I am having a very basic confusion on how we calculate the height of atmosphere when we assume that the density does not change with altitude(density remains 1.29 kg/m$^3$).

I want to know why we say that in this case that the atmospheric pressure is $$P=\rho gh$$

I think my confusion here is if there is any pressure exerted on the atmosphere itself!

Because we can say in this question that $$ P_{\text{on the atmosphere}}-P_{\text{by the atmosphere}}=\rho gh $$

Then if pressure on the atmosphere is 0 then my confusion is gone :) Otherwise please explain the same.

By the way the answer to the above question is about 8Kms.


1 Answer 1


Perhaps you're looking for the formula $$\nabla p = \rho \mathbf g$$ that relates the pressure gradient at any point with an acceleration field, which is usually taken to be gravity alone in many practical cases. If both $\rho$ and $\mathbf g$ are constant, then $\mathbf g$ comes from a potential $V=-\mathbf g\cdot\mathbf r$, so that $p= p_0+\rho\mathbf g\cdot\mathbf r$, where $p_0$ is any integration constant. Since $\mathbf g$ is pointing downward, the inner product $\mathbf g\cdot\mathbf r$ reduces to just $-g\Delta h$, where $\Delta h$ is assumed to be positive when measured w.r.t to a "ground" level and going upward. Hence the pressure profile at constant density and gravity is $$p = p_0 - \rho g h,$$ where $p_0$ is the pressure at $h=0$ (e.g. the pressure at sea level). As you go up, $h$ increases and therefore the pressure decreases, which physically corresponds to the fact that there is a shorter column of fluid at higher heights. Clearly, when there is no more fluid on top the $p$ must be zero, which leads to the equation $$p_0 = \rho g h$$ in $h$, and the solution is the total height of fluid.

  • $\begingroup$ Thanks a lot...i understood:) Basically i was wondering if there is any pressure the atmosphere itself. $\endgroup$
    – geek101
    Jan 10, 2015 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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