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.
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.
Now there are some ambiguities here:
- Do negative masses behave mathematically consistent(we can use $F=m a$), which I have assumed to be the case.
- 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.
There are many fundamental concepts that both you and Trimok have misunderstood.
First of all, you can completely ignore the mass energy. The kinetic energy in no way "compensates" for the loss in mass energy because no mass energy is lost in the example that you gave. Sure, the mass of the rocket decreased as it ejected propellant to move forward, but that mass hasn't disappeared. The propellant is still out there, floating around in space. You can't just ignore its mass energy because it's not in the rocket anymore, it still exists. In other words your total $E_m$ term is always the same throughout this situation, at no point is matter converted into another form of energy, like kinetic energy as you seem to suggest.
Second, this is conceptually wrong :
The gravitational attraction between me and the earth will significantly decrease and after a certain distance it will cease to act as the inertia of my rocket would be bigger than the gravitational force between my rocket and the earth.
You can't compare the inertia of a system with a force (they're not even measured in the same unit). One can't be "greater" than another. Regardless of the mass of an object, however small a force you apply on it it will still have an impact. There's no cutoff. This doesn't matter much because indeed, you can get to a point where this is small enough that you consider yourself to be in a situation where the potentially energy is now maximal (you seem to think that because this is 0 the potential energy has now disappeared, but this is not the case, it is sometimes chosen to be 0 at infinity but keep in mind that this is still an increase because in such conventions this potential energy is negative at lift off). Potential energy increases as you get further from the earth. This is because, intuitively enough, now that you are further from the earth you now have the potential to gain more kinetic energy by falling back towards the earth through a longer distance, gaining more speed in the process.
Finally, let's work out the actual energy distribution throughout all this. Like I said we can ignore mass energy because the total mass is conserved. Instead what we actually need to consider is :
$E_k$ : the kinetic energy of the rocket.
$E_c$ : the chemical potential energy stored in the propellant. It is this energy that will propel the rocket, not mass energy as you seem to think.
$E_g$ : the gravitational potential energy.
In the beginning, the rocket is sitting on the surface of the earth, and $E_c$ is maximal, $E_g$ is minimal, and $E_k$ is just $0$. As you start burning propellant, you release the energy stored in your propellant and $E_c$ begins to decrease as $E_k$ increases. To completely compute $E_k$ you actually have to take into account both the kinetic energy of the rocket and the kinetic energy of the ejected propellant. If you do, you will notice that this is actually not enough to make up for the loss in $E_c$. Indeed, the difference between the two corresponds to the gain in $E_g$ as the rocket goes further and further away from the earth, and indeed we will always have conservation of $E_k + E_c + E_g$.
Best Answer
Your thought experiment of dropping an iron ball and a feather need not be in water; in fact, it is more commonly considered in air, but the pertinent facts are the same.
All objects, regardless of their mass or composition, are accelerated identically by gravity.
But within a particular medium, the acceleration of particular objects might be impeded by greater resistance than that of other objects. In ordinary air, your feather will fall more slowly than an iron ball, because of air resistance.