Time taken by a air bubble on the sea depth to reach sea level

There is an air bubble at a depth of s meters above sea level.

Suppose the air is an ideal gas and Temperature is constant.

I would like to calculate the time it takes for it to reach sea level.

The bubble is subjected to Archimedes' force, meaning F = V * d(H2O) * g, and also to its weight, call Fp = m * g.

As the bubble travels, it sees its water pressure decrease, so its volume increase, receiving thus an increased Archimedes' force.

Call Fris the net force which the air bubble is subjected to.

Then by my calculations, I have written this differential equation, (keep in mind I am a DE noob) having assumed

$F_archimedes = V * dH2O * g$

$pV = nRT$

$p = dH2O * g * h(t)$

(h is the depth level in meters with respect to time), m = the mass of the gas bubble, the resulting acceleration in function to time is:

$a(t) = (nRT/m) * (1/h(t)) - g$.

which becomes, being the coefficient constant,

$\frac{d^2h}{dt^2} = \frac{D} {h(t)} - E$

I have a doubt on a minus sign on the last equation I have written. From this how do I calculate the time it takes to approach sea level and the velocity it has?

I only know the depth $d$ by which the bubble starts.

Quick solution: The key is to think about the bubble as an object of negative mass in water. What you have is essentially a blob of mass $-\rho_{water}V(h)$ in the gravitational field of earth, the important thing here is to realise that even if the mass of an object increases during it's fall (the volume of the bubble expands), unless you account for drag this will not affect the object's acceleration. Thus the bubble will accelerate to the surface with constant acceleration $g$.

So if the bubble forms at depth $h_0$ it will take the bubble $t = \sqrt{\frac{2h_0}{g}}$, neglecting all drag effects which may become relevant for all but very small speeds.

Formal solution:

Since you are assuming the bubble is an ideal gas we know that: $$PV = NRT,$$ it may be more useful to rewrite this in terms of the initial volume $V_0$ and pressure $P_0$ at the initial depth $h_0$: $$P_0 V_0 = PV,$$ this can be then further simplified by assuming the form of the hydrostatic pressure as $P_{buoy}=\rho_{water}gh(t)$, thus we can write: $$V(t) = \frac{P_0V_0}{P(h)} = \frac{\rho_{water}gh_0V_0}{\rho_{water}gh(t)}=\frac{h_0}{h(t)}V_0.$$

The buoyant force acting on the gas bubble can be written as: $$F_{buoy}=\rho_{water}gV(t) = \rho_{water}g\frac{h_0}{h(t)}V_0,$$ but also more generally by second Newton's law as: $$F_{buoy} = \rho_{water}V(t)\frac{d^2h}{dt^2} = \rho_{water}g\frac{h_0}{h(t)}V_0.$$ From this equation you can simply see that $$\frac{d^2h}{dt^2}=g,$$ i.e. in our reference frame the bubble accelerates to the surface at $g \approx 9.81$$m$/$s^2$.

• where you wrote Fbuoy = pwater *V(t) * second derivative of space versus time you are writing Fbuoy = mass * acceleration, where mass is calculated by density * volume. – f126ck Dec 10 '17 at 22:37
• shouldn't so the density of the mass be the air density instead of water? I'm confused – f126ck Dec 10 '17 at 22:38
• This is the tricky step. I think the idea is that as the air accelerates upwards the same volume of water accelerates downwards. Don't forget that, strictly, the Archimedes principle applies to static situations, and that's not what we have here. – Philip Wood Dec 10 '17 at 22:45
• Yes, Philip is right. In principle you need no force to move the bubble, because it has inherently zero mass, however you need force to move the water above it down to take the bubble's previous position so that the bubble moves up. In reality what is really happening here is a movement of water, the bubble is just the volume that contains none. For the bubble to move, the water around it must relocate. – Akerai Dec 10 '17 at 23:05
• But of course, there is viscosity. – Pieter Dec 10 '17 at 23:17