2V=546914.836030647918
T=5476680,1-2D=867939647.9552314515,1-2V=551458.1101653286365
T=5476740,1-2D=853941439.733579603,1-2V=555986.337524742902
T=5476800,1-2D=840368475.301228603,1-2V=560484.4563337365155
T=5476860,1-2D=827252613.128671878,1-2V=564935.7096771743685
T=5476920,1-2D=814627000.542593402,1-2V=569321.6142772057405
T=5476980,1-2D=802525933.1015610725,1-2V=573621.966371284859
T=5477040,1-2D=790984669.315663018,1-2V=577814.8937732030945
T=5477100,1-2D=780039196.629838596,1-2V=581876.9633317254875
T=5477160,1-2D=769725945.6779551135,1-2V=585783.3524949840085
T=5477220,1-2D=760081451.3965974,1-2V=589508.0923553990765
T=5477280,1-2D=751141961.6949159185,1-2V=593024.387226889725
T=5477340,1-2D=742942996.995704328,1-2V=596305.01239282572
T=5477400,1-2D=735518867.0185585275,1-2V=599322.7871633229035
T=5477460,1-2D=728902154.5210790985,1-2V=602051.114943836055
T=5477520,1-2D=723123179.1235852825,1-2V=604464.5759700433865
T=5477580,1-2D=718209457.5209524245,1-2V=606539.552217658005
T=5477640,1-2D=714185178.987149016,1-2V=608254.858418504328
T=5477700,1-2D=711070716.7471742805,1-2V=609592.348863596693
T=5477760,1-2D=708882196.208790332,1-2V=610537.4674891308845
T=5477820,1-2D=707631139.987776514,1-2V=611079.70920976198
T=5477880,1-2D=707324207.0449097115,1-2V=611212.9638887180785
T=5477940,1-2D=707963039.1783741025,1-2V=610935.720636088349
T=5478000,1-2D=709544222.864903068,1-2V=610251.118805712845
T=5478060,1-2D=712059368.459305039,1-2V=609166.8422453304225
T=5478120,1-2D=715495302.5913432985,1-2V=607694.863925355939
T=5478180,1-2D=719834363.8143278005,1-2V=605851.057845719971
T=5478240,1-2D=725054786.6790505495,1-2V=603654.7030468749305
T=5478300,1-2D=731131155.8214382935,1-2V=601127.909878490852

### #13 Janus

Janus

Understanding

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• 407 posts

Posted 10 June 2006 - 01:09 PM

I think Janus is right, though I calculate 10% to much too small a change to cause the Earth’s orbit to intersect the surface of the Sun.

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Note that I said "decrease the Earth's velocity to about 10% of its present value" not " decrease the Earth's velocity by about 10% of its present value".