Estimating the Size of The Bre'el Moon

[Mr. Oragahn]

As for the moon's size, another possible pointer towards a large diameter and or mass is that the moon at a distance of 500 km on it's perigee made the tides rise 10 meters

...

Does anyone know a good way to calculate the size of a body based on it's distance and it's tidal effects?
-Mike

Well, tidal effects depend a lot on how the water is distributed. Peak water levels in certain selected locations on Earth (eastern CA) can vary by 16-17 meters.

That said, if we're assuming that this is talking about a ten meter bulge on a typical coastline bordering a very large body of water... that we can very roughly approximate.

In general, the tidal forces are going to be approximately proportionate to mr^-3, but we're much too close for that approximation to be too accurate. That would give our asteroid proportionately 218,000 times the tidal force per unit mass.

The radius of an Earthlike planet is much larger than 500 km. Numerically - throwing it into the calculator - I get only about 39% of the the above using Earth's radius + 500 km as the center-to-center distance - so we'd expect 1e18 kg of mass for tidal forces equal to that of Earth's moon. To have ten times the tidal forces - (average 54 cm tides for a water world the size of Earth orbited by our moon) - we're talking 1e19 kg, or around ten quadrillion tons - putting it in 22nd place among the moons of our system and around 18th place for asteroids.

Which, as I've pointed out, means a ridiculous amount of energy in the warp field. Ignoring everything else in the solar system, that changes the moon's gravitational potential by around ... oh... 6e26 joules. Which is a hair much, in my opinion.

We can play with the parameters a good bit, of course, but that's about what we'd be looking at.

May we get a bit more details about the formulas and how you used them please?

[Jedi Master Spock]

Since I made the mistake of comparing 1/r^2 and 1/R^2 in a numeric approximation when I should have been comparing the crucially important near-side figure of 1/(R-r)^2 and 1/R^2 as exact values, yes. Always press! Always question!

Tidal force is the difference in gravitational force between the edges and center of mass of an object, so we're trying to match the difference Gm/R^2-Gm/(R+/-r)^2; after warning of inappropriate approximations, I went ahead and used one myself.

The r/R^3 approximation does work well for long distances, though. Bear that in mind for normal scenarios where tidal forces come into play.

So... we can hit down into the e16 kg range (~4e16) on our first-order estimate, which is more reasonable, even if the warp field energies still look fearsome, but the whole situation remains quite complex, strongly dependent on how coastlines are arranged, how large the planet is, etc. There's a lot of "unknown" floating in the mix - I wouldn't commit to anything more specific than 10^(16.5+/-1.5) kg based on the tidal effects.

[Roondar]

Except of course that if weapons yields where in the thousands of teratons region we'd see very different results of said weapons against unshielded targets than we actually do.

Why?
SW weapons are supposed to be Giga and Teratons weapons as well, yet their effects on unshielded crafts are no different then those in ST...

Although Mr. Oragahn more or less said what I meant, I still feel the need to add a few things:

1) I don't agree on the Giga/Teraton level weapons for SW any more than I do for ST. In fact, almost all uber-yield figures in Sci-Fi are quite suspect, usually missing elements you'd expect for such yields, such as the utterly blinding explosions that'd occur.

2) The effects on unshielded objects (like asteroids and the like) we see fall far short of said yields.

3) They didn't name the size of the fragments they'd have left - this is a problem because the yield-calculators around assume things about fragmentation that are obviously not correct.

(No ST explosion I've ever seen comes to '10 meter fragments' so those figures are right out. They all are either really small fragment (sub-meter) or really large ones)

[Mr. Oragahn]

Yes, I was about to say that yesterday. It's too conveniant to consider figures were debris would be 10 meters wide.

[Mike DiCenso]

Actually, thousands of gigatons for a Camilla sized moon would be if you assume 10 meter fragments. The only information that could even allow us to guess at the size of the fragments comes from Data's response to Riker about there being possibly "thousands" of fragments. For Data, that kind of a statement is unusually imprecise as that could mean anything from 2,000 to 9,999.9 fragments.

As things stand right now, the 10e16 kg mass places us well below 100 km right now. Depending on the dimensions and density of the moon, it could be anywhere from 20 to 30 km. Basically the size of Mars' smaller moon Phobos. Still fairly decent sized here.
-Mike

[Mr. Oragahn]

This is more what I had in mind.

A 10 meters wide debris wold have a volume of 523.59 m³.
A 10 km wide moon, assuming it's spherical (same for debris) would have a volume of 523.59 e9 m³. We'd obviously be well beyond the thousands of fragments, even if a good amount of the asteroid was melted or vapourized.

[Mike DiCenso]

Assuming Data is not just trying to emulate humans in being so imprecise, also assuming an average between the low and higher end (5000 fragments), and we use Phobos as our model (5,700 km^3 in volume). We would have fragments of approximately 1 km or slightly larger in size.

Of course there are problems with this.

-Data is being unusually imprecise here. Why not give a more accurate estimate, since he should be able to easily calculate the Bre'eel moon's physical characteristics, and what the E-D's weapons could do to it?

- How long were they expecting to bombard the moon? The impact was calculated at 29 hours, but presumably they would start a bombardment well before then, and the implication is that it might be completed well before the initial estimate, too.

-Which weapons would used in the bombardment of the moon? Just phasers. Just photon torpedoes? A mix of both?
-Mike

[l33telboi]

The reason Data was so imprecise was probably because the number of fragments would heavily depend on when, where, what, for what duration and at what pace the moon is fired upon. To get an exact figure you'd have to know all those factors, none of which Data could possibly know by that point.

Besides, calculating something like this would be nigh impossible without knowing every single piece of information on how exactly the moon is built on the inside. Where will it be likely to form cracks, etc?

Data is undoubtedly smart, but calculating how many fragments would result from the bombardment would require him to know things he can't possibly know just by looking at it. Not even using scanners would solve all the questions.

[Mr. Oragahn]

Not even God could. They simply didn't knew how they'd even try to dent that asteroid.
It requires an amount of precognition I'm sure no one would claim Data has.
Not that he has any anyway. :)

[Roondar]

You know, I just noted something very interesting about this whole episode.

When they are considering blowing up the moon, the following exchange takes place:


Riker: 'Could we blow it into pieces?'
Data: 'The total mass of the moon would remain the same, commander'


On the face of it, this is a very reasonable thing to say: when you blow stuff up, you don't remove matter.

However, I do believe we've just 'un-canonized' the NDF effect - phasers are always said to make stuff 'disappear' in some wonky fashion but Data's statement makes it rather clear that Starfleet weapons don't do that.

Secondly, since we now know the moons most likely mass, we can finalize the E-D's maximum tractorbeam output.

To make an object weighing 1 * 10^16 KG move by 92 m/sec requires 4,23 * 10^19 J, or in the ten seconds they took, 4,2 * 10^18 watt - aka ~1 GT/sec.

Next up, determining maximum atainable impulse speeds assuming 4,2*10^18 watts as top-end output. But that's for another thread ;)

[l33telboi]

However, I do believe we've just 'un-canonized' the NDF effect - phasers are always said to make stuff 'disappear' in some wonky fashion but Data's statement makes it rather clear that Starfleet weapons don't do that.

Interesting indeed. However, keep in mind that the "NDF effect" doesn't and has never existed outside visuals, even the term NDF which comes from the tech manuals isn't the same as what we see in the visuals. As such it should come as no surprise that dialogue doesn't factor in flawed visuals.

On the other hand the whole 'stuff goes away' thing is still there, we do know that the visuals show us this, be it VFX error or not. So I think the better question is "why didn't they phazorise away the moon?" I'm guessing this too would have quite severe and long-lasting effects on the planet below, but then the same holds true when it comes to blasting it to pieces.

Whatever the case, speaking from a visuals-is-God point-of-view, Data seemed to be talking about a conventional attack, photon torpedoes or the exotic part of the phasers turned off.

[Mike DiCenso]

Not even God could. They simply didn't knew how they'd even try to dent that asteroid.
It requires an amount of precognition I'm sure no one would claim Data has.
Not that he has any anyway. :)

Then we can't take what Data said as factual, if there is that much margin for error. For all that we know, he could be vastly underestimating the number of fragments by several orders of magnitude.
-Mike

[Mike DiCenso]

The reason Data was so imprecise was probably because the number of fragments would heavily depend on when, where, what, for what duration and at what pace the moon is fired upon. To get an exact figure you'd have to know all those factors, none of which Data could possibly know by that point.

Besides, calculating something like this would be nigh impossible without knowing every single piece of information on how exactly the moon is built on the inside. Where will it be likely to form cracks, etc?

Data is undoubtedly smart, but calculating how many fragments would result from the bombardment would require him to know things he can't possibly know just by looking at it. Not even using scanners would solve all the questions.

But don't you think that Data could, if he was so inclined, download all available information from Bre'el about their moon's structure and makeup? Certainly Data should know about the weapons' yeilds for the E-D fairly precisely, and he should then be able to extrapolate a reasonable range for the size and number of debris that would be generated by the moon's destruction.
-Mike

[Mike DiCenso]

Next up, determining maximum atainable impulse speeds assuming 4,2*10^18 watts as top-end output. But that's for another thread ;)

In the tractor beam thread, I noted how the E-D could be reducing her mass by mass by as much as 4 million metric tons. That is the E-D goes from some 6.5 million tons down to just 1.625 tons! Now that may not be a perfect estimate, it probably assumes that a certain amount of power has to be thrown into the warp field in order to not only encompass the object, but to be able to shunt a certain amount of mass into the subspace domain.
-Mike

[l33telboi]

But don't you think that Data could, if he was so inclined, download all available information from Bre'el about their moon's structure and makeup? Certainly Data should know about the weapons' yeilds for the E-D fairly precisely, and he should then be able to extrapolate a reasonable range for the size and number of debris that would be generated by the moon's destruction.
-Mike

Well I'm sure he could get some rough details on the composition and structure of the moon, but nothing in the region where he could calculate the exact number of fragments that would be generated. I mean things like that would even change with time and we are talking about stuff that’s really really tiny here.

But could he acquire enough info to get a general idea on how many fragments would result? Sure. That might be what the ‘thousands’ part is meant to show, after all.

However, I do think it more likely that he simply decided to go with a phrase that's basically meant to mean 'many' fragments, because making an exact calculation in that case would still require him to set up an exact method of attack and factor in everything he knows on the moon, it'd be needlessly excessive (even for Data) and it probably wouldn't even be his place to speculate on exact methods of bombardment without consulting the Captain and the rest of the crew.