homework and exercises - What are the forces acting between two air bubbles in water?

The exact question is

Two air bubbles in water

1. attract each other


2. repel each other


3. do not exert any force on each other


4. may attract or repel depending upon the distance between them.

The chapter is about gravitation. The given answer is

A lighter body inside a denser medium behaves like negative mass as far as gravitational force is considered. Two air bubbles i.e. two negative masses will attract each other.

What is negative mass in this context and how can it be applied to such macroscopic objects? How would it result in attraction?

My reasoning is:

Consider the bubble A in the above image. The air particles forming the bubble A would be attracted more to the left(away from B) as there are more dense particles towards that side-the air particles making up bubble B are less dense than the medium and they will attract the air from bubble A to a smaller extent than if the volume of bubble B was filled with the medium.

A similar case would apply to B due to lesser density of particles forming A and the bubbles would be (indirectly) repelled.

So what is happening in this case?

Answer

Just consider the water, due to gravitational attraction(which I'm not sure how effective it is in this case); water molecules like to be as close to each other as possible. This means that they like to push the bubbles as close to each other as possible. Since the air has negligible mass, its gravitational forces can be neglected compared to the water ones.

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The other way to go through this reasoning is by what has been suggested in the question, i.e. assuming the bubbles have negative mass. This solution has few steps, as following:

• 
  

  Say we have a huge spherical lump of water(with the radius $R$), without any bubbles inside. The gravity potential of this sphere is $\frac{-3GM^2}{5R}$, but we are not going to use that.

  
  


• 
  

  Now, say we don't remove a small spherical part(radius $r$ and mass $m$) of the water and replace it with a same-sized sphere with density $\rho'$(or mass $m'$) but rather add it to the current sphere(a ghost like sphere which can only interact through gravity with the world). To calculate the gravitational force acting on this sphere using Shell's theorem, we also need to know the distance from the center; assume it's $x$. Since the sphere had been in equilibrium before, the new net force will be(note the force should be proportional to the mass of each object):

  
  

$$ -\frac{GM'(x)}{x^2}m' \tag{1}$$ where $M'(x)$ is the mass of water inside a sphere with radius $x$.

• The next step is to do the same with another small sphere of water. To make relations more simplified, I will put this one at $-x$. Now the net force on the first and second sphere will be:

$$F_1=-\frac{GM'(x)}{x^2} m' - \frac{G m' m'}{(2x)^2} \\ F_2=\frac{GM'(x)}{(x)^2} m' + \frac{G m' m'}{(2x)^2}$$

Or the accelerations:

$$a_1=-\frac{GM'(x)}{x^2} - \frac{G m' }{(2x)^2} \\ a_2=\frac{GM'(x)}{(x)^2} + \frac{G m' }{(2x)^2}$$

• In the case of our problem $m' = -m$, therefore:

$$a_1=-\frac{GM'(x)}{x^2} + \frac{G m }{(2x)^2},$$

($M'(x) \gg m$)which is towards the center(and the other sphere). So it looks like the two spheres are attracting each other.

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Now there are some ambiguities here:

1. Do negative masses behave mathematically consistent(we can use $F=m a$), which I have assumed to be the case.


2. Is this attraction duo to the big sphere of water or the other sphere? Looking at equation $(1)$, it seems to be the first case; unless my previous sign convention is wrong which simply means that the big sphere of water will try blow the bubbles away, although an attraction force between them(if they are closer than a certain distance they will attract).

Also I should point, if the bubbles get in touch; they will immediately collapse into a single bubble. This is due to the surface tension, not the gravitational effects for sure.