Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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? – alvoutila Feb 12 '13 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 '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 '13 at 7:31

Check the formula $P= hdg$ where $h$ is the height, $d$ the density and $g$ the acceleration of gravity.

share|cite|improve this answer
Pappu: any references for this equation( for example wikipedia)? – alvoutila Feb 12 '13 at 18:37
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 '13 at 19:10
the keyword is hydrostatic pressure – Andre Holzner Feb 14 '13 at 7:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.