Planet Cracking

[CNBA3]

Quick question, what is the necessary TNT needed to crack a planet's surface? say like the lithsphere (Earth's crust and upper mantle)?

 

[IkaniL]

I don't know if we exactly have an exact amount for that specific question. This might require a calc.

You could possibly request a calc group member this question. But remember to do so nicely.

 

[Amelia_Lonelyheart]

It has a lot of variables. The size and depth of the cracks and how they're made, for example.

 

[IkaniL]

Which is why a calc might be in order. Since we would need an idea of those variables to come up with a possible value.

 

[CNBA3]

It has a lot of variables. The size and depth of the cracks and how they're made, for example.

The method is the explosion breaking the crust 30 miles thick and explosion radius roughly 17 by 22 by 18 km

 

[Assaltwaffle]

Wait, you say radius and then list theee values. I am confused. If it is a hemisphereic or sphereical crater it would only need one radius value. Also you give three dimensions, so even that would cover the height, length, and depth of a non-sphereical explosion, but none of those match the km conversion from 30 miles. I am a bit confused.

 

[IkaniL]

This could be a good practice calc for me to try. Can I do it?

 

[Assaltwaffle]

Well I'm not sure what is actually being described. If it is a sphereical explosion with a radius of 30 miles that could work. If it is a rectangle with the other dimensions that can also work.

If it is specificied as to what size this thing is then sure, go right on ahead.

 

[IkaniL]

I'll try it with the radius of 30m. Then if he specifies something different, I can revise the calc.

Have you seen my message on the post we were conversing on? o.o

 

[CNBA3]

Wait, you say radius and then list theee values. I am confused. If it is a hemisphereic or sphereical crater it would only need one radius value. Also you give three dimensions, so even that would cover the height, length, and depth of a non-sphereical explosion, but none of those match the km conversion from 30 miles. I am a bit confused.

Apologies, I did not meant to add a third number value, the thickness remains at 30 miles

 

[IkaniL]

So the crust is 30 miles thick, but the explosion radius is only 17 by 22?

 

[IkaniL]

Okay, here is the calc.

M = Miles

R = Radius in kilometers

km = Kilometers

Y = Yield

R = 30m | 30m = 48.2803 km

48.2803 = Y^(1/3)*0.28

Y = 5,126,674.992146 Kilotons

 

[CNBA3]

@IkaniL, Thank you

 

[IkaniL]

To be exact though, I had to multiply by 0.5

5,126,674.992146x0.5 = 2,563,337.49607 kilotons

That is the exact energy from the blast.

 

[IkaniL]

You're welcome. Credit Assault though for the guiding help.

 

[Assaltwaffle]

Idk what when wrong there, but something definitely went wrong. That blast shouldn't be more than 1 Megaton (1,000 kilotons). Also, if the Earth was cracked by this, then we use Destruction, not explosions.

Edit: Wait nevermind. The explosion part is correct, but the the type of calc used isn't. You calced correctly, but I think calcing the hole in the crust is going to be better than calcing how much energy an explosion of that size would use.

 

[IkaniL]

So I will need to use the LxWxD formula then?

 

[Assaltwaffle]

I mean he still hasn't clarified what shape this hole is and which value is the "third value" that is wrong.

I'd use hemisphere formula with 30 miles as the radius.

 

[IkaniL]

Okay, if it turns out to be the hemispheric shape. Then using the formula for volume is the best?

https://ncalculators.com/area-volume/hemisphere-calculator.htm

 

[Assaltwaffle]

Yes.

 

[IkaniL]

(2/3)*¤Ç*48280.3^3/3 = 7.8568279e+13

r = 30m

 

[Assaltwaffle]

Why'd you divide by three at the end? A hemisphere's formula is just (2/3)xpixR^3. Using that formula I got 2.355x10^14 meters. You then covert cubic meters to cubic centimeters by adding six zeros, so total volume is 2.355x10^20 cc. Then we apply that to our destruction values. V-Frag is 69x, and Pulverization is 214.35.

The results would be 1.62x10^22 joules and 5.048x10^22 joules, respectively.

 

[IkaniL]

Thought the 3 under the formula mean't to divide it.
2.355x10^20x214.35 is 5.0479425e+22 or 5.0479425x10^22 when I put it into the Google Calculator. What makes it 5.048 though?

 

[Assaltwaffle]

I'm not sure why calculator soup has that /3, but it is wrong. Also I just round the 79 into 8.

 

[IkaniL]

Is there a specific rule regarding rounding with decimals like these? I only thought you did it when a five was involved.

 

[Assaltwaffle]

Like I said before, if the number is 5 or higher, you can round up to add a 1 to the next digit. So 0.0015 can be turned into 0.002, or 0.1368 can turn into 0.137.

 

[IkaniL]

Ohhh. I must've have missed the higher part of what you said. Got it now.