Boiling Away a Planet's Atmosphere

[Jake]

I wanted to find the energy required to boil away Earth's Atmosphere. My reasoning was that to do this, all of the atmosphere's air particles would have to be accelerated to escape velocity. To do this, I used the average molecular velocity equation with 11,200m/s (escape velocity) to find the temperature, which came out to be 1.458E8K. The equation is
v^2=sqrt[(3RT)/MM], where R is 8.314J/mol*K, MM is 28.97g/mol, and T is the temperature. The two main portions of the atmosphere are the troposphere and stratosphere, which contain 80% and 19.9% of the air respectively. To find the temperatures of the troposphere and stratosphere, I used the graph here: http://www.eoearth.org/article/Atmosphere_layers
and took an average of each to get Tt=253K and Ts=243K. I then found the heat capacity of dry air to be 1005J/kg*K. I looked up the mass of the atmosphere and found it to be 5.27E18kg. Finally, I found the energy to boil away the atmosphere from the equation Q = .8*mc(1.458E8-Tt)+.2*mc(1.458E8-Ts) to get an energy of 7.722E29J. Is this right? I'm kind of unsure because this value is insanely large.

[Norade]

3.2 x 1026 J: Energy required blow off Terra's atmosphere = 77 petatons

That's from Atomic Rocket which is a well regarded source. The energy required to boil it off would probably be more because doing that is less efficient than simply forcing it away from the planet at a high speed.

[Jake]

What I have in mind is definitely boiling, and said boiling is a side affect of what I am looking at, so I'm using it as a lower estimate in firepower.

[Bottlestein]

What I have in mind is definitely boiling, and said boiling is a side affect of what I am looking at, so I'm using it as a lower estimate in firepower.

The mechanism of how the boiling occurs is key here. Remember, in any fluid, heat transfer will have a significant convection component. If its a laser, depending on its wavelength, the height of the atmosphere where the heating occurs will determine how much convection plays a role. If its a nuclear device, again the height of detonation will determine how much convection will affect your energy determination.

Again, the important thing to remember is the height above ground level, as well as area, that you expect the "heating" to occur. Depending on this, you will see if convection helps raise molecules to escape vel. (making your estimate an upper bound on energy deposited), or if convection hinders molecules traveling away from the center of the earth, making your estimate a lower bound on energy deposited.

[Jake]

It's superheated plasma hitting the ground, so the height will be at the very bottom of the atmosphere. I don't know much about convection, so I'm not sure how this would affect the energy required to burn away the atmosphere.

[Samuel]

It's superheated plasma hitting the ground, so the height will be at the very bottom of the atmosphere. I don't know much about convection, so I'm not sure how this would affect the energy required to burn away the atmosphere.

So basically you are treating it as a projectile that, when it hits unloads heat? How accurate a representation is this?

[Jake]

It should be pretty accurate, because in universe it is somehow magnetically contained until it impacts its target.

[Jake]

Also, its not just one projectile, multiple plasma bolts are fired throughout the planet continuously and it takes around a day for the atmosphere to burn off.

[Samuel]

It should be pretty accurate, because in universe it is somehow magnetically contained until it impacts its target.

That isn't really possible. I don't think the mechanism really matters though- as long as all of the plasma gets to the atmosphere all of its heat will be dumped into the atmosphere.