# Effect of gravity on large tower with reduced air pressure

Suppose I built a 5km high building. The pressure of the air at the top would be around 50% compared to the pressure at the ground (I assume that the exact values don't matter in my case).

Now consider a tube that is going from the bottom up to the top. The tube is initially closed at both ends. Now I would use a vacuum pump to reduce the pressure inside the tube (at the bottom) to about 50%.

What happens if I open the tube at the top? Will gravity draw air from the top of the building to the ground and restore "normal" pressure?

If so, how long would it roughly take for that to happen? Or more generic: How can I compute how long it will take? What are the parameters I'd need to take into account?

• One detail about this question is ill-posed. If gravity is to play a significant role, there will be a gradient in pressure in the tube with higher pressure at the bottom of the tube. So, what does it mean to reduce the pressure inside the tube to about 50%? Are you trying to say that half of the air molecules are removed? – user3814483 Oct 14 '14 at 19:35
• Hm ... I'm not absolutely sure. Is removing half of the air molecules equivalent to reducing the pressure "at the bottom" (I'm going to add that to the question) to 50%? – Daniel Jour Oct 14 '14 at 19:48
• Regardless of the specific change, the pressure at the top will be less than the external pressure at that height. You'll therefor have a pressure difference and a tube diameter. The speed of refilling will depend on the size of the tube. – BowlOfRed Oct 14 '14 at 20:54

When you open the tube, yes, air will be drawn into it. This is indirectly due to gravity, in the sense that what will drive the flow is a pressure gradient. The pressure in the tube is lower than the pressure outside (after you pump air out and let things settle down, this is true no matter where you open the hole), so air is drawn in. Gravity has a role in maintaining the pressure gradient, as air is drawn into the tube, it is replaced and the pressure outside remains $\sim$ constant because there is much more air outside the tube than in, and gravity maintains a pressure gradient in the atmosphere.
If you want a full blown solution (e.g. to run a simulation), you'll be solving the Navier-Stokes equation: $$\rho\frac{D\mathbf{v}}{Dt} = -\nabla p + \nabla\cdot\mathbf{T}+\mathbf{f}$$ You'd probably set this up as a tube in a simulation box where the box boundaries are fixed to a model of the atmospheric pressure, density, etc. gradients. You could also model just the tube to simplify away how air flows into it from the region around the hole; in this case your boundary condition is simply to fix the pressure, density, etc. to the external values at the altitude of the hole. This is probably a good starting point to try in 1D.
Be sure to find a good book or other reference on fluid dynamics that covers this equation before attempting to solve it as it can be a bit tricky - for instance the general case cannot be solved without resorting to numerical approximations. Also make sure you're familiar with the notation, e.g. the $\frac{D}{Dt}$ is a material derivative.