l33telboi wrote:Mr. Oragahn wrote:It's why I said it's in between. There's obviously work that needs to be done, but we're looking at an explosion on the surface of a rubble pile asteroid in zero g.
What you did was divide the figure with two - and that was it. If you want to say it's in any way accurate, then show me a source that validates that methodology.
What the initial figure is about is the energy necessary to shatter the entirety of an asteroid of x meters. It comes as logical that removing half of the matter would make the figure lower for the energy required to move the mass. Now, we're talking about a crater, a part of the volume of the crater will not escape into space, but be pressed into the asteroid. The rubble pile model is that the structure is fragile and brittle. A shockwave would shatter it, but the pressure would pancake the material. With further reading about the composition and behaviour of porous asteroids, it appeared that craters also form with considerable pressure, increasing the density of matter around "ground zero" by getting rid of a large quantity of the porosity, without involving melting. Ejecta also occurs.
It's a very crude figure, but I never settled on it as a firm value, just to get a quick rough idea of where we were heading for.
Also, I noticed that it's often forgotten that torpedoes can slam into their naturally composed targets, and detonate inside. Their speed and momentum alone would allow them to smash at great speeds into these asteroids.
No such thing is mentioned. Neither do I believe a photonic torp could burrow into an asteroid. Perhaps a centimeter or two, but that's just about it.
Later torpedoes, like in TNG, have been seen entering a dying red star and an episode did involving torpedoes slamming into the crust and somehow going deeper. It was an episode with a planet having another structural issue.
Also, the degree of penetration depends on the variety of the trigger. The Cardassian moon destroyer was said to use a kinetic trigger, I think considered something rather crude, so torpedoes possibly come with high quality proxy fuses.
The presence of a strong casing and shields largely supports the idea that torpedoes could penetrate their target unless stopped by particular alloys or defensive fields.
Photonic torps don't have shields. Not even the ships during this era have shields.
The Klingon's ships have shields, and the Terrans already had hull polarization. They were handed torpedo technology iirc.
Nevertheless, the torpedoes glow, so there's obviously a strong field of some kind and I see no reason to consider them unshielded.
Most formulas I've seen involving detonating a nuclear explosion on the surface of a typical rock/dust soil.
The formula I'm talking about let's you adjust what type of composition the asteroid is made of, and it assumes zero gee. You can't find a better suited formula because there quite simply is none. It fits every single parameter we need it to fit.
Because it seems to have all necessary parameters does not mean it's correct, nor the only one that would work.
I actually retrived the original thread wherein the formula was first presented.
http://forums.spacebattles.com/showthre ... 610&page=2
I happened to post in it for a brief time, but either was too focused on Star Wars and missed the Star Trek part, or didn't pay much attetion to that thread at that time beyond my quick jump (seems to be the case).
Let me quote the exchange between you and higbvuyb:
l33telboi wrote:Oh yes. Always. Isynic dug up a formula for how big a carter would be created by a given yield bomb back in the “NX-01 vs. Whitestar†thread. Scaling laws for surface explosions give a value of 5x10^-3 meters/J(1/3), and in this case the diameter is 3000m, so dividing the diameter with this value we get 600,000J^(1/3) which is 2.16e17J, or 51 megatons.
higbvuyb wrote:From my research, from modelling and empirical evidence, crater depth varies as the one-quarter(th?) power as yield goes above 1 megaton due to significant gravitational effects.
l33telboi wrote:Could you elaborate a little? Assuming we're talking about 600,000J^(1/4), it would mean the yield we increase even further. A little too high.
higbvuyb wrote:http://glasstone.blogspot.com/
This is a very comprehensive resource for nuclear weapons in general.
Other revisions have been discussed in a variety of blog posts, particularly
http://glasstone.blogspot.com/2006/0...last-wave.html which explains that the entire cratering model used in Glasstone and Dolan is obsolete for high yields, because it ignores the conservation of energy. This was only discovered in 1987. The yield-dependent scaling for crater dimensions (radius and depth) transitions from the cube-root of yield scaling at low yields (below about 1 kt or so)
to fourth-root at high yields (above 10 Mt or so) because of gravity. At low yields, the fraction of cratering energy used to physically dump ejecta out of the crater against gravity (to produce the surrounding lip and debris) is trivial compared to the amount of energy used to physically break up the soil. This hydrodynamic effect means that at very low yields, crater dimensions scale (like blast overpressure ranges) with the cube-root of yield. But at higher yields than about 10 kt, the deeper crater produced means that a relatively larger fraction of the energy of cratering is employed to do work against gravity. The work energy you need to do is simply E = mgh where E is the gravitational potential energy needed to be overcome by work, m is the crater mass, g is the acceleration due to gravity (9.8 ms^-2) and h is the average vertical height that the material is moved from the crater to the height of the lip and ejecta layer surrounding the hole (this height is approximately half the depth of the crater). Since m is equal to the density of the cratered soil multiplied by the crater volume (which is approximately proportional to the cube of the depth of the crater, if the crater depth to radius ratio is approximately independent of yield), it follows that the energy used overcoming gravity in cratering is proportional to the fourth power of the crater depth. Hence, the fraction of energy used in overcoming gravity gets very big for deep craters from high yield nuclear weapons, and the crater depth (and radius) scale approximately as the one-fourth power (i.e. the fourth-root) of the weapon yield, for very high yields.
Inbetween about 1 kt and 100 kt there is a transition from cube-root to fourth-root scaling, where the average scaling law is roughly the 0.3 power of yield, and this is what Glasstone and Dolan as well as DNA-EM-1 (up to 1984) used. However, in 1987 it was realised that this empirical approximation was in error (the biggest Nevada cratering test was Sedan, 104 kt) because it ignored the reduced scaling of crater dimensions at high yields, where the immense problem of doing work against gravity kicks in and curtails the purely hydrodynamic cratering physics.
For example, in the case of a 10 Mt surface burst on dry soil, the 1957, 1962, and 1964 editions of Glasstone's Effects of Nuclear Weapons predicted a crater radius of 414 metres (the 10 Mt Mike test in 1952 had a radius of over twice that size, but that was due to the water-saturated porous coral of the island and surrounding reef, which is crushed very easily by the shock wave at high overpressures). This was reduced to 295 metres in Glasstone and Dolan, 1977, when the scaling law was changed from the cube-root to the 0.3 power of yield. The 1981 revision of Dolan's DNA-EM-1 brings it down to 145 metres, because of the tiny amount of energy which goes into the bomb case shock for a modern, efficient 10 Mt class thermonuclear warhead (Brode and Bjork discovered this bomb design effect on cratering in 1960; high-yield efficient weapons release over 80% of their yield as X-rays which are inefficient at cratering because they just cause ablation of the soil below the bomb, creating a shock wave and some compression, but far less cratering action than the dense bomb case shock wave produces in soil). Then in 1987, the introduction of gravity effects reduced the crater radius for a 10 Mt surface burst on dry soil to just 92 metres, only 22% of the figure believed up to 1964.
As you can see, its actually 10 MT, not 1 MT which is what I said before.
l33telboi wrote:...That would bump the yield of the photonic torpedoes to insane levels. Still, interesting research. I guess the 51 megaton result would be a super extreme lower limit.
Oh wait. Gravity. Actually, wouldn't the figure be accurate considering there's no gravity in space? There'd be no need to account for it.
higbvuyb wrote:If it was in space, then, yes, your calculation works.
I beg to disagree with both hig's conclusion and your claim that the equation has all that's needed for space detonations.
I did read the blog in question.
The post links to another page, posted by the author much earlier on, and highlighting actually an element which the equation you refer to doesn't take into account.
http://glasstone.blogspot.com/2006/03/s ... dolan.html
(My comments in orange.)
It’s fascinating that, despite the best scientific brains working on nuclear weapons effects for many decades - the Manhattan Project focussed a large amount of effort on the problem, and utilised the top physicists who had developed quantum mechanics and nuclear physics, and people like Bethe were still writing secret papers on fireball effects into the 1960s - such fundamental physical effects were simply ignored for decades. This was due to the restricted number of people working on the problem due to secrecy, and maybe some kind of ‘groupthink’ (psychological peer-pressure): not to upset colleagues by ‘rocking the boat’ with too much freethinking, radical questions, innovative ideas.
Well, as far as nuclear explosions were concerned, perhaps, but more refined equations for impacts against asteroids and rocks already took care of local gravity, as in Öpik (1969), Gault's (1974), and Dence's (1977) and Croft (1977), as described in the following document: Comparison of Six Crater-Scaling Laws ,SAO/NASA Astrophysics Data System (ADS). It will be used later on.
The equation E = mgh isn't a speculative theory requiring nuclear tests to confirm it, it's a basic physical fact that can be experimentally proved in any physics laboratory: you can easily measure the energy needed to raise a mass (the amount of electric energy supplied to an electric motor while it winches up a standard 1 kg mass is a simple example of the kind of physical fact involved). In trying to analyse the effects of nuclear weapons, false approximations were sometimes used, which then became imbedded as a doctrine or faith about the ‘correct’ way to approach or analyze a particular problem. People, when questioned about a fundamental belief in such analysis, then are tempted respond dogmatically by simply referring to what the ‘consensus’ is, as if accepted dogmatic religious-style authority is somehow a substitute science, which is of course the unceasing need to keep asking probing questions, checking factual details for errors, omissions and misunderstandings, and forever searching for a deeper understanding of nature.
For example, in the case of a 10 Mt surface burst on dry soil, the 1957, 1962, and 1964 editions of Glasstone's Effects of Nuclear Weapons predicted a crater radius of 414 metres (the 10 Mt Mike test in 1952 had a radius of over twice that size, but that was due to the water-saturated porous coral of the island and surrounding reef, which is crushed very easily by the shock wave at high overpressures). This was reduced to 295 metres in Glasstone and Dolan, 1977, when the scaling law was changed from the cube-root to the 0.3 power of yield. The 1981 revision of Dolan's DNA-EM-1 brings it down to 145 metres, because of the tiny amount of energy which goes into the bomb case shock for a modern, efficient 10 Mt class thermonuclear warhead (Brode and Bjork discovered this bomb design effect on cratering in 1960; high-yield efficient weapons release over 80% of their yield as X-rays which are inefficient at cratering because they just cause ablation of the soil below the bomb, creating a shock wave and some compression, but far less cratering action than the dense bomb case shock wave produces in soil). Then in 1987, the introduction of gravity effects reduced the crater radius for a 10 Mt surface burst on dry soil to just 92 metres, only 22% of the figure believed up to 1964!
‘It is shown that the primary cause of cratering for such an explosion is not “airslap,†as previously suggested, but rather the direct action of the energetic bomb vapors. High-yield surface bursts are therefore less effective in cratering by that portion of the energy that escapes as radiation in the earliest phases of the explosion.
[Hence the immense crater size from the 10 Mt liquid-deuterium Mike test in 1952 with its massive 82 ton steel casing shock is irrelevant to compact modern warheads which have lighter casings and are more efficient and produce smaller case shocks and thus smaller craters.]’ - H. L. Brode and R. L. Bjork, Cratering from a Megaton Surface Burst, RAND Corp., RM-2600, 1960.
The equation presented on the page ignores the impact of the mass of the weapon's casing, for the reason provided above. This won't affect our results much for torps, because I doubt they even weight 1 tonne, which wouldn't make much of a difference. But it will matter absolutely if you consider heavier devices, like for what our Cardassian missile-ship would do as it slams into a moon, for which impact related formulas would actually be just as good, if not better suited. For the same yield, it may leave a much greater crater, due to the material mass, than if we simply considered its yield --we assume the design is not stupid up to the point of soaking too much of the energy and partially neutering the device itself!
It's also very a interesting source of information, for any event that involves a heavily metal clad bomb or plasma-fusion reactor blowing up from inside a craft or station near the ground or on the ground.
Besides, the Glasstone-blog equation doesn't incorporate the momentum of the device itself, in that there would be a difference between detonating a bomb placed on the ground, and detonating a bomb hitting the ground at high speed. The author of the blog himself, considering its subjets, may have wished to consider than his discussion on the shortcoming of using nuclear ordinance against surface and underground targets, the devices could well be delivered by MIRVs and dropped on target as part of a whole propulsion capable system.
But these parts aside, we should clearly deal with the meat and potato of the question, and why I disagree with you.
The fourth-root part of the equation is only relevant so far as gravity matters much.
But you'll easily find a good number of large asteroids with low bulk densities and gravities being low multiples of e-2 or e-3.
I see that the blog's author, Nigel, deemed the former equation false because it didn't consider gravity for yields greater than 1 Kt.
While it is right, as long as the formula was used for nuclear or high explosisves below or around 1 Kt, at a gravity of 9.8 m/s², the predictions worked (although the formulas was akwardly thought to work even for yields such as 100 Kt).
The old formula was precisely empirical in origin, based on the observations of completed tests
on Earth, and was created to predict the logged results in a table and for curves for sub or very low kiloton explosives. You would not need to add an addendum based on height, mass and gravity, when the first part would already assume the existence of an 9.8 gravity, because the formula was created to predict figures based on the values obtained from the existing reliable but very few tests.
Gravity is assumed, but not visible in the old formula. One of the main reason is that, first, several other parameters don't show up either, and secondly, it's a purely empirical short formula derived from a table of numberse. It's really meant to be that simple, really, but it does not mean gravity wouldn't matter at all.
It's still there and works for low yields. It is not enough anymore for greater yields. Hence the addendum, if you want to use that equation for greater yields used
on Earth.
I cannot emphasize that part enough, since the blog's author updated the formula solely for nuclear blasts on Earth and nowhere else.
You could say that the second part of his equation, mgh, solved for potential energy, would compensate for the initial equations's shortcomings at yields above 1 Kt, but still in a 1g gravity.
Indeed, if you use the new equation in such a context of 9.8 m/s², it's good enough, and that was the point of the author's post.
Still, I don't see any introduction of atmospheric pressure either in the second term, and yet there's an entire column of air to deal with above the soon-to-be crater's volume.
But notice that in the case of this equation, g is a constant:
It is not meant to be used for a gravity other than Earth's.
This is also my problem with hig's conclusion, because the old equation was empirical, and based on Terran experiments.
To me, it's painfully obvious that a 1 kiloton explosion test would have produced entirely different effects if it had happened on the surface of an asteroid, and we're not even talking about porosity yet!
That's why the first term of the equation, the old part, can not be used for an explosions elsewhere than on Earth's surface.
I also noticed that you thought that simply because the author mentionned higher yields would require the change from the cube-power to fourth-power, this applied to the whole formula, but this is not exact. It only applies to the second term, the first one remaining unchanged.
The (mX)-term is proportional to the cube of the crater depth (because m is the product of volume and density, and volume is proportional to depth-cubed if the crater radius/depth ratio is constant), while the (mgh)-term is proportional to the fourth-power of the crater depth because m is proportional to the density times the depth cubed (if the depth/radius ratio is constant) and h is always directly proportional to the crater depth (h is roughly half the crater depth), so the product mgh is proportional to the product of depth cubed and depth, i.e., to the fourth-power of crater depth. So for bigger craters and bigger bomb yields, a larger fraction of the total cratering energy then gets used to overcome gravity, causing the gravity term to predominate and the crater size to scale at most by W1/4 at high yields. This makes the crater size scaling law transition from cube-root (W1/3) at low yields to fourth-root (W1/4) at higher yields!
Only [mgh] is concerned by this. Why the link with higher yields then?
When you use greater yields, you use this equation, with the added term, and it's this added term that involves the fourth-power.
It's rather clear what both terms correspond to:
(My comments in orange.)
Early theoretical studies of crater formation, even using powerful computer simulations, employed explosion dynamics that ignored gravitation. Almost all of the books on the ‘effects of nuclear weapons’ in the public domain give nonsense for megaton surface bursts. It was only in 1986 that a full study of the effects of gravity in reducing crater sizes in the megaton range was performed: R. M. Schmidt, K. A. Holsapple, and K. R. Housen, ‘Gravity effects in cratering’, U.S. Department of Defense, Defense Nuclear Agency, report DNA-TR-86-182. In addition to secrecy issues on the details, the complexity of the unclassified portions of the new scaling procedures in this official treatment cover up the mechanisms, so here is a simple analytical explanation which is clearer:
If the energy used in cratering is E, the cratered mass M, and the explosive energy needed to physically break up a unit mass of the soil under consideration is X, then the old equation E = MX (which implies that crater volume is directly proportional to bomb yield and hence crater depth and diameter scale as the cube-root of yield) is completely false, as it omits gravitational work energy needed to shift soil from the crater to the surrounding ground.
E = MX is the "old" equation, which still finds its way into the new equation. A very close version of the old equation is found in 1977 The Effects of Nuclear Weapons, 3rd ed. (36.8 Mb), in Characteristics of Surface and Shallow Undeground Bursts, 6.09 (p.235) and 6.72 (p.253-255). It is abundantly clear that the formula was simple, and solely made to fit the observed results with minimal parameters. It was presented as a rough method, and totally extrapolated for yields higher than 1 Kt.
The following is necessary to understand the context of the equation.
This gravitational work energy is easy to estimate as ½ MgD, where M is the mass excavated, g is gravitational acceleration (9.8 m/s& ), D is crater depth, and ½ is a rough approximation of the average proportionof the crater depth which displaced soil is vertically moved against gravity in forming the crater.
Hence the correct cratering energy not E = MX but rather E = MX + ½MgD. For yields well below 1-kt, the second term (on the right hand side) of this expression, ½ MgD, is insignificant compared to MX, so the volume excavated scales directly with yield, and since the volume is proportional to the cube of the average linear dimension, this means that the radius and depth both scale with the cube-root of yield for low yields.
But for very large yields, the second term, ½MgD, becomes more important, and this use of energy to overcome gravity in excavation limits the energy available for explosive digging, so the linear dimensions then scale as only the fourth-root (or quarter-power) of yield. Surface burst craters are paraboloid in shape, so they have a volume of: p R2 D/2 = (p /2)(R/D)2 D3, where the ratio of R/D is about 1.88 for a surface burst on dry soil. The mass of crater material is this volume multiplied by the density, r , of the soil material: M = rp(R/D)2 D3 /2.
Hence, the total cratering energy is: E = MX + ½ MgD = r (p /2)R2 D(X + ½gD).
The density of hard rock, soft rock and hard soil (for example granite, sandstone or basalt) is typically 2.65 kg/litre (2,650 kg per cubic metre), wet soil is around 2.10 kg/litre, water saturated coral reef is 2.02 kg/litre, typical dry soil is 1.70 kg/litre, Nevada desert is 1.60 kg/litre, lunar soil is 1.50 kg/litre (for analysis of the craters on the moon, where gravity is 6 times smaller than at the earth’s surface), and ice is 0.93 kg/litre.
The change over from cube-root to quarter-root scaling with increasing yield means that old crater size estimates (for example, those in the well-known 1977 book by Glasstone and Dolan, U.S. Department of Defence, 1977, The Effects of Nuclear Weapons) are far too big in the megaton range, and need to be multiplied by a correction factor.
The correction factor is easy to find. The purely explosive cratering energy efficiency, f, falls as gravity takes more energy, and is simply f = MX/(MX + ½MgD) = (1 + ½gD/X)-1.
Because gravity effects are small in the low and sub kiloton range, the correct crater radius for small explosions indeed scales hydrodynamically, as R ~ E1/3, so the 1-kt crater sizes in Glasstone and Dolan should be scaled by the correct factor R ~ W1/3(1 + ½ gD/X)-1/3 instead of by the empirical factor of R ~ W0.3 given by Glasstone and Dolan for Nevada explosion data of 1-100 kt. Glasstone and Dolan overestimates crater sizes by a large factor for megaton yield bursts. (The Americans had been mislead by data from coral craters, since coral is porous and is simply crushed to sand by the shock wave, instead of being excavated explosively like other media.
In megaton surface bursts on wet soft rock, the depth D increases only as W1/4, the ‘fourth root’ or ‘one-quarter power’ of yield scaling. Obviously for small craters, D scales as the cube-root of yield, but the correction factor (1 + ½ gD/X)-1/3 is only significant for the megaton range anyway, so a good approximation is to put D in this correction as proportional to the fourth-root of yield in this correction factor formula. The value of X for any soil material is a constant which may be easily calculated from the published crater sizes for a 1 kt surface burst, where gravity is not of importance (X is the cratered mass divided by the energy used in cratering, the latter being determined by an energy balance for the explosion effects).
The crater is made by two processes: the shock wave pulverisation of the soil (the energy required to do this is approximately proportional to the mass of soil pulverised) and the upward recoil of pulverised soil in reaction (by Newton’s 3rd law) to the downward push of the explosion (the energy required to do this excavation depends on gravitation, since it takes energy MgD to raise mass M a distance D upward against gravity acceleration g).
Have you even looked at what you'd obtain with the second term of the equation alone?
It works with crater depth, and considers the mass of matter lifted upwards against gravity.
So we have to make some assumptions here about what the radius would be. In some crater cases, matter falls back into the hole. Therefore the crater shape would be more of a flat lens, and thus the ratio depth:radius would not be 1:1. This is precisely addressed by the author ("the ratio of Radius/Depth is about 1.88"). Which means he works from a final depth perspective.
But to obtain a bigger number and assume all the mass is ejected into space, I'll use an hemisphere.
We'll leave aside the problem of using the second term alone, since then it doesn't take into account the fact that to have a crater in an environment where gravity is strictly superior to 0, and get something like an escape velocity, you have to blow matter sideways and not just lift it up, since a positive gravity shall bring matter back into its original spot.
Now let's go beyond the use the author made of this term, and apply it in a case of low gravity.
If we take it in a simplistic way, where depth equals radius, we're therefore dealing with a 1500 meters radius hemisphere.
For the mass M, let's take a density of 2.2 g/cm³, higher than a conservative value for porous structures. It's also slightly superior to wet soil (2.1 g/cc).
As for surface gravity, we'll pick e-2 m/s² instead of the inferior and also common e-3 value.
V = 4/3 * pi * R³ = 7.068 e9 m³
M = 7.068 e9 * 2.2 e3 = 1.55496 e13 kg.
E = mgh = 1.55496 e13 * e-2 * 1.5 e3 = 233.244 e 12 J, or
55.75 kilotons.
That's for the term of the equation which is supposed to dictate the final result, and yet was obtained from non conservative parameters.
It obviously adds little to nothing to what you think is a by default correct lower end value (51 MT).
See, that's why I think the other document I provided, which in the end, argued in favor of a yield around the 1 digit megaton range, more or less, was closer to reality.
Now, if we use this term according to the Terran parameters it was meant to be used with, we multiply the result by 980, and we get
54.635 megatons.
We realize that if we add the 51 megatons obtained from the first term of the updated equation, and the 54 megatons from the second term, we get
109 megatons.
This will be important for the next point: proving my intuition is correct.
Let's look at several asteroids first.
http://www.solarviews.com/eng/asteroid.htm
http://en.wikipedia.org/wiki/433_Eros
http://en.wikipedia.org/wiki/45_Eugenia
http://en.wikipedia.org/wiki/90_Antiope
http://en.wikipedia.org/wiki/253_Mathilde
http://en.wikipedia.org/wiki/951_Gaspra
http://en.wikipedia.org/wiki/2_Pallas
Now, let's use the crater scaling formula from Dence et al. (1977), extended with term (gE/g)^3/16 for comparative measures (as defined in the document provider earlier, p.1):
D = 1.71 e-2 * E^0.294 * (gE/g)^3/16 , when D > 2.4 10^5 cm. D stands for the crater's diameter.
An equation that is not much different than the one quickly used for small yield HE/nuclear detonation below or at 1 kT, but which includes gravity.
We can notice that if we use the equation for an impact on Earth, the term "(gE/g)^3/16" would return 1 and thus disappear from the equation.
It would appear that this equation would be rather reliable and more than close enough, even if primarily designed for impact with solid objects. That said, at a certain level of energy, the effects are similar.
So we can try to obtain an energy figure from the equation, for a low gravity asteroid.
Let's use this formula with
asteroid 253 Mathilde:
Code: Select all
Dimensions 52.8 km (66×48×46 km)
Mass 1.033(±0.044) e17 kg
Mean density 1.3 g/cm³
Equatorial surface gravity 0.0025 m/s²
Therefore, energy is obtained:
E^0.294 = D / 1.71 * e2 / (gE/g)^3/16
E^0.294 = D / 1.71 * e2 / (9.8/0.0025)^0.1875
E = [3000 / 1.71 * 100 / (9.8/0.0025)^0.1875]^1/0.294
E =
3,511 e12 J, or
839.15 kilotons. This would fit rather well with the Groumall case's parameters and the ENT battle between the NX-01 and a Klingon ship through an asteroid field (which should provide nice numbers, as the Klingon weapons were hitting asteroids, breaking them and also taking down the NX-01's defenses down, as reported by the technician in charge of the defensive systems).
Now, this is where it's important to remember the former result, with the old updated formula.
Let's try Dence's extended formula for an explosion on Earth, to see if a change of gravity matters.
E_earth = ( 3000 / 1.71 * 100 )^1/0.294
E_earth =
687,322 e12 J, or
164.27 megatons.
Roughly
195.75 times the first figure for the low gravity example. The difference is absolutely huge!
Compare this with the
109 megatons obtained earlier on, for a detonation on Earth, when using the updated formula found on the blog and adding the results of its two terms. It's not terribly far from it, and it's even superior to the result with that updated "old" formula.
I think the demonstration is clear that the old formula cannot be arbitrarily used for impacts in space without considering the gravity part of it.
Also, to show that a difference of gravity doesn't only matter in that equation, let's use another one. That is, the most complete gravity adaptable equation proposed in the document is a slight modification of Gault's formula (1974), and offers energy, gravity (with gM being the Moon's gravity), density parameters for both target and projectile (ρt and ρp respectively), and even angular impact (θ), which we'll put at 90°.
D = 2.7 e-2 * ρp^1/6 * ρt^(-1/2) * E^0.28 [1-0.095 (1-cosθ)] * (gM/g)^3/16 , with D > e5 cm.
The perpendicular impact taken out of the equation, we're left with:
D = 2.7 e-2 * ρp^1/6 * ρt^(-1/2) * E^0.28 * 0.905 * (gM/g)^3/16
With E isolated, the formula becomes:
E = [D / 2.7 * e2 / ρp^1/6 / ρt^(-1/2) / 0.905 / (gM/g)^3/16]^1/0.28
The moon has the following properties:
Code: Select all
Mass 7.347 7 × 1022 kg (0.012 3 Earths)
Mean density 3,346.4 kg/m³
Equatorial surface gravity 1.622 m/s² (0.165 4 g)
Trying to see what happens with a change in gravity means we don't need to care about projectile density.
But in any case, I'd suggest using a very low density projectile of 1 g/cc or below.
Now, the only term that is ought to change in the formula is "(gM/g)^3/16", which becomes [(gM/g)^3/16]^1/0.28 once E is isolated.
So let's try to compare the factors (F_mathilde and F_moon).
With the Mathilde asteroids (0.0025 m/s²), we obtain:
F_mathilde = [(gM/g)^3/16]^1/0.28 = (gM/g)^3/4.48 = 76.403
With the Moon, we obtain:
F_moon = [(gM/g)^3/16]^1/0.28 = 1
So we can see that there's already a huge difference between both premises, and it's not as huge as if we had compared Earth to 253 Mathilde with the same term: the difference would be little, since the final exponents in the two formulas are 1/0.28 and 1/0.0294.
So, in this case, the asteroid related yield would be
76.403 times weaker than that of the Moon's.
Using that ratio on the 36 MT figure presented in
this post, obtained via the Moon related formula E = 4*e15*D^3, we see that the yield for the asteroid blast would have been
471.185 kilotons for a 3 km wide crater.
Twice less than the
839.15 kilotons found earlier on, which can be explained by the fact the results are not from the same equation and were calculated for Lunar parameters in this case.
But both still are under the megaton range.
And that returns 4.9, 127.5 and 1.3 megatons respectively for granite (2330 kg/m³), nickel-iron (7870 kg/m³) and ice (900 kg/m³), all for a 1.5 km depth. And that's assuming he meant the crater created by the explosion, in contrast to the final crater that will be obviously larger than it's deep.
Crater depth, not width. The calculator we used gauged width. It's also much more recent the Wong's formula. It's also in widespread use.
Perhaps, but not for space related calculations.
Also, "The coupling of energy to asteroids and comets" (1994) is cited in several papers, including NASA documents, which are post 2000.
Now, as for Wong's equation. Here's a schematic of what it suggests, when purely considering the depth:
That's the theoretical representation of what it would be if you dug such a crater, according to its depth.
Now, it's only theoretical. I find it terribly unlikely that if you managed to dig such, the pointy edges would not collapse. Schematic cross section views of craters with long cracks starting from the bowl shaped surface of the crater, and radiating through the material for several meters. It's already obvious that on this schematic, the pointy edges would be fragile, as they couldn't count of the vast bulk of the remaining asteroid to maintain structural strength, notably because dampening is not part of the estimation either (dampening due to porosity is precisely what would prevent edges from cracking).
It's easy to imagine that in case of more fragile material, the weaker broken rim of the crater would stand farther, and extend to match the asteroid's width very closely.
OK. For some reason I remembered the evil NX-01 firing beams as well. Pulses. That's also what the E-NIL was firing, right? Because it didn't look like pulses. Although it looked like quantum torpedoes.
No, the E-NIL fires normal beam-like phasers. The blue balls were torpedoes.
And these torpedoes are shielded?
EDIT: corrections brought to the first Dence extended formula used in the post, and to the observation.
I noticed the mistake while checking the post with a fresh mind and a recharged cells. Consider this:
There is a Dence equation for smaller diameters, which you could pretty much consider to be like the old formula once used "on Earth":
D = 1.9 e-3 * E^0.333 * (gE/g)^3/16 , with D < 2.4 e5 cm.
Obviously, aside from the difference in some constants, gravity will, again, play a major role in the final result. The whole observation being that there's no reason that the mechanical forces dictating such differences in the yields, would not apply to explosives.
