Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If you blow a bunch of soap bubbles outside, and a gust of wind hits them, will the bigger ones be more or less accelerated by the wind than the smaller ones?

Intuitively, and maybe from remembered experience, I expect the small ones to be accelerated more. But I'm not sure.

Based on theory I would think they'd be affected about the same, because:

$$a = \frac{F}{m}$$

$F$ proportional to cross sectional area, which is proportional to $r^2$

$m$ proportional to surface area, which is also proportional to $r^2$

I'm assuming that bigger bubbles tend to have about the same thickness as smaller ones, but I don't know that.

Similarly, damping due to air friction should affect big and small bubbles about the same, right?

share|improve this question
    
How the mass will scale depends on the size of the bubble. For a sufficiently large and thin bubble, the mass will be dominated by the enclosed air and scale as r^3. –  Anonymous Coward Apr 12 '11 at 17:59
    
@Anon, good point about the enclosed air (which @Georg also brought up). –  LarsH Apr 13 '11 at 11:16
1  
Anyway this is a pure academic problem since the inertia of a bubble is negligible; all the bubbles would be just flying with a wind flow. –  mbq Apr 15 '11 at 10:40
    
@mbq , that is the thing some answeres did refuse to believe. –  Georg Apr 15 '11 at 13:57

4 Answers 4

up vote 0 down vote accepted

F=ma. Which means $a = \frac{F}{m}$. To find out any variations in 'a' versus bubble size, we need to find how F/m varies with bubble size.

The force on the bubble due to wind is somewhere between $k r^2 v $ and $k r^2 v^2 $. In either case, for any given wind velocity of interest, it is proportional to $r^2$: $F = k_1 r^2$.

The mass of the bubble is made of two components: The mass of the bubble film (MF) and the mass of the enclosed air (MA). The mass of the film (MF) is proprtional to bubble surface area which is proportional to $r^2$: $MF = k_2 r^2$. The mass of the entrapped air (MA) is proportional to bubble volume which is proportional to $r^3$: $MA = k_3 r^3$. So...

$$F/m = \frac{k_1 r^2 }{MF + MA} = \frac{k_1 r^2 }{ k_2 r^2+k_3r^3} = \frac{k_1}{k_2+k_3 r}$$

So, acceleration due to wind is proportional to $\frac{k_1/k_3}{(k_2/k_3)+r}$, which means acceleration is INVERSELY proportional to $(k2/k3)+r$. So, as bubble size goes up acceleration of the bubble goes down: The velocities of bigger bubbles are less affected by the wind.

share|improve this answer
    
Jeff, this sounds good so far... still thinking it through. You have an initial typo (F = m a, not m/a), but I don't think that affects your answer. –  LarsH Apr 13 '11 at 19:35
    
@Jeff, "The mass of the film (MF) is proprtional to bubble surface area which is proportional to r^2" - this assumes that the thickness of the film is independent of r. Do we know that? I guess it's a reasonable assumption... –  LarsH Apr 13 '11 at 19:36
    
No, I assumed this; and seems valid in that thickness is likely governed by surface tension, which I would expect to remain fairly constant regardless of bubble size. If thickness does vary with r, then film mass varies with r^3, which means F/m would vary with 1/r rather than 1/(k+r), making my final conclusion more valid. –  Jeff_the_Engineer Apr 13 '11 at 19:44
    
Jeff, I went thru your answer and it makes sense to me. As I understand it, you're coming to a similar conclusion to Georg's, while sidestepping the controversial question of what drag law to use, and explaining your reasoning more clearly to this layman. On the other hand I don't know how to reconcile your answer with @Martin's... –  LarsH Apr 13 '11 at 20:12
    
@Jeff, regarding thickness, I agree that your conclusion remains valid if thickness varies with r. Even if thickness varies inversely to r, your conclusion is still valid. –  LarsH Apr 13 '11 at 20:15

Yes this is right. Friction is also proportional to cross-section which is r^2. So the acceleration should be approximately invariant. (The friction here is nothing but the wind in the opposite direction)

Here is another argument that makes the result much more clear:
Since these things are floating in the air, the mass of the bubble is irrelevant (ie the density is approximately that of air).
Second, any of these bubbles are very large compared to the microscopic properties of air.

Therefore we can treat the air as a continuous medium, and you have a completely scale-invariant situation where the size of the sphere have no influence.

share|improve this answer
    
Sounds plausible. But not concrete enough for me to verify... –  LarsH Apr 13 '11 at 20:28

This is a response to Georg:

The condition of applicability of the Stokes formula $F_d=6\pi\mu Rv$ is:

$$Re=\frac{\rho_aRv}{\mu}<<1$$ $\mu$ is the dynamic viscosity of air

From $mg=6\pi\mu Rv$; $m=4\pi R^2l\rho_w$ we get:

$$v=\frac{2\rho_wglR}{3\mu}$$ and $$Re=\frac{2\rho_a\rho_wglR^2}{3\mu^2}<<1$$ and

$$R<<\sqrt{\frac{3\mu^2}{2\rho_a\rho_wgl}}$$ Now $\mu=1.8*10^{-5}$ Pa*s, $\rho_a=1.2$ kg/m^3, $\rho_w=10^3$kg/m^3, $g=10$m/s^2, $l=10^{-7}$m

This gives: $$R<<0.1cm$$ We're not talking about such bubbles, i think.

share|improve this answer
    
If You read the whole thread, You will see that this was discussed in detail already. And: Your first "un"equation (...<< 1) is wrong. Stokes is applicable up to Re 10 at least. Turbulence onset is around 300 (but such values depend on geometry a lot, a sphere or some tubular flow do not have the same "critical" Re's) –  Georg Apr 15 '11 at 9:53
    
@Georg: If $Re=10$ then $R=0.3cm$. Are we talking about such bubbles? –  Martin Gales Apr 15 '11 at 10:44
    
Do not make a fool of yourself, Georg. Stokes flow holds at condition $Re<<1$. See wikipedia for ex. –  Martin Gales Apr 15 '11 at 10:53
    
""Are we talking about such bubbles? "" Yes, we are. Please note, that such values in dimensionless analysis are very coarse! And: You have to put in v and mass, both are not known! What drag law did You use this time? –  Georg Apr 15 '11 at 11:13

This is not a trivial question and it makes sense to consider at first the simplest case where a bubble falls freely in static air to find the bubble's terminal velocity.

$R$-radius of a bubble
$\rho_w$-density of water
$\rho_a$-density of air
$l$-thickness of the bubble

The force on a bubble due to gravity:

$$F_g=4\pi R^2l\rho_wg$$ Because it is difficult to decide when the actual drag force is proportional to $v$ or $v^2$ then to avoid Reynolds numbers lets choose a different way.

Let $v$ be the terminal velocity of the bubble. If the bubble falls distance $h$ then it transmits to the air mass $m=\pi R^2h\rho_a$ amount of energy $E=F_dh=\frac{mv^2}{2}$. From this drag force:

$$F_d=\pi R^2\rho_a\frac{v^2}{2}$$ So actual drag force is proportional rather to $v^2$, i think.

Finally from $F_g=F_d$ we get the terminal velocity of freely falling bubble:

$$v=\sqrt{\frac{8\rho_wgl}{\rho_a}}$$ Thus as a first approximation, terminal velocity of a bubble does not depend of it's radius but depends of it's thickness. So in general, answer to the question in title, i think, is: No

share|improve this answer
1  
Do You know the story when Baron Münchhausen pulled himself out of the swamp gripping his own hair? This is the "moral" of Your deduction. –  Georg Apr 13 '11 at 10:15
2  
@Georg: Show exactly where i am wrong. –  Martin Gales Apr 13 '11 at 11:07
    
@Martin, thanks for this insightful and careful answer, +1. @Georg, you told me to "look at the right places." Give us reason to think that your answer and comments are "right places." –  LarsH Apr 13 '11 at 11:21
    
What "drag law" was used to determine that "terminal velocity"? You alway get, what You feed in. –  Georg Apr 13 '11 at 12:24
    
@Georg, I can't tell whether you didn't read @Martin's answer carefully, or you know something that you're not telling us. Martin says he is taking a different way (i.e. not using a drag law) and it looks to me like he really is not using a drag law. I don't know enough to completely validate his derivation of Fd, but he's given me more reason to trust his answer than you have. –  LarsH Apr 13 '11 at 19:17

Your Answer

 
discard

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.