Can I use gravitational binding energy for non-spherical objects?

[Arceus0x]

can I?
I have a cylindrical planet-sized object that blows up

can i use GBE for it? And if so then how?

 

[Greenshifter]

Use this as a reference: https://vsbattles.fandom.com/wiki/U...10_(Alien_Force):_Planet_Petropia_Destruction

 

[DontTalkDT]

Use this as a reference: https://vsbattles.fandom.com/wiki/U...10_(Alien_Force):_Planet_Petropia_Destruction

Don't actually, since you can't just use the GBE formula for spherical stuff for non-spherical stuff.

GBE for non-spherical stuff is possible, but requires integrals to derive an altered GBE formula. For a cylindrical structure doing so is possible, but I don't think you can in practice do it for the clock thing in the video above... it's just too complicated of a shape with non-uniform density and stuff.

 

[Arceus0x]

GBE for non-spherical stuff is possible, but requires integrals to derive an altered GBE formula. For a cylindrical structure doing so is possible, but I don't think you can in practice do it for the clock thing in the video above... it's just too complicated of a shape with non-uniform density and stuff.

is there any other way to calculate it's destruction then? Cause if you use the classic explosion formula it gives you high 6-A...and im pretty sure that something that has been calculated to be larger than earth exploding into an explosion multiple times its size isn't only high 6-A

 

[Greenshifter]

KE?

 

[Greenshifter]

Don't actually, since you can't just use the GBE formula for spherical stuff for non-spherical stuff.

GBE for non-spherical stuff is possible, but requires integrals to derive an altered GBE formula. For a cylindrical structure doing so is possible, but I don't think you can in practice do it for the clock thing in the video above... it's just too complicated of a shape with non-uniform density and stuff.

I see, thx. Can you send the integrals I need to solve for Petropia?

 

[Arceus0x]

KE?

I guess its really the only way. I'll calc the mass of an 80% hollow cylinder made of brass (pocket watches are usually made of brass and Nova's a clockwork star) and I may try KE...idk what else i can try.

 

[Arceus0x]

or maybe like 50-40% hollow since its jam packed with stuff

 

[Greenshifter]

Seems like a good plan.

 

[Arceus0x]

ill make a calc once i have the chance. KE could work since we see multiple pieces flying off (besides the massive things surrounding the nova)

 

[DontTalkDT]

I see, thx. Can you send the integrals I need to solve for Petropia?

That depends on how you want to model the shape, which looks complicated in that case...

General procedure (if I spontaneously don't forget something):

1. Figure out where the center of mass is.
2. Figure out the volume of the parts of the object that are between r and r+h meters away from the center of mass (I call that part a shell), for any r > 0 and h > 0.
3. Using that volume and the density of the object figure out the mass of the shells depending on r and h.
4. Now for the first integral. You want to calculate the energy necessary to move the shell an infinite distance away from all the mass that is within that shell. What's useful here is that you can assume all that mass within to lie in the center of pass. That's a potential energy calculation or you're basically calculating the gravitational energy. What helps is that you can assume that the center of mass of your shell lies in r for this. (I think...)
5. Now you do this for all shells, with thickness h. I.e. for a shell from 0 to h, from h to 2h, from 2h to 3h and so on. (technically to infinity, but since your object ends at some point you will only get 0 for shells from there).
6. Now sum up the gravitational energy results for all those shells. By that you get an approximation of the result.
7. To get from the approximation to the precise result you now want to let h pass the limit to 0. (h -> 0) You will notice that doing so gives you infinite terms to add up. So this is where the second integral comes into play. If g(r,h) is the function that gives you the energy for the shell starting at distance r with thickness h, then you want to look at g(r,h)/h and figure out its value for h = 0. (Useful: Due to L'Hopital's rule that is equal to g(r,h) derivated for the variable h and evaluated at 0 i.e. g(r,h)/h = g'(r,0) ) Then you want to integrate that result relative to r from 0 to infinite. That's your final result.

 

[Arceus0x]

anyways this can be closed since i've no more questions

 

[Greenshifter]

That depends on how you want to model the shape, which looks complicated in that case...

General procedure (if I spontaneously don't forget something):

1. Figure out where the center of mass is.
2. Figure out the volume of the parts of the object that are between r and r+h meters away from the center of mass (I call that part a shell), for any r > 0 and h > 0.
3. Using that volume and the density of the object figure out the mass of the shells depending on r and h.
4. Now for the first integral. You want to calculate the energy necessary to move the shell an infinite distance away from all the mass that is within that shell. What's useful here is that you can assume all that mass within to lie in the center of pass. That's a potential energy calculation or you're basically calculating the gravitational energy. What helps is that you can assume that the center of mass of your shell lies in r for this. (I think...)
5. Now you do this for all shells, with thickness h. I.e. for a shell from 0 to h, from h to 2h, from 2h to 3h and so on. (technically to infinity, but since your object ends at some point you will only get 0 for shells from there).
6. Now sum up the gravitational energy results for all those shells. By that you get an approximation of the result.
7. To get from the approximation to the precise result you now want to let h pass the limit to 0. (h -> 0) You will notice that doing so gives you infinite terms to add up. So this is where the second integral comes into play. If g(r,h) is the function that gives you the energy for the shell starting at distance r with thickness h, then you want to look at g(r,h)/h and figure out its value for h = 0. (Useful: Due to L'Hopital's rule that is equal to g(r,h) derivated for the variable h and evaluated at 0 i.e. g(r,h)/h = g'(r,0) ) Then you want to integrate that result relative to r from 0 to infinite. That's your final result.

Thanks, I’ll give it a look when I have a lot of time kek. You got any references for this? Seeing it in formula form will probably make it easier.

 

[DontTalkDT]

Thanks, I’ll give it a look when I have a lot of time kek. You got any references for this? Seeing it in formula form will probably make it easier.

Not really, no. Especially not in form of a single formula (a formula on this would be incredibly convoluted...).
You can look at the derivation for a sphere on wikipedia as an example, if it helps you, although they do it differently from me 'cause... physicists.

 

[Greenshifter]

Ok, thx