Two helium balloons meet a cocktail party. Do you think they might find each other attractive?

Question

Here we are talking about the mutual force between the two balloons. This is in the context of Newton's law of gravity (so electrostatics etc. are ignored). You can assume that the cocktail party is held in a room that contains air. It is not necessary to assume that the Earth's gravitational field is present, but if it were present, the helium balloons would tend to rise to the ceiling.

[Edit] Question was re-posted: Is the force due to gravity between two objects immersed in a fluid correctly given by$ -G(M_1-m_1)(M_2-m_2)/r^2$? ($m$ = mass of displaced fluid)

Better present your question: how are positionned the 2 balloons relatively to the gravitationnal field? — dan, Oct 31 '20 at 17:10
@ dan The magnitude of the attraction depends only on the relative separation of the balloons and on their sizes and is independent of the Earth's gravitational field. I'm sensing some scepticism, so I'll prepare a proof of the formula and edit the answer later this afternoon. — Roger Wood, Oct 31 '20 at 18:45
@ dan The only relevant details are that there is air in the room and that the balloons are lighter than air. Should I add that information to the question? — Roger Wood, Oct 31 '20 at 18:56
@ uhoh this question/answer might interest you. This line of enquirey was prompted by your own question on Keplerian orbits. https://physics.stackexchange.com/questions/453460/which-mass-distributions-guarantee-two-bodies-have-non-keplerian-orbits-which-n/587819?noredirect=1#comment1324761_587819 — Roger Wood, Nov 01 '20 at 02:56
This question can be reopened if stated more clearly. However there is a post closure duplicate : https://physics.stackexchange.com/questions/591103/is-the-force-due-to-gravity-between-two-objects-immersed-in-a-fluid-correctly-gi — my2cts, Nov 03 '20 at 22:47
@David White Opposites attract in electrostatics, but repel in gravitostatics — Roger Wood, Feb 24 '21 at 19:35

Roger Wood · Answer 1 · 2020-10-31T23:22:23.403

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The force due to gravitational effects between two spherical objects, A and B, immersed in a fluid is
-G(M_A-m_A)(M_B-m_B)/r² where M is the mass of an object and m is the mass of the fluid it displaces.
For two helium balloons, both (M_A-m_A) and (M_B-m_B) will be negative, so the balloons will find each other attractive. If one of the balloons is filled with CO2, they will find each other quite repulsive. If both balloons are filled with CO2, they will again be attracted to each other.
[It's probably bad form to answer one's own question, but I wasn't aware of this result and I thought it was rather cute]
I'm editing this to add a proof since there has been some scepticism expressed about the formula:
Proof:

The attraction between the two balloons is measured as the change in total force on the first balloon, A, when the second balloon, B, is introduced.
The introduction of the second balloon changes the mass distribution and the gravitational fields and the pressure gradients in the air.
The change in mass distribution is such that a spherical volume of helium with mass, M_B, replaces a spherical volume of air with mass, m_B. The change in mass is M_B - m_B.
As a result of the change is mass, the gravitational field in the surrounding air also changes. The change in field is $\Delta$g = -G(M_B-m_B)/r², where r is the distance from the center of the sphere.
Note: the change in field also causes a change in the pressure gradients in the air.
As a result, there is a change in force on the first balloon, A. The force has two components, one arising from the change in gravitational force on balloon A and the other arising from the change in pressure gradient across the balloon A's surface (i.e. it's buoyancy).
According to Newton, the first force is $\Delta$g.M_A. According to Archimedes, the second force is -$\Delta$g.m_A where, again, m_A is the mass of the displaced air. The net change in force is $\Delta$g(M_A-m_A).
Substituting $\Delta$g, gives us the final version of the formula. The force between the two (spherical) helium balloons in air is -G(M_A-m_A)(M_B-m_B)/r² operating along the line joining their centers.

edited Oct 31 '20 at 23:22

answered Oct 31 '20 at 00:28

Roger Wood

2,393

2

the gravitational force is always attractive – Oct 31 '20 at 00:41
I don't know where you got this formula from, but its just wrong. Gravitational force between two objects is always attractive, and its magnitude is independent of intervening medium. – dnaik Oct 31 '20 at 02:13
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@dnaik When an object is immersed in a fluid, the net field around the object is g = -G(M-m)/r^2. Furthermore, the force on an object is immersed in a fluid due to a field is g(M-m). That is why a helium balloon rises in air (but falls in a vacuum) – Roger Wood Oct 31 '20 at 02:55
@dnaik buoyancy is a force due to gravity. I'll modify the answer to say "force due to gravity" rather than "gravitational force". Do you agree that the formula correctly describes the net force between the two objects? – Roger Wood Oct 31 '20 at 03:26
No. Your formula only stands if the balloons are placed vertically one above the other, in an external gravitational field like that of the earth. If they are placed in a fluid in no external gravity, or if they are placed in a fluid on the earth, but horizontally next to each other, then they will move towards each other irrespective of what thay are filled with. – dnaik Oct 31 '20 at 03:35
@dnaik do you agree that g(M-m) correctly describes the net force on an object in a fluid in a field, g? – Roger Wood Oct 31 '20 at 03:42
yes that is correct – dnaik Oct 31 '20 at 04:38
@dnaik how about -G(M-m)/r^2 as the change in field when the object is introduced and the corresponding fluid displaced? – Roger Wood Oct 31 '20 at 05:00
Let us continue this discussion in chat. – dnaik Oct 31 '20 at 05:01
@ Wolphram jonny It's a matter of perspective. A bubble of air surrounded by water will appear to have a repulsive gravitational field radiating from its center. A nearby grain of sand will be repelled from the bubble, although it's probably more correct to say that it is attracted towards the greater bulk of the surrounding water. Nevertheless, the field appears to radiate from the bubble. – Roger Wood Oct 31 '20 at 23:28
buoyancy acts vertically, and as Dnaik said, there is no horizontal effect. Otherwise please provide a demostration (I mean prove it) – Nov 01 '20 at 14:39
@RogerWood Physics SE has some shortcomings compared to smaller sites and it often spectacularly disappoints! Chemistry, Matter Modeling, Space Exploration, Astronomy and Earth Science are all much better just for example, and Electronics is even worse. Across sites I've seen a strong positive correlation between size and angst. – uhoh Nov 03 '20 at 01:48
See this answer in Physics meta and Is there such a thing as an SE angst-o-meter? Some attempt to quantify on a site-by-site basis? – uhoh Nov 03 '20 at 01:48
The antagonistic commenters haven't been introduced to solid state physics yet. The negative mass is an unintuitive concept and in these situations, a more familiar picture is found by considering a positive charge with a positive mass. A beauty of Stack Exchange and simultaneously its Achilles heel is that anybody on the internet has a 20% vote in closing a question and shutting down a discussion, and some people do it as sport. – uhoh Nov 03 '20 at 01:52
@ uhoh Hah, thanks for the moral support :-) I'm certainly not going to wade into that Physics meta discussion, but I can see how much effort you have put in! I'm actually enjoying the interactions though folks around here do seem a little grumpy. I'll persevere in asking light-hearted questions that have some serious underlying physics. https://physics.stackexchange.com/questions/591363/can-i-destroy-bennu-if-i-shoot-it-with-my-electron-gun – Roger Wood Nov 03 '20 at 05:07

Two helium balloons meet a cocktail party. Do you think they might find each other attractive?

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