Starship Asterisk*

iampete wrote:
Can you provide a link [for this 10 h period]? My (admittedly very cursory) search has come up with a lowest value of ~20 hours based on sedimentary rock layering evidence.

Unfortunatedly not. As far as i remember the 10 h period was mentioned within a program in the 'Horizon' series on British television (BBC). Some of these programs are made in collaboration with the station (or license holder) WGBH in the USA. These programs are usually accurate, though sometimes controversial.

The value of 10 hours as mentioned in my post should have had the restriction 'the order of magnitude'. Approximatedly was not the correct formulation, since it suggests a more precise value than order of magnitude. The 10h period dated about 3.5 Gy ago. It was based on a mathematical-theoretical model.

In a similar program on 'snowball earth' the aforementioned period of ≅ 20h was presented, refering to the time about 1 Gy ago. The proof on snowball earth was based on geological data and the value confirms your 20 hours.

iamlucky13 wrote:

orin stepanek wrote:I figured that old Sol may be a 2nd or 3rd generation star; but never thought of it as being formed from a neutron star collision's debris field. I guess it's building blocks had to come from somewhere! Could it be that the sun and solar system may have happened to come into the clutter that the super nova left behind; might explain all the junk orbiting the sun.
Orin

I figured actually debris from multiple supernovae and possible neutron star mergers.

Regarding gold lodes, according to wikipedia, this is owed partially to its generally non-reactive properties of gold. Materials surrounding may be easily oxidized and carried away over the millenia, leaving tiny bits of gold. Because it is so maleable, these bits can cold-weld together, forming flakes and nuggets. There's also reserach suggesting that certain microbes may accellerate this process.

http://en.wikipedia.org/wiki/Gold#Occurrence

I don't think oxidation (or the lack there of) is a significant factor in Au lodes other than the convenience Au is not prone to oxidation allowing it to remain stable and in place, hydrothermal leaching from the parent rock, causing Au to precipitate out when the pressure and temperature are reduced is the geologic explanation. I guess my point is that magma currents and hydrothermal leaching are responsible for the deposition of Au, and is not related to the rotation of the Earth.

Oliver Manuel has great respect for the scientific method, and all his papers prove this. It is the mainstream scientific community that hates him because he is an outsider and shows that mainstream is wrong about some very basic things. The clique of mainstream hates outsiders who show them they are wrong, like Copernicus, Galileo, Birkeland, Alfven, Jurgens and now Manuel.

kovil wrote:Oliver Manuel has great respect for the scientific method, and all his papers prove this. It is the mainstream scientific community that hates him because he is an outsider and shows that mainstream is wrong about some very basic things. The clique of mainstream hates outsiders who show them they are wrong, like Copernicus, Galileo, Birkeland, Alfven, Jurgens and now Manuel.

We're talking about the same Manuel who proposes that the Sun has a continuous hollow shell of iron, just beneath the chromosphere, that's somehow magically levitated and structurally supported (despite the surface temperature of the Sun being above the vaporization temperature of iron) around the plasma and neutron star core?

At the predictable rate of collisions, I don't see how colliding neutron stars could significantly increase the % of heavy metals in the universe. "Nucleosynthesis processes, involving the capture of neutrons or protons, and radioactive decays, happen in situations through out most of the Universe. Slow neutron capture can occur later in the lives of Sun-like stars, before they become white dwarfs. Proton capture is a result of a white dwarf or neutron star cannibalizing gas from a companion star, rapid neutron capture, can take place during the catastrophic stellar collapse which occurs just before the dramatic explosion of a Type II supernova." - A much simpler and elegant explanation for heavy metals that is not dependent on highly "exotic" situations.

henk21cm wrote:

iampete wrote:That would have been one heck of a spin rate.

How fast? Lets try to counterbalance the centrifugal force by the equitorial gravitational force.

ω² R = g ................................ (Eq. 1)

with g the acceleration of gravity. With ω = 2 π/T this leads with (1) to:

T ≅ 2 √ (R)

if we approximate π² by g. T is de axial rotation period of the earth. With R = 6.5E6 m, T ≅ 5100 s: 1h 25m.

A spherical earth rotating this fast (period ~ 1h 25m) would tear itself apart by centrifugal force.
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Long before that point however (period ~ 3h 50m) the earth would elongate into a prolate spheroid [like the large Kuiper belt object "Santa" (period ~ 3h 55m)] and then tear itself apart.

http://en.wikipedia.org/wiki/(136108)_2003_EL61
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Phase locked satellites elongate and tear apart at even slower rotation rates (period ~ 5h 25m) because differential gravity gradients are equally important (to centrifugal forces).

Neptune's Naiad (period = 7 h 4 min)
http://en.wikipedia.org/wiki/Naiad_%28moon%29

Mars Phobos (period = 7 h 39 min)
http://en.wikipedia.org/wiki/Phobos_%28moon%29
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Only very small solid asteroids or moons can rotate faster:
1998 KY26 (period = 10.7 minutes)
............................................................
http://spacewatch.lpl.arizona.edu/1998ky26.html

<<The asteroid 1998 KY26 was discovered by Tom Gehrels on the night of May 28, 1998 (UT) during a routine scanning session. Although appearing as a small dot on the screen like any other asteroid, this near-Earth asteroid has some interesting properties. First, it passed close enough to the Earth for radar observations to be taken. From light curve measurements, its rotation period was calculated to be 10.7 minutes - the fastest rotating asteroid known. In fact, it is the fastest known rotating body in the solar system! >>

neufer wrote: A spherical earth rotating this fast (period ~ 1h 25m) would tear itself apart by centrifugal force.

Why? It is not that i ask this question that i am convinced that such an extreme situation might exist. In my illustration i used the fact that the gravitational pull on the earth (equatorial) surface has the same magnitude as the centrifugal force at the earth (equatorial) surface. This was to place the centrifugal force within a reference frame, which makes it easier to understand.

Under "tearing apart" i envision a situation in which the centrifugal forces are (much) larger than the gravitational forces.

BTW, for Jupiter the centrifugal equatorial force is roughly 0.1 of the equatorial gravitational pull. (Reference: Handbook of Chemistry and Physics, 52nd edition)

neufer wrote: Long before that point however (period ~ 3h 50m) the earth would elongate into a prolate spheroid [like the large Kuiper belt object "Santa" (period ~ 3h 55m)] and then tear itself apart.

The prolate spheroid: i agree. Tearing apart, Art, please explain and convince me with the numbers and formulae.

henk21cm wrote:

neufer wrote:A spherical earth rotating this fast (period ~ 1h 25m) would tear itself apart by centrifugal force.

Why? It is not that i ask this question that i am convinced that such an extreme situation might exist. In my illustration i used the fact that the gravitational pull on the earth (equatorial) surface has the same magnitude as the centrifugal force at the earth (equatorial) surface. This was to place the centrifugal force within a reference frame, which makes it easier to understand.

Under "tearing apart" i envision a situation in which the centrifugal forces are (much) larger than the gravitational forces.

BTW, for Jupiter the centrifugal equatorial force is roughly 0.1 of the equatorial gravitational pull. (Reference: Handbook of Chemistry and Physics, 52nd edition)

While the centrifugal forces have no polar component the gravitational forces have a tremendous polar component that will squash the earth like a pancake lacking counter pressure from the other (radial) dimensions (due to canceling centrifugal forces). The earth is slightly oblate and Jupiter noticeably oblate precisely to get balancing counter pressure from the other (radial) dimensions.

henk21cm wrote:

neufer wrote:Long before that point however (period ~ 3h 50m) the earth would elongate into a prolate spheroid [like the large Kuiper belt object "Santa" (period ~ 3h 55m)] and then tear itself apart.

The prolate spheroid: i agree. Tearing apart, Art, please explain and convince me with the numbers and formulae.

I am using the Roche limit formulae:
http://en.wikipedia.org/wiki/Roche_limit
Using the assumption:

1) that all the densities are all equal
2) that the centrifugal component & gravity gradient components
. tearing the satellites apart are qualitatively & quantitatively equal

Roche's rigid limit: d^3 = 2 x R^3
Roche's fluid limit: d^3 = 14.5268 x R^3

Using Kepler's 3rd law: P^2 = A x d^3

Roche's rigid limit: P^2 = [2 x A x R^3]
Roche's fluid limit: P^2 = 7.2634 x [2 x A x R^3]

Hence, in the case of asteroids:
Fluid limit Period = SQRT(7.2634) X Hank's rigid limit Period of 5100 s. ~ 3h 50m

For actual satellites the Periods are SQRT(2) longer still since the centrifugal force [= C/(P^2)] is augmented by an equal gravity gradient force.

I admit it is a quite lot of hand waving (and some fudging)
but I think that it is basically correct.

G'day Art,

Thanks for your detailed reply. I like that!

neufer wrote: Using the assumption:

1) that all the densities are all equal
2) that the centrifugal component & gravity gradient components
. tearing the satellites apart are qualitatively & quantitatively equal

Roche's rigid limit: d^3 = 2 x R^3
Roche's fluid limit: d^3 = 14.5268 x R^3

Using Kepler's 3rd law: P^2 = A x d^3

Where A = 4π²/(GM) with M the combined mass of the primary and secundary and G the gravitation constant (6.67E-11 N m²/kg²). Assuming the secundary has much less mass than the primary:

A ≅ 4π²/(G x M_Earth)

neufer wrote:Roche's rigid limit: P^2 = [2 x A x R^3]
Roche's fluid limit: P^2 = 7.2634 x [2 x A x R^3]

Then you apply R = R_earth, so:

P² = (2.44)³ [ 4 π² x (R_earth)³ / (G x M_earth)]

Applying the well known Newtonian gravitation law:

g = G x M_earth / (R_earth)² ,

this leaves for the period:

P² = (2.44)³ [ 4 π² x R_earth/g]

or:

P = 2.44 x 2 x π √ (2.44 x R_earth/g)

neufer wrote:
Hence, in the case of asteroids:
Fluid limit Period = SQRT(7.2634) X Hank's rigid limit Period of 5100 s. ~ 3h 50m

Filling in the numbers:
R_earth = 6.37E6 m
g = 9.8 m/s²

P ≅ 19300 s => 5h 20m.

which is a slightly different period.

neufer wrote: I admit it is a quite lot of hand waving (and some fudging) but I think that it is basically correct.

That is not my main point of discussion. Your approach is based on the fact that the unfortunate liquid (slushy) object approaches the planet from deep space. When coming into reach of the planet, the object is then subjected to a differential gravitational field, which has a tendency to elongate the object. The wiki refers to 'spaghettification', which is predicted to occur near the observation horizon of a black hole.

Where your approach can be debated is that for an object (boulder or slushy snowball) on the surface of planet earth there is no reason of elongation of the boulder, since it is supported by the earths crust. The crust is rather stiff. An average piece of rock has a stiffness of the order of 10 MPa, depending on the type of rock. Soft soils, yes, you might be right, their stiffness is much less, of the order of hunderd kPa, so more strain. The approach using Roche's theory is valid for an unsupported object.

From an observational point of view: a boulder on the surface of the earth is well within the La Roche limit. Nevertheless it does not fall apart into tiny grains of sand by differential gravitational forces. Yes, it does fall apart, due to biological activity within natural cracks and crevices and freeze-thaw cycles.

Looking forward for your next reply in this intellectual debate!

henk21cm wrote:G'day Art,
Where your approach can be debated is that for an object (boulder or slushy snowball) on the surface of planet earth there is no reason of elongation of the boulder, since it is supported by the earths crust. The crust is rather stiff. An average piece of rock has a stiffness of the order of 10 MPa, depending on the type of rock. Soft soils, yes, you might be right, their stiffness is much less, of the order of hunderd kPa, so more strain. The approach using Roche's theory is valid for an unsupported object.

From an observational point of view: a boulder on the surface of the earth is well within the La Roche limit. Nevertheless it does not fall apart into tiny grains of sand by differential gravitational forces. Yes, it does fall apart, due to biological activity within natural cracks and crevices and freeze-thaw cycles.

Looking forward for your next reply in this intellectual debate!

In the recent debate about what is a planet and what isn't a planet one of the suggestions was that objects small enough to resist being deformed into a spheroidal shape by their own self gravity should not qualify as planets. The earth is big enough & mushy enough to qualify as a planet under this definition (as are Mercury, Pluto, Ceres and Vesta).

A hundred years ago the debate raging was on whether the earth was big enough & mushy enough to be deformed into an oblate spheroid with an equatorial diameter 1/300th larger that the pole to pole distance (as would indeed be the case for a fluid earth in equilibrium):

http://www.wku.edu/~smithch/wallace/S115-116.htm

The earth was found to be just such an oblate spheroid (as Newton had predicted by the equilibrium fluid model).

For solid objects smaller than 500 km in diameter rigidity does become more and more of an important factor in maintaining both shape and structural integrity; however, for objects larger than 500 km in diameter the fluid approximation assumption generally works pretty well (especially if they are gaseous or have liquid cores/mantles of water or molten rock).

neufer wrote:A hundred years ago the debate raging was on whether the earth was big enough & mushy enough to be deformed into an oblate spheroid with an equatorial diameter 1/300th larger that the pole to pole distance (as would indeed be the case for a fluid earth in equilibrium):

http://www.wku.edu/~smithch/wallace/S115-116.htm

Read it. Indeed, the faster an object is rotating, the more non-spherical it will be. The correct word -as far as the dictionary provides me with intelligible information- is oblate: flattened at the poles. Regarding the discussion between the distinguished gentlemen in 1886: the direction of the plumbline does not point to the center of the earth, due to these centrifugal forces, although the deviation, α, is small. The maximum deviation

tg(α_max) = 0.5 ω² R / g

is found at 45° lattitude. Its value: 0.1°, i.e. 6'.

neufer wrote: however, for objects larger than 500 km in diameter the fluid approximation assumption generally works pretty well (especially if they are gaseous or have liquid cores/mantles of water or molten rock).

Molten, that is correct. Mountains higher than about 10 km on Earth are believed to sink into the earths liquid core, since the stresses and strains in the rock at the base of the mountain are assumed to be so large (product of stress and strain is proportional to dissipated energy) that the rock will melt. The liquid state of the earths interior, well i do not dispute that.

IMO the La Roche approach is not valid for this calculation, since the basic condition of the secundary object being unsupported is flawed. Nevertheless, you have a point that at a rotational period of 1h 50 m the shape of the earth is definitely not the same as nowadays at 23h 56 m. What will happen, when we increase the rotational frequency of the earth, is that the earth will be more and more oblate. The polar radius will decrease, whereas the equatorial radius will increase, in doing so conserving the volume. My origal assumption of R = 6.37E6 m will not hold, it must be larger. Since by an increase in radius, the gravitational forces decreases twice as fast as the centrifugal force increases, the predicted value of 5100 s is too fast.

My main problem is that i can not calculate the degree of oblateness as a function of the rotational period. What i can do is make a rough assumption: the new equatorial radius, R_N, will increase to twice its original value, R_O,: 1.274E7 m. Again counterbalancing the gravitational pull by the centrifugal force at the equator:

ω² R_N = G x M / (R_N)²

with G x M = g x (R_O)²

T ≅ 2 x {(R_N)/(R_O)} x √ (R_N)

where:
R_N is the new equatorial radius of the earth (1.274E7 m), 2 x R_O
R_O is the old equatorial radius of the earth (6.37E6 m)
g is the accelleration of gravity at the equator of the old earth: 9.8 m/s² and approximated bt π²

Filling in the numbers, they result in:

T ≅ 14275 s, i.e. 3h 58m, more than twice as slow as the original value.

Note that this answer is based on the assumption that with a rotation period of 4h the earth is so oblate that it results into an equatorial radius of 12 740 km, in stead of the present 6370 km.

henk21cm wrote:Filling in the numbers, they result in:

T ≅ 14275 s, i.e. 3h 58m, more than twice as slow as the original value.

Note that this answer is based on the assumption that with a rotation period of 4h the earth is so oblate that it results into an equatorial radius of 12 740 km, in stead of the present 6370 km.

Which (although wrong for so many reasons) is pretty close to my prior calculation of 3h 50m:

I am using the Roche limit formulae:
http://en.wikipedia.org/wiki/Roche_limit
Using the assumption:

1) that all the densities are all equal
2) that the centrifugal component & gravity gradient components
. tearing the satellites apart are qualitatively & quantitatively equal

Roche's rigid limit: d^3 = 2 x R^3
Roche's fluid limit: d^3 = 14.5268 x R^3

Using Kepler's 3rd law: P^2 = A x d^3

Roche's rigid limit: P^2 = [2 x A x R^3]
Roche's fluid limit: P^2 = 7.2634 x [2 x A x R^3]

Hence, in the case of asteroids:
Fluid limit Period = SQRT(7.2634) X Hank's rigid limit Period of 5100 s. ~ 3h 50m

But I see now that you are correct in that for a prolate ellipsoid moon the gravitational gradient pulling along the long axis will be MORE THAN(i.e., about twice as effective as) the centrifugal force alone. Hence:

Fluid limit Period = SQRT(14.5268/3) X Henk's rigid limit Period of 5100 s. ~ 3h 7m.

Which explains why the large (1,960×1,518×996 km) Kuiper belt prolate ellipsoid "Santa" (with a rotational period ~ 3h 55m)] is not as close to tearing itself apart as I had previously thought.

http://en.wikipedia.org/wiki/(136108)_2003_EL61

Your theory would have "Santa" being an unstable oblate spheroid.

G'day Art,

neufer wrote: Which (although wrong for so many reasons) is pretty close to my prior calculation of 3h 50m:

I added italics for the words which triggered my special attention. I'm always eager to learn from my mistakes and errors. Would you be so kind to point me what is wrong and for which reasons? Is it the assumptions? Is it the numbers, which i did not add up correctly?

Next you focus the attention on the Kuiperbelt object named Santa, since it is an example of a fast rotating object.

The wiki reports:
M = 4.2E21 kg
r1 = 1.96E6 m
r2 = 1.518E6 m
r3 = 0.996E6 m
g = 0.44 m/s²
T = 14075 s

Unfortunately the wiki does not report around which axis Santa is rotating.

First calculate the accelleration of gravity at the surface of Santa, when it is not rotating. This is application of Newtons law of attraction between two masses.

a1 = GM/(r1)^2 = 6.67E-11 x 4.2E21/(1.96E6)^2 = 0.0729 m/s²
a2 = GM/(r2)^2 = 6.67E-11 x 4.2E21/(1.518E6)^2 = 0.1215 m/s²
a3 = GM/(r3)^2 = 6.67E-11 x 4.2E21/(0.996E6)^2 = 0.2823 m/s²

Neither of these values are equal to 0.44 m/s² as mentioned in the wiki.
Since r3 is about half the size of r1, the accelleration at r3 is four times as large as at r1. The numbers proove it, so not likely an numerical error. Since it does not reproduce the value of 0.44 m/s² in the wiki, there must be an error in my formula or in the data.

Assuming Santa is an ellipsoid with three different axes, its volume is

V = 4 π x r1 x r2 x r3 / 3.

Filling in the numbers: V = 1.241E19 m³ Suppos Sante is made of solid ice, than its mass would be 850 times its volume: 1E22 kg. Nevertheless it is assumed to have a soli d rock core, so it must be havier. With the wiki mass, its density is 340 kg/m³ and that is a lot lighter than water. Other references mention that Santas mass is 32% of Pluto, which produce a value in accordance with the value mentioned in the wiki, e.g. http://www.gps.caltech.edu/~mbrown/2003EL61/

Now calculate the centrifugal accelleration for the three radii, when Santa is rotating:

c1 = ω²r1 = 4 π² x 1.96E6 / 14075² = 0.39 m/s²
c2 = ω²r2 = 4 π² x 1.518E6 / 14075² = 0.30 m/s²
c3 = ω²r3 = 4 π² x 0.996E6 / 14075² = 0.20 m/s²

In this case c3 is half the value of c1, since ci is proportional to the radius. The centrifugal accellerations are considerable compared to the accelleration of gravity at the surface of Santa.

The application of the formula for the surface acceleration of gravity produces 9.8 m/s² for the earth, which is quite natural. So i trust this formula. It is also reported in http://en.wikipedia.org/wiki/Surface_gravity.
In the formula for the calculation of the centrifugal force isn't an error either. It is a textbook result, see e.g. http://en.wikipedia.org/wiki/Centrifugal_force

Combination of ai and ci lead to the following result:

• If Santa is rotating around r3, c1 > a1 and c2 > a2. An object placed at the equator of Santa would launch itself. It should be anchored to stay on the surface.
  If Santa is rotating around r2, c1 > a1 and c3 < a3. At the equator at the most elongated point an object will be launched, at the least elongated point it is stable.
  If Santa is rotating around r1, c2 > a2 and c3 < a3. At the equator at the most elongated point (1518 km) an object will be launched, at the equator at 996 km it is stable.
  Using the value of the wiki (0.44m/s²) an object will remain on the equatorial surface.

So Art, please be so kind and point me where i made an error in these calculations

By the way, it is not my intention to launch a theory of any kind, i just like do some simple numerical excercises to get a grip on the numbers involved.

henk21cm wrote:So Art, please be so kind and point me where i made an error in these calculations

http://en.wikipedia.org/wiki/(136108)_2003_EL61

1) Mean density of Santa: 2.6–3.3 g/cm³ = 3 times that of pure ice.

2) Gravity for a prolate ellipsoid does NOT drop off with the inverse square of the distance from the center of mass.

I was under the impression an Australian scientist in Queensland established several years ago that gold is the excretion of nano bugs and that life at the nano scale comprised the majority of the bio-mass on Earth. 8)