# At what depth in the water atmospheric pressure is 100 times greater than on the ground?

At what depth in the water atmospheric pressure is 100 times greater than on the ground? This question comes from the fact that average pressure in Earth( 1000 mbar) is 100 times greater than in Mars( 7 mbar).

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The atmospheric pressure at STP is 101325 N/m$^2$, so 100 times this is 1.01325 $\times$ 10$^7$ N/m$^2$. You just have to work out the height of a column of water with a 1 m$^2$ base and weighing 1.01325 $\times$ 10$^7 /g$ kg, where $g$ is the acceleration due to gravity. I make it about 1.03 kilometers, though note that it will vary slightly with temperature because the density of water varies with temperature.

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P= hdg, where h=height ~ unknown, d=density of liquid ~ $1000 kg/m^3$ and g=gravitation constant ~ $9,81 m/s^2$ $h=P/dg = \frac{100 \cdot 10^5 Pa}{1000 kg/m^3 \cdot 9,81 m/s^2 }$ = $1019 m$ ( Note. $100 \cdot 10^5 = 10 \cdot 10^6 = 10^7 = 10000000 Pa$) Any comments on this? –  laovultai Feb 12 at 19:00
It looks as if you have done the same calculation as me, except that you took 1 atmosphere to be 10$^5$Pa and it's 1.01325 $\times$ 10$^5$Pa. That's why my figure is 1.325% higher than yours. –  John Rennie Feb 13 at 10:18
to be picky, at the surface, you already have one atmosphere of pressure, so one should add/subtract that (and you'll get about 9.81m less depth)... –  Andre Holzner Feb 14 at 7:31
Check the formula $P= hdg$ where $h$ is the height, $d$ the density and $g$ the acceleration of gravity.
Welcome, pappu. We have the MathJax rendering engine active on the site which means that you can write your mathematics in a LaTeX alike language. I'm going to simply put your existing math in that form. You could use \rho to get $\rho$ for the density if you wanted. Further, while your answer is correct it would be a lot better with some discussion of why this makes sense and if the OP needs to worry about variable density in the overlying material. –  dmckee Feb 12 at 19:10