Solar and Multi-Solar System level energies

[Rib78]

The information linked here contradicts our current listing for solar system busting energy. This would have ramifications for our multi-solar system busting energy too (because it was based on the ss busting figure IIRC).

Also, the Attack Potency currently lists the top of large star level and the bottom of solar system level as different values.

How do we deal with this/these?

 

[ClassicGameGuys]

Just use 11.4 foes as multi-solar system level since solar system level begins at 5.7 foes. Simple enough imo

 

[Antvasima]

As a minimum we want to use the level necessary to destroy two solar systems at a distance of a few lightyears from each other at the same time, since our closest solar system is around 4 lightyears away, if I remember correctly.

 

[Antvasima]

Previously we used a cylindrical calculation, but a spherical discharge may be necessary to keep the values sufficiently separate.

 

[Rib78]

Didn't the cylindrical calculation use the normal multi-solar value as a way of determining required energy per unit of area, though? or did it use the star level value? if it used the former then surely redoing the multi-solar calculation with the new value would keep them distinct an seperate.

I'm unclear on how the cylindrical value was determined actually; was the distance between solar systems the height of the cylinder or the diameter, because if it was the diameter I think a lower value would be more appropriate.

 

[The_Living_Tribunal1]

it so turns out that i was looking for the amount of enery to destroy the solar ssytem, and i came across that

also, i then realized that it was the same info which was used to disprove spc being solar system level

 

[DontTalk]

So if we use 2.12E+54 erg as Solar System Level (2.12E+47 J) than we should use the same Method for multi-Solar System Level.

So the distance to our closest Solar System is 4.2 Lightyears, which is about 3.97362E+16 m.

So if the explosion happens between them the distance to each of them ist about 1.98681E+16m.

So the Energy of a spherical explosion (which we would use after the method of the link) would be dispersed on an area of 4*¤Ç*(1.98681E+16m)^2 = 4.960466699197374266E+33m.

Now I will take the energy that is required to destroy the sun from the linked website. That is 2.30E+48 erg = 2.3E+41 J.

The cross-sectional area of the sun is ¤Ç*(695800000m)^2 = 1.520963253150300027E+18m

That means 1.520963253150300027E+18m / 4.960466699197374266E+33m = 0.0000000000000003066 = 0.00000000000003066%

So 2.3E+41 J is 0.00000000000003066% of the total energy. So 100% of the energy are 7.5016307893020221787345075E+56 J. (I hope that is right, I am kinda confused right know out of various reasons)

 

[The_Living_Tribunal1]

energy to bust our solar system is based off the binding energy of neptune keep in mind


also, not all solar systems are the same

there are solar systems of large stars which have many (probably hundreds) mercury sized planets

 

[Rib78]

Maybe we should base it of the sun though, becaue some systems may have planets who's ratio of size and energy required to bust is greater than Neptunes, and others may not have anything like Neptune at all. All Solar Systems have Stars, and while they may not be like our own, it works better for a minimum estimate than the Neptune does.

 

[DontTalk]

I think Neptune makes sense for the Solar system busting feat. There it is assumed that the explosion is in the center of the solar system, in other words the sun. Because of that the energy required to destroy the sun is smaller than the energy to destroy just Neptune, since Neptune is further away from the center of the explosion.

For Multi-Soler System the Sun makes more sense. Her GBE is bigger than Neptunes and since the center of the explosion is supposed to be outside of the solar system the distance between center and Neptune and Center and Sun is more or less the same (we actually don´t know which is bigger). So here the Sun is a good estimation.

Using a smaller star is of course worth a thought, but a lot of variation is possible when it comes to such things, so when we actually used a very small star we would end up with results which for almost all cases in fiction are too low, so I think our sun is fine.

 

[Rib78]

Yeah, that does make more sense.

Applying the method you used to get e+56 for a sphere up there, I get e+53 for a disk.

Sol to Proxima Centuari (Roughly) = 40131561472800000m
Sun to Pluto (Which is how the linked website measured the solar system = 5906376272000m

Surface Area of the Disk 3.14*d*h
3.14*40131561472800000*5906376272000
=744280801665541621281024000000m^2

Energy Per Unit Area 2.3e+41/1520963253150300027
=151219958485921182315788.11666271Jm^-2

Energy E=744280801665541621281024000000*151219958485921182315788.11666271
E=1.1255011192973134113763094138624e+53
=1.1255e+53

One problem with this is that it assumes both solar systems are as large as atleast as large as ours, a smaller diameter would yield a far smaller total energy. Not a massive problem I guess because 99.999% of potential actual multi-solar busting will have a blast with atleast these dimensions, but I can image feats that I feel should be multi-solar that wouldn't meet this.

 

[The_Living_Tribunal1]

@rib: for your earlier reply: yes, we shud base it off our solar system

 

[The_Living_Tribunal1]

also, neptune is taken as a measure, because it has a very large orbital radius (so most of the orbit of empty), and it needs a lot of power to be destroyed, so if the blast coming from the center can destroy neptune, then it wud be a solar sstem buster (which is very impressive as its energy by then has massively spread out and most of it is wasted anyway)

 

[Antvasima]

I also think that we should base the calculation off two systems equivalent to our own, at a distance of 4.2 lightyears from each other.

 

[The_Living_Tribunal1]

oh wait

damn

we might have to re calculate what universal destruction wud be then, i mean withthe large amount of energy getting wasted and spread out

 

[Antvasima]

We will only use an energy scale up to galaxy level. Anything above that level is likely more readily apparent, and does not need to be calculated.

 

[The_Living_Tribunal1]

oh thats good, keeps things simple