4
$\begingroup$

There seems to be general disbelief over this formula, so I'm challenging someone to show me the fallacy in the proof below.
The assumptions are that the two objects are rigid and spherically symmetric and that the fluid is of uniform density and incompressible. Note that the gravitational fields induce pressure gradients in the fluid and these affect the force between the two objects.
Proof:

  1. The mutual force between the two objects is measured as the change in total force on the first object when the second object is introduced.
  2. The introduction of the second object changes the mass distribution and the gravitational fields and the pressure gradients in the fluid.
  3. The change in mass distribution is such that a spherical volume with the second object's mass, $M_2$, replaces a spherical volume of fluid with mass, $m_2$. The change in mass is $M_2- m_2$.
  4. As a result of the change in mass, the gravitational field in the surrounding fluid also changes. For Newtonian gravity, the change in the field is $\Delta g = -G\frac{M_2-m_2}{r^2}$, where r is the distance from the center of the sphere. (Note: the change in the field also causes a change in the pressure gradients in the fluid.)
  5. As a result, there is a change in force on the first object. The force has two components, one arising from the change in gravitational force on the object and the other arising from the change in the pressure gradient across the object's surface (i.e. a change in buoyancy force).
  6. According to Newton, the first force is $\Delta g M_1$. According to Archimedes, the second force is -$\Delta g m_1$ where, again, $m_1$ is the mass of the displaced fluid. The net change in force is $\Delta g(M_1-m_1)$.
  7. Substituting $\Delta $g, gives us the final version of the formula. The mutual force between two spherically symmetric objects immersed in a uniform fluid is $-G\frac{(M_1-m_1)(M_2-m_2)}{r^2}$. The force acts along the line joining their centers.
$\endgroup$
7
  • $\begingroup$ I do now see a likely flaw in the 'proof''. Archimedes was obviously thinking about uniform pressure gradients. I have no idea what happens to the buoyancy component if the object is sitting in a radial pressure gradient. So it seems likely the formula is only an approximation that's useful when the objects are far apart compared with their size. $\endgroup$
    – Roger Wood
    Commented Nov 2, 2020 at 0:07
  • $\begingroup$ Perhaps surprisingly, according to Lima and Manteiro, Archimedes principle remains true in a nonuniform gravitational field, so I think the equation is correct even for small separations. researchgate.net/publication/… $\endgroup$
    – Roger Wood
    Commented Nov 2, 2020 at 4:39
  • 1
    $\begingroup$ I think that there is a semantic issue in this question that needs to be clarified. Do you consider the 'force due to gravity' of M2 from M1 to be the net change in gravitational forces acting on M2 after replacing a volume of fluid (at distance r) by the equal volume of the body M1, or the net change in gravitational force acting on M2 after replacing an 'empty' volume of space (at distance r) by the equal volume of the body M1. The former description matches your equation, but it doesn't represent the total gravitational interaction of M2 with M1. $\endgroup$
    – Penguino
    Commented Nov 2, 2020 at 21:16
  • 1
    $\begingroup$ I think you're perfectly correct, the only quibble is in defining what "the mutual force" means. I think your choice of definition is perfectly sensible, too. $\endgroup$
    – knzhou
    Commented Nov 4, 2020 at 3:50
  • 1
    $\begingroup$ I think this proof would be much easier to understand if there were a few pictures $\endgroup$ Commented Nov 25, 2020 at 6:53

3 Answers 3

6
$\begingroup$

I think there is a lot of confusion here. First of all, by Newtonian gravity, the gravitational force between two distinct objects depends only on their masses and the distance between them. That's it. There are GR corrections of course but we can consider those negligible for now.

Things already go a bit wrong for me in 2). From the question it seems like you are talking about the gravitational influence of one spherical object on another object, but then in 2) you're talking about the fluid, which has no influence on the gravitational influence between the two objects. Does the fluid exert gravitational influence? If it has energy of course it does, but it in no way affects the gravitational influence between the two spherical objects.

Consider this, you are essentially talking about 3 objects here: Spheres A and B, and the fluid C. There are three gravitational forces here, $F_{AB}$, $F_{AC}$, and $F_{BC}$. Your question asks about $F_{AB}$, which as I said must be $F_{AB} = \frac{G m_A m_B}{r^2}$, regardless of the fluid.

Now the gravitational force from the fluid does depend on its density and distribution, but it in no way relates to the equation proposed. Let me know if you have any more questions.

$\endgroup$
5
  • $\begingroup$ It's semantics of course, but, in saying, "force due to gravity", I'm intending that pressure gradients due to gravity be included. Otherwise, you are absolutely correct. $\endgroup$
    – Roger Wood
    Commented Nov 1, 2020 at 23:59
  • $\begingroup$ Hmm sorry if I'm confused but I still don't follow. You're using Newtonian gravity in your original proof no? There is no gravitational contribution from pressure in Newtonian gravity. Also I'm confused about your calculation of the gravitational influence of the fluid. That is a complicated formulation that depends completely on the density and distribution of the fluid, you would need to use Gauss's Law. I'm not sure how you calculate it here. $\endgroup$ Commented Nov 2, 2020 at 0:40
  • $\begingroup$ @ CuriousHegemon I'm assuming this is in static equilibrium, The presence of a gravitational field causes a pressure gradient equal to field times mass density. The pressure gradient in turn causes a force on surface of the the object. The application of Gauss's Law results in Archimedes principle which equates the force with the volume of the object times the mass density of the fluid times the gravitational field. The force is sometimes called buoyancy. If you switch the gravitational field off, the force goes away, so it's important that it be included.. $\endgroup$
    – Roger Wood
    Commented Nov 2, 2020 at 3:14
  • 1
    $\begingroup$ So if I understand you're talking about the net force on the spheres? Well obviously that's not equal to $\frac{G m M}{r^2}$ since the fluid is acting a force on the spheres. I'm a bit confused because either we're talking about the isolated gravitational force between the objects, which is completely unaffected by the liquid's interference, or we're talking about the net force on one of the spheres, which is obviously not just $F_G$. If it's the latter case, it would be good to clarify your question, it's very confusing if that's what you're trying to talk about. $\endgroup$ Commented Nov 2, 2020 at 12:59
  • $\begingroup$ @ CuriousHegemon Good point. I'll add the sentence: "Note that the gravitational fields induce pressure gradients in the fluid and these affect the force between the two objects" $\endgroup$
    – Roger Wood
    Commented Nov 2, 2020 at 20:46
3
+100
$\begingroup$

The formula $$ F = -G ~ \frac{\Delta M_1\Delta M_2}{r^2}$$ is correct. The force is the derivative of the energy with respect to the separation $r$. By displacing one of the immersed masses, say $M_1$, effectively only a mass $M_1-m_1$ is displaced. Indeed a volume of fluid with mass $m_1$ is displaced oppositely.

A very similar line of reasoning based on potential energy goes as follows. When a massive objects is added to the system, a mass $m_1$ is removed so effectively the mass added is $\Delta M_1$. Therefore an additional potential $V_1=-G\frac{\Delta M_1}{r}$ is added. Adding another object $M_2$ only adds mass $\Delta M_2$ which feels $V_1$, resulting in the potential energy $$ V_{12} = -G ~ \frac{\Delta M_1\Delta M_2}{r}$$

$\endgroup$
3
  • $\begingroup$ @ my2cts I like your answer. I realize now, I specified the objects in the question to be spherical but forgot to say that should also have uniform density (like the fluid). There's maybe a another interesting formula for arbitrarily shaped objects (again with uniform density): F_fluid = F_vacuum *((M1-m1)(M2-m2))/(M1M2) assuming that you keep both objects in the same orientation and with the same separation when you measure the force in a vacuum. $\endgroup$
    – Roger Wood
    Commented Nov 5, 2020 at 3:43
  • $\begingroup$ @RogerWood Then you have to integrate over these arbitrary volumes using the mass difference with the medium. $\endgroup$
    – my2cts
    Commented Nov 5, 2020 at 11:00
  • $\begingroup$ @ my2cts No need for integration for "F_fluid = F_vacuum *((M1-m1)(M2-m2))/(M1M2)", Isn't this valid for any shape as long as the mass densities in both the object and the fluid are uniform? So the difference between the objects in and out of the fluid also has constant mass density. This equation no longer says anything about the separation, r, and we wouldn't expect the 1/r^2 behavior to apply. $\endgroup$
    – Roger Wood
    Commented Nov 5, 2020 at 19:42
1
$\begingroup$

The formula is certainly correct (thanks for the endorsement @my2cts). To reiterate: it is important to appreciate that gravity also affects pressure gradients in the fluid and thus the forces on the objects.

The formula can be viewed as a generalization of Newton's Law of gravity or as a generalization of Archimedes principle or as a combination of the two. It is fascinating in that the forces produced are sometimes repulsive and that the inherent quadratic behavior becomes very apparent.

Referring back to "Two helium balloons at a cocktail party", the answer is 'yes', they do find each other attractive. The force is about a picoNewton (not huge). Interestingly, two CO2 ballons experience a similar attraction, even though in this case they are heavier than air. Perhaps most fascinating is that there is a similar magnitude force between one helium and one CO2 balloon, but in this case the force is repulsive!

This repulsive force can, of course, be interpreted as the CO2 balloon being attracted towards the greater bulk of the fluid (away from the He balloon) and the He balloon being pushed away by the buoyancy force that the CO2 balloon induces.

The formula does not require that the objects have uniform density but it does require that the objects (and the density profiles) have spherical symmetry. It does require that the fluid be incompressible and have uniform density but does not seem to require that both objects be immersed in the same fluid.

There is a variation on the formula that can be used on objects of arbitary shape. This relates the force seen when the objects are immersed in the fluid to the force seen when the objects are in a vacuum (note, it no longer says anything about how the force changes with separation).
Ffluid = Fvacuum (M1-m1)(M2-m2)/(M1M2)
This formula requires that both objects have uniform density and the fluid has uniform density (this ensures that the forces on the object and on the displaced fluid are simply scaled versions).

I don't know how to deal with compressible fluids (such as air) but in many cases the compression and density is already set by other factors. The tiny variations in density caused by the gravitational fields from the objects themselves can probably be ignored.

The formula is probably more of a curiosity, but maybe it could find application in the movement of mascons in the Earth's mantle and lithosphere, for example.

$\endgroup$

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.