Forum adverts like this one are shown to any user who is not logged in. Join us by filling out a tiny 3 field form and you will get your own, free, dakka user account which gives a good range of benefits to you: No adverts like this in the forums anymore. Times and dates in your local timezone. Full tracking of what you have read so you can skip to your first unread post, easily see what has changed since you last logged in, and easily see what is new at a glance. Email notifications for threads you want to watch closely. Being a part of the oldest wargaming community on the net. If you are already a member then feel free to login now.

I don't have the math background to know what equations to use to find the answers, so calculations with explanations would be appreciated, with Earth serving as the example planet.

(I'm working on a crossover so this seems a good starting place as each faction can roast planets. With this I can calculate their big weapons, and from there calculate their shields, all of which would make it easier to determine how much damage each ship can deal and take relative to the others)

According to this, you're looking at 3.6x10^27 joules of energy to boil off the oceans of earth and eliminate all ground water.

I would suggest starting with small scale tests in the backyard

A whole lot.

I'd suggest reading this comic:

https://what-if.xkcd.com/13/

 Konrax wrote:
I would suggest starting with small scale tests in the backyard

The 4th of July isn't too far off. "Celebrate the birth of your country by blowing up a small piece of it!"

This site does include a fair bit of maths in comparing Star Trek and Star Wars weapon energies, including considering the energy requirements for blowing up asteroids (ignoring gravity) and planets. A more focused section dealing with planetary bombardment can be found here.

 Furyou Miko wrote:
According to this, you're looking at 3.6x10^27 joules of energy to boil off the oceans of earth and eliminate all ground water.

Thank you! exactly the sort of thing I'm looking for!

So we have 3.6e27 Joules of energy to boil off all of the water on earth. That's over 8.6x10^17 Tons of TNT, or 860 Petatons.
this


*How do you make it say "this" instead of being the whole URL?
edit: Thanks!

Next step, is how much energy would it take to melt the outer 5 meters of Earth's crust after the water's removed?

 Bill1138 wrote:

(http://convert-to.com/conversion/energy/convert-j-to-tn.html)
*How do you make it say "this" instead of being the whole URL?

Its [*url=http://www.example.com]This[*/url] without the asterisks.

What formula would I use to calculate how much energy is needed to melt a planet's surface (ignoring water)?

To work out the energy needed to melt something, you need two pieces of information;

First, the specific heat capacity, which is how much energy is required to raise the temperature of a substance by one degree. Use that to work out how much energy you need to raise the substance from its initial temperature to its melting point.

Then you need the latent heat of fusion, which is the energy required to actually melt the substance (equivalently, the energy released as it freezes).

Once you have those numbers, you know how much energy is required to melt a kilogram of that substance. Now you need to work out how much stuff you're melting. To simplify, work out the surface area of the planet, and multiply that by the thickness of the surface layer you want to melt. then multiply that resultant volume by the density of the substance you're thinking about.

It can be calculated. I had watched a video on this calculation with Earth before but I forget the exact details. The calculation was able to figure out what it took to crack Eath in half to what it would take to blow it into complete bits. The numbers are insanely huge. We could gather all our nukes during their height, all our bombs and ammunitions and let it off in one go. We'd probably create another massive extinction level event but it wouldn't be enough to even end Earth's cycle of life let alone crack the planet. Earth would repair itself and life would continue it's evolutionary ways with the new environment as it has with even more powerful extinction level events that has occured in Earth's past. The numbers are staggering but not unimaginable or unknowable.

So to get rid of all the water we have:
3.6e27 Joules of energy, which is over 8.6x10^17 Tons of TNT, or 860 Petatons.

According to this, the density of the crust's silicate rocks is 2.2 g/cm^3. According to this, Silicon has a density of 2.3290 g/cm^3.

So if I understand correctly, if we calculate the answer with Silicon, the calculated value should be slightly higher than the actual low-end estimate. Correct?


0.0356055615887202 mol = 1 g
1 mol = 28.0855 g

Surface Density of 2.3290 g/cm^3
Silicon has a Molar heat Capacity of: 19.789 J/mol·K
Silicon has a melting point of: 1687 K
Silicon has a heat of fusion of: 50.21 kJ/mol
Earth's average surface temperature is It's 287 K,


So check my math on this...

My Molar Heat Capacity and Heat of Fusion are in mols, so I have to convert the surface density into mols
0.0356055615887202 x 2.329 = 0.08304 mol/cm^3

So I multiply that by the molar heat capacity? That should give me the amount of energy in J to raise the cm^3 by 1 K, correct?
0.08304 x 19.789 = 1.643278 J to raise 1cm^3 of Silicon by 1K.

If Earth's average surface temperature is 287K, and I need to raise it to 1687K, then I need to raise it by 1400K.
1.643278 J x 1400K = 2300.5892 J per cm^3

Then I use Silicon's Heat of Fusion to calculate the amount of energy needed to melt the heated material
0.08304 mol/cm^3 x 50.21 kJ/mol = 4.1694384 kJ/cm^3 or 4169.4384 kJ/cm^3

So the total amount of energy to melt each cm^3 of the Earth's crust from its current state is
2300.5892 J + 4169438.4 J = 4171738.989 J/cm^3

If Earth's Radius is 6371Km, then the surface area is (2 x Pi x R) = 40030.17359Km^2 or 4003017359cm^2

I can multiply the total J/cm^3 by the Surface area and multiply by how deep I want to go to get an estimate of how much energy would be required to do so. Correct? I don't think I can get any more precise than this because of the limitations of significant digits of the numbers I'm working with.
4171738.989 J/cm^3 x 4003017359cm^3 = 1.669954359x10^16J for 1cm deep

1.669954359x10^16J x 500cm = 8.349771795e18J

Add this value to the energy required to vaporize all of the water on earth:
8.349771795e18 + 3.6e27 ~ 3.6e27J?

Can that be right? Is the energy required to vaporize all of the water on earth so much more that that required to melt all of the continents into slag?

Yes, it actually makes sense if you think about it.

Seventy one percent of earth's surface is ocean.

That means that only 29% of it is land mass. Further, you only have to melt maybe half a mile of the very topmost layer of the surface of that landmass to have 'melted the continents into slag'.

Furthermore to that, the average temperature of the oceans is 16 degrees and water has a relatively high enthalpy - the amount of energy it can store before changing state. That means that to affect a tiny change in the ocean's temperature, you need an insane amount of energy. Also bear in mind that the deeper you go into the ocean, the colder it gets: while the average is 16 degrees, at the bottom of the ocean its more like two degrees once you get more than a few metres from a thermal vent.

The ground, however, gets hotter as to go deeper. near the surface, you're looking at about 15 degrees in the UK, but below the surface that rises by a massive 2 degrees per 100 metres - which means that a mere kilometre under ground, you're looking at a rock temperature of thirty five degrees.

This is why thermal vents exist, incidentally - where the bottom of the ocean meets the rock, you get these vents that constantly pump hot gas into the water at about 400 degrees. This is a really good way to illustrate just how hard it is to heat up the oceans: That's four times the nominal boiling point of water, and it is all lost once you stray a few metres away from the vent, the water temperature is back down to just above freezing.

The end result is that while you need to add a hell of a lot of energy to boil the oceans, the continents already contain a lot of the energy you need to get them turning over.

By the time you are boiling off the dregs of the oceans, the continents are already toast. Assuming you are talking about a constant orbital barrage over time, and aren’t doing it all in one pass.

Love this thread, by the way.

Water is insanely good at capturing, storing and dissipating the energy across it's volume so I'm not surprised it would take more energy to vaporize our oceans then it would to melt the Himalayan Mtn range considering how much of the planet is water.

So how deep would you have to melt on average to make the mountains disappear? Would it have to be the full height of the mountain? More? Less?

Atomic Rocket I found to be quite informative.

Also useful table of various energy requirements for specific actions:
Energy required to vaporize all the oceans of Terra and dehydrate the crust is 7.0 × 10^27 Joules or 2 Exatons
Energy required to melt the (dry) crust of Terra is 2.9 × 10^28 or 7 Exatons
Energy required blow off Terra's oceans into space is 1.0 × 10^29 or 24 Exatons
Earth's rotational energy is 2.1 × 10^29 or 50 Exatons

Boiling off the oceans would probably envelope the entire planet in superheated steam and sterilize the entire surface rather easily. Those pesky survivalists in deep bunkers will need a second dosage, apparently.

Miradorm wrote:
Atomic Rocket I found to be quite informative.

Also useful table of various energy requirements for specific actions:
Energy required to vaporize all the oceans of Terra and dehydrate the crust is 7.0 × 10^27 Joules or 2 Exatons
Energy required to melt the (dry) crust of Terra is 2.9 × 10^28 or 7 Exatons
Energy required blow off Terra's oceans into space is 1.0 × 10^29 or 24 Exatons
Earth's rotational energy is 2.1 × 10^29 or 50 Exatons

7 Exatons? That must be melting the entire thickness of the crust, because it is drastically higher than what I calculated for the top few meters.

Is there anywhere I can see the math for this? Typically I prefer to know and understand why a number is what it is so I can re-create the process if the need arises.

I honestly don't know how they came up with those numbers I just copy pasted from the linked table. Outside of asking the person who maintains it where he got that data point at no clue.