Talk:Hill sphere

Diagram request

I added the image from Lagrangian points article to here aswell, because the plot shows the Hill spheres also. Not sure if a specific diagram would be better. --Petteri Aimonen (talk) 12:06, 29 September 2009 (UTC)

Roche lobe

Most of the sources I can find seem to indicate that the roche lobe and the hill sphere are really the same thing. (the latter being a spherical approximation of the former?). They both have similar definitions and represent the same concept as far as my understanding goes. However the article seems to say that the two must not be confused. Is anyone able to clarify this? 69.157.226.139 16:55, 18 April 2007 (UTC)

Hill sphere of moon

Does anyone know if the Moon can support a stable orbit? A quick calculation gives it a Hill sphere of ~60,000 km with respect to Earth, and ~350,000 km with respect to the Sun, however it's not really a 3-body but a 4-body problem in this case. Deuar 21:53, 9 June 2006 (UTC)

Comments, verification

Article states "It is also called the Roche sphere because the French astronomer Édouard Roche independently described it." My information points to Roche describing it mathematically and the concept merely being borrowed intellectually for the specific purposes intended by Hill, in which event it may indeed be called a Hill sphere or a Roche sphere, but it was not "independently described" by Roche - it was more a "child" of his work. Please let's discuss!

24.215.19.163 11:22, 18 December 2006 (UTC)

Hill sphere and Distance a (part 1)

In using the Semi-Major Axis in the formula, a literal interpretation would mean distance to the barycenter (shared by the two bodies), not the distance between the actual bodies themselves. Does the author mean the distance to the barycenter (which could be many km outside the larger body)? Or, does the author mean the distance between the two bodies (center to center)? I assume that it is the latter (since the barycenter is the result of the shared orbits). Tesseract501 10:48, 24 February 2007 (UTC)

Well I'm not the author, but I would surmise that the shortest distance between the large and smaller body is most critical. The moment of closest approach is the time when the third orbiting body is most perturbed. Of course since the large body in question is usually the Sun, the difference is insignificant anyway. Especially since we have a ≈ in the formula, indicating that the true practical distance at which bodies orbiting the smaller mass may begin to be lost differs by many percent with respect to the value obtained from the formula. So, the details of whether it's distance to the barycenter or what is lost anyway in general slop and uncertainty. Deuar 14:58, 26 February 2007 (UTC)

Hill sphere and Distance a (part 2)

Another queston: The formula in this article indicates the value for "a" as the Semi-major Axis. I've seen other sites use the distance of closest approach (Periapsis) in the assumption that the Hill Sphere size changes based on the distance of the two objects. Which is correct? The formula on this page incorporates the eccentricity of the orbit. Does that support using the SMA? What if the eccentricity is high and the smaller, comet-like object is much closer at Periapsis than at Apapsis? If the Hill Sphere does, in fact, shrink at Periapsis, using the SMA in the formula may show a Hill Sphere that is too large to accommodate stable orbits at closest approach. Tesseract501 10:52, 24 February 2007 (UTC)

The formula with the eccentricity is the correct one. Since the loss process involves an acumulation of small kicks at each orbit, it is useless to talk about the Hill sphere changing with location in the orbit. If the body is too far out at any time at all, it will eventually be lost after making enough orbits. This is why the moment when perturbations are strongest (perihelion) is the relevant time at which to calculate the Hill radius. Deuar 14:58, 26 February 2007 (UTC)

Hill sphere and Distance a (part 3)

One last question, regarding the reference to the moon having to be LESS THAN 7 months within the Hill Sphere. That sentence seems to pop out with no explanation as to what or why (then again, it may be ignorance). Does the sentence tie in to the use of SMA (instead of using the distance of closest approach)? To me the sentence implies a safety period for a stable orbit that is temporarily outside of the Hill Sphere? If necessary, may someone clarify the meaning of the 7-month statement? Tesseract501 10:58, 24 February 2007 (UTC)

This seems to be just someone's way to phrase the fact that all stable orbits around the Earth must have an orbital period of 7 months or less. It does not refer to orbits around any other object. Should be rephrased since it's evidently causing confusion. Deuar 14:58, 26 February 2007 (UTC)

Finally getting back to you. Thanks for all the feedback. You (Deuar) are one of the most helpful folks in Wikipedia-space ... in my fuzzy opinion at least. Your input on talk pages is consistently clear, respectuful, and to the point. I fail that test, especially the last part. I had great fun with the Hill Sphere logic, after I got my head around it. That, along with the logic for orbital mechanics, L-points, and a bunch of other suprise rules that crepted (or screamed) up ... I've designed (in data tables that is) a funny little system made up of three planets (well four, since one is its own sub-binary), and a mass of shared and unshared satellites. 32 co-orbiting bodies together (not to mention the system being part of a tri-star binary system as well. The whole thing in a highly improbable, but with equal parts of imaginative and insanity. All the 32 bodies ultimately share the same principle barycenter. So, its like rush-hour traffic, along 3 axes. In theory at least, the whole dance should hold together. I guess I have too much time on my hands! Wish I could put it in software and see it in action. I did it all in boring data-tables. Tesseract501 (talk) 02:02, 9 March 2009 (UTC)

I too found this remark confusing as it seems to come out of nowhere with no explanation. I've tagged it "vague" for now but would appreciate an explanation or removal for clarity. MikeEagling (talk) 10:55, 28 March 2010 (UTC)

Derivation

In the section on derivation, the derived formula has a "3" missing, when compared with the formula quoted at the start of the article (and in many other sources). I can't see how the derivation argument can be modified to produce the "3", so am doubting the validity of the derivation. There at least needs to be some explanation of why the "3" is missing.

87.127.140.166 (talk) 18:50, 14 April 2010 (UTC)

SoI (astrodynamics)

On the Talk:Sphere of influence (astrodynamics) page back in April of 2009, an editor raised the question about the similarity between the Hill sphere and the SoI (astrodynamics) pages. Are these candidates for a merge?
 —  Paine (Ellsworth's Climax)  17:00, 19 June 2010 (UTC)

No. After looking more closely at the two articles, I see differences between the Hill sphere and the SoI, so I no longer see the need to merge the two articles. There does seem to be an anomaly, though. (more to come)
 —  Paine (Ellsworth's Climax)  22:52, 25 June 2010 (UTC)

---

The Hill sphere for Earth, as given in this article, has radius of 1.5 million km, which is about 932,000 miles. This article also states that it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius. The other article on the SoI gives a radius value of 925,000 km, which is about 575,000 miles, or about 62% of the radius of the Hill sphere. So:

• Hill radius = 1,500,000 km or 932,000 miles
• SoI radius = 925,000 km or 575,000 miles

Now comes the truly interesting part. Editors please note the derivation from the SoI article:

The general equation describing the radius of the sphere $r_{SOI}$ of a planet:

$r_{SOI}=a_{p}({\frac {m_{p}}{m_{s}}})^{2/5}$

where

$a_{p}$ is the semimajor axis of the planet's orbit relative to the largest body in the system, usually the Sun

$m_{p}$ and $m_{s}$ are the masses of the planet and Sun, respectively.

And now, please note the derivation from this, the Hill sphere article:

$R_{H}=a\left({\frac {M_{\mathrm {planet} }}{M_{\star }}}\right)^{1/3}$

Now, I'm no math wizard, yet it seems to me that these formulas tell us that the SoI radius is greater than the Hill radius for any given mass relationship. Am I missing something? Isn't the SoI supposed to be smaller than the Hill sphere?
 —  Paine (Ellsworth's Climax)  04:52, 5 July 2010 (UTC)

Guys the reason why the two are so similar is because they're identical. The SOI is just another name for the Hill Sphere or even the Laplace Sphere. And the article actually derives the Laplace Sphere NOT the Hill Sphere, which is a 0.4 power of the ratio of the masses, not the 1/3 power referenced. I just need to track down an actual proper reference before I can do a proper edit.
 —Qraal (talk) 11:41, 2 September 2010 (UTC)

Why isn't Roche mentioned in the first para?

For Pete's sake! A common alternate name is usually incorporated into the first paragraph, in brackets, along the lines of (also known as a Roche Sphere). What's going on here? The first mention of Roche isn't until the middle of para 2. It feels as if a particularly pedantic editor has been at work. Blitterbug (talk) 07:41, 2 April 2011 (UTC)

Perhaps it's because the necessary parenthesis "not to be confused with the Roche limit" seems too 'heavy' for the lede. —Tamfang (talk) 00:28, 3 April 2011 (UTC)

Tidal forces

Why "tidal forces" are mentioned in the 2nd paragraph? May be I'm wrong, but IMHO tidal forces don't influence at all the described process. These are pure, simple gravitational forces, that rule the process, as long as we consider "a[ny] third object", regardless of its size.

Oh.... it seems that I've realized. The article considers tidal forces influencing not the third object alone, but the system "second object - third object". That's correct. — Preceding unsigned comment added by 213.87.192.62 (talk) 13:13, 12 September 2011 (UTC)

Further Examples confusion

"An astronaut could not orbit the Space Shuttle (with mass of 104 tonnes), where the orbit is 300 km above the Earth, since the Hill sphere of the shuttle is only 120 cm in radius, much smaller than the shuttle itself. In fact, in any low Earth orbit, a spherical body must be 800 times denser than lead in order to fit inside its own Hill sphere, or else it will be incapable of supporting an orbit. A spherical geostationary satellite would need to be more than 5 times denser than lead to support satellites of its own; such a satellite would be 2.5 times denser than osmium, the densest naturally-occurring material on Earth. Only at twice the geostationary distance could a lead sphere possibly support its own satellite; since the moon is more than three times further than the 3-fold geostationary distance necessary, lunar orbits are possible."

The space shuttle example makes sense: (radius of earth + 300 km) * ((104000 kg)/(3 * mass of earth))^(1/3) = 1.2 meters (paste the LHS into google to evaluate). But I don't buy the latter examples, perhaps I'm doing something wrong? e.g. a 10cm radius sphere merely twice as dense as lead in LEO: (radius of earth + 300 km) * ((4/3 * pi * (10 cm)^3 * 22 g/cm^3)/(3 * mass of earth))^(1/3) = 11.5 cm, outside its own radius. Or a lead sphere in GEO, with an Hill sphere around 50cm radius.

Solar Hill sphere?

• I am wondering what size is the solar hill sphere to the Milky Way? Reddwarf2956 (talk) 15:37, 22 April 2012 (UTC)
  • The page for astronomical unit lists it as 230,000 AU. Nikki (talk) 23:27, 5 June 2012 (UTC)

I calculated it myself as ~266,000 AU using 26,000 ly as the Solar mean orbital radius and 237.5 Ma (mean value of 225-250 Ma) as the Solar orbital period:

• c = 299,792,458 m/s
• G = 6.67384 × 10⁻¹¹ m³/kg/s²
• 1 d = 86,400 s
• 1 a = 365.25 a
• 1 ly = c(1 a)
• m_⊙ = 1.98892 × 10³⁰ kg
• r_μ⊙ = 26,000 ly
• P_⊙ = ½(225 + 250) Ma ≈ 238 Ma
• v_μ⊙ = 2πr_μ⊙/P_⊙ = √(GM²/(r_μ⊙[M + m_⊙])) ≈ 206 km/s
• M = 2π(2πr_μ⊙³ + √(Gm_⊙P_⊙²r_μ⊙³ + π²r_μ⊙⁶))/(GP_⊙²) ≈ 7.88 × 10¹⁰ m_⊙
• r_H⊙ = r_μ⊙∛(m_⊙/(3M)) ≈ 3.978 × 10¹⁶ m ≈ 265,900 AU ≈ 4.205 ly

This is very close to the orbit of Proxima Centauri (Alpha Centauri C at 4.243 ly from Sol) around Alpha Centauri AB (at 4.366 ly from Sol). More likely the practical Hill radius for Sol corresponds the maximum extent of the Oort Cloud (at ~100,000 AU or ~1.58 ly).

Using instead an instantaneous orbital speed of 251 km/s and a corresponding orbital radius of 27,000 ly yields a central mass equivalent of 1.21 × 10¹¹ m_⊙ and a more reasonable Solar Hill radius of ~3.78 ly (~239,000 AU), which is likely the rounded value given on the astronomical unit page.

Nikki (talk) 02:00, 6 June 2012 (UTC)