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 $F = V d_{\mathrm{H_2O}}$ g, and also to its weight, call $F_p = m g$.

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

Call $F$ 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

$F_{\text{Archimedes}} = V d_{\mathrm{H_2O}} g$

$pV = nRT$

$p = d_{\mathrm{H_2O}} g h(t)$

($h$ 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:

$$a(t) = \frac{nRT}{m} \frac{1}{h(t)} - g.$$

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.

  • $\begingroup$ Do not forget that increasing drag force will stop speed increase of a bubble after it reaches terminal velocity. $\endgroup$ Dec 6, 2023 at 12:13
  • $\begingroup$ I edited out the $*$'s originally used to indicate multiplication. Use \times for the symbol when formatting. $\endgroup$
    – Jono94
    Dec 27, 2023 at 14:54

2 Answers 2


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$.

  • $\begingroup$ 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. $\endgroup$ Dec 10, 2017 at 22:37
  • $\begingroup$ shouldn't so the density of the mass be the air density instead of water? I'm confused $\endgroup$ Dec 10, 2017 at 22:38
  • $\begingroup$ 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. $\endgroup$ Dec 10, 2017 at 22:45
  • $\begingroup$ 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. $\endgroup$
    – Akerai
    Dec 10, 2017 at 23:05
  • $\begingroup$ But of course, there is viscosity. $\endgroup$
    – user137289
    Dec 10, 2017 at 23:17

Akerai's answer captures the main idea clearly, but the effective inertia (added mass) of the rising bubble should be corrected from $\rho_\text{water} V$ to $\frac{1}{2}\rho_\text{water} V$, as I shall derive below. $\newcommand{\bvec}[1]{\boldsymbol{\mathrm{#1}}}$

Bernoulli equation

Consider the incompressible Navier-Stokes equation $$ \rho\left(\bvec{v}\cdot\nabla + \frac{\partial}{\partial t}\right)\bvec{v} = -\nabla P + \rho \bvec{g} + \mu\nabla^2 \bvec{v}. $$

For inviscid fluid this is: $$ \rho \left((\bvec{v}\cdot\nabla)\bvec{v} + \frac{\partial\bvec{v}}{\partial t}\right) + \nabla P - \rho \bvec{g} = 0. $$

Invoking a vector calculus identity, we can write $$ (\bvec{v}\cdot\nabla)\bvec{v} = \frac{1}{2}\nabla(\bvec{v}\cdot\bvec{v}) - \bvec{v}\times(\nabla\times\bvec{v}). $$

Then irrotationality ($\nabla \times \bvec{v}=0$) allows us to introduce a potential $\phi$ such that $\bvec{v}=\nabla\phi$, and $$ \rho\left(\frac{1}{2}\nabla(\nabla\phi\cdot\nabla\phi) + \frac{\partial (\nabla\phi)}{\partial t}\right) + \nabla P - \rho \bvec{g} = 0, $$ Letting $\bvec{g}=-g \bvec{\hat{z}}$ gives $$ \nabla\left( \frac{\partial\phi}{\partial t} + \frac{1}{2}\left|\nabla\phi\right|^2 + \frac{P}{\rho}+gz \right) = 0, $$ $$ \frac{\partial\phi}{\partial t} + \frac{1}{2}\left|\nabla\phi\right|^2 + \frac{P}{\rho}+gz = B(t). $$ This is the unsteady Bernoulli equation.

Solving for $\phi(r,\theta,t)$

Now we solve for the explicit form of $\phi$ for a sphere of radius $R$ moving through originally stationary fluid.

By incompressibility of fluid we have $\nabla\cdot\bvec{v}=0$, so $$ \nabla\cdot(\nabla\phi) = \nabla^2\phi = 0. $$ Thus $\phi$ satisfies the Laplace equation in spherical coordinates. The general axisymmetric solution is $$ \phi(r,\theta) = \sum^\infty_{l=0}\left(A_l r^l + \frac{B_l}{r^{l+1}}\right)P_l(\cos\theta). $$

Assume the sphere is moving along the $z$ axis at velocity $\bvec{u}(t)=u_z(t)\bvec{\hat{z}}$. In the rest frame of the sphere, this implies the boundary conditions $$ \phi(r,\theta,t) \to 0 \text{ as } r\to\infty, $$ $$ (-\bvec{u})\cdot\bvec{\hat{r}} = \bvec{v}\cdot\left.\bvec{\hat{r}}\right|_{r=R}. $$

Only the term $B_1$ remains subject to the B.C.'s, so $$ \phi(r,\theta,t) = u_z(t)\frac{R^3}{2r^2}\cos\theta, \quad r\ge R. $$

Force on the object

Only non-zero component of $\bvec{F}$ is $F_z$: $$ F_z = \int \mathrm{d}a_z \text{ }P_\text{surface}. $$ The integration can be carried out explicitly in spherical coordinates: $$ \mathrm{d}a_z = \cos\theta\mathrm{d}a = 2\pi R^2\sin\theta\cos\theta\mathrm{d}\theta, $$ $$ \frac{\partial\phi}{\partial t} = \dot{u_z}\frac{R^3}{2r^2}\cos\theta, $$ $$ \nabla\phi = -u_z\frac{R^3}{2r^2}\left(2\cos\theta\bvec{\hat{r}} + \sin\theta\bvec{\hat{\theta}} \right). $$

Then the Bernoulli equation gives $$ \begin{align} P &= B(t) - \rho\left.\left(\frac{\partial\phi}{\partial t} + \frac{1}{2}\left|\nabla\phi\right|^2 + gz\right)\right|_{r=R} \\ &= B(t) - \frac{1}{2}\rho\left( \dot{u_z}R\cos\theta + \frac{1}{4}u_z^2\left(1+3\cos^2\theta\right)+gz \right). \end{align} $$ Upon integration we have $$ F_z = -\frac{2}{3}\pi R^3 \rho \dot{u_z} = -\frac{1}{2}\rho V \dot{u_z}. $$

This is the extra resistance force experienced by a sphere moving with acceleration $\dot{u_z}$.

The quantity derived above is the added mass $$m_\text{added} = \frac{2}{3}\pi R^3 \rho $$ of a sphere moving in a fluid, with the physical intuition being that as an object accelerates in a fluid, so too must the fluid around it, so the force needed to accelerate the object should be corrected by $$ F = \left(m+m_\text{added}\right)a. $$

Back to the original problem, we now have $$ \begin{align} \bvec{F}_\text{B} &= \frac{4}{3}\pi R^3\rho g\bvec{\hat{z}} = - \left(\frac{4}{3}\pi R^3\rho_\text{air}g + \frac{2}{3}\pi R^3\rho g \right)\ddot{h}(t) \bvec{\hat{z}} \\ & \approx - \frac{2}{3}\pi R^3\rho g \ddot{h}(t) \bvec{\hat{z}}, \end{align} $$ since air is about $830$ times less dense than water. Finally, $$ h(t) = h_0 - gt^2, $$ $$ T = \sqrt{\frac{h_0}{g}}. $$

Note that this result probably isn't close to the actual value at all, as water is viscous. When considering viscosity, not only do we need to add a viscous resistance term, Bernoulli equation also doesn't apply anymore.

  • $\begingroup$ Awesome answer. Thanks. Unlikely to get a lot of upvotes as there's a lot of algebra to walk through and the question is ancient. But at least one person appreciated it (me). Correct me if I'm wrong - this answer seems to assume the volume of the bubble doesn't change much. If a bubble is rising to the surface of a liquid the pressure will change radically and the size of the bubble will also change radically. But I think you can just change $m_{added}$ to $m_{added}(t)$ and get the right answer. $\endgroup$
    – AXensen
    Dec 28, 2023 at 18:38
  • $\begingroup$ Yes you are correct. Assuming an always spherical shape, the $R$ term in the added mass should be $R(t)$, which can be determined by the equation of state of the gas inside the bubble. $\endgroup$
    – Jono94
    Dec 29, 2023 at 4:46

Your Answer

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

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