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Enceladus, one of the Saturn's moons that is known for water geysers, is estimated to contain a body of water as big as a ball $220 ~\mathrm{km}$ in radius. That's about $\sim 4.46 \times 10^{7} ~\mathrm{km}^{3}$ of water!

https://businessinsider.com.pl/international/these-ocean-worlds-reveal-just-how-little-water-we-have-on-earth/s4vp2pm

If Mars was a perfect sphere, then spilling all that water on its surface would give us about 308m deep layer of water

Solve $\left( R + h \right)^{3} - R^{3} = 220^{3},\, R = 3390\, \text{for}\, h$

this gives us approx. $h = 0.308$

Now, Mars is obviously NOT a perfect sphere. I don't know its topography too well, but I heard that there are places as deep as $4$ kilometres on its surface. Spreading all this water on the Mars would probably result in some nice ocean or two, just perfect for terraforming

https://mars.nasa.gov/mgs/sci/mola/dec10-99rel/global_cover.gif

The question is, how can I somehow calculate or estimate what percentage of Mars' surface would be under the water, if we dropped so much of water on its surface? ($\sim 4.46 \times 10^{7} ~\mathrm{km}^{3}$ of water)

Is there some free software that allows for this kind of simulations?

Or maybe someone was already pondering about bringing some of the Saturn's moons to the Mars, for the purpose of its terraforming, and already did the calculations? Any ideas about how can I approach this problem?

(this is not a homework or anything like that, I was just casually taking a stroll and started wondering about terraforming the Mars. Just a day like everyday 🙂)

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