# Balloon on Earth

1. Mar 11, 2013

### Gh778

Imagine a balloon with nothing in it except two masses M1 and M2, like first drawing. The gas of atmosphere apply a force to move up Fgas, and the weight apply a force to move down Fm1+2. Weight of M1+M2 is equal to the force from gas (opposite direction). Now, rotate M1 and M2, like that the weight is lower because gravity is a law like 1/d². Fgas>Fm1+2. The balloon attrack Earth to up ? I though it was the diametrical atmosphere (where balloon would be) but we can put a second balloon too without rotate masses (like second drawing). So, can you explain how forces works on Earth ?

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Last edited: Mar 11, 2013
2. Mar 12, 2013

### Gh778

Maybe it's easier to understand my problem, watch the drawing please. A circular solid SOLID has around it GAS (it's gravity that attrack gas around it). I create a hole HOLE1 at right, this give a force F1. The force F1 come from the pressure of gas. I create another hole but now in the SOLID at left. Normally, H3 gas with gravity must compensate F1, but like a hole exist in the solid, there is less gravity, so the force F2 is lower than F1. Where is the force that compensate 100% of F1 ?

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3. Mar 12, 2013

### Staff: Mentor

Each force has an opposing force of the same magnitude. That is just Newton. There cannot be any unbalanced force. Note that F1 is just an effective force - it is a difference between forces from air (if it would be present) to the forces in this setup (without air).

The balloon and the solid mass both pull on the string with the same force, for a total force of 0.

I really don't see the point in your 10th thread with nearly the same question.

4. Mar 13, 2013

### Gh778

I'm trying to explain with an easier example : take a circle with very thin wall, put in it only a mass M but not at the center. The mass of wall of the circle can be consider like 0. Put around the circle a gas. For me, the gas will be like the drawing is showing. But the law of attraction is like 1/d². At external of gas, the pressure is near 0. At the surface of the circle, the pressure is P. In this case, how is possible for the pressure of gas to be the same, because d1 different of d2 ?

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5. Mar 13, 2013

### Staff: Mentor

The pressure of a gas is not affected by the shape of its container and differences in gravitational force are only significant for large differences in d.

6. Mar 13, 2013

### Gh778

Each molecule of gas receive a force Fx like drawing is showing. The number of molecules at left is more important than at right, each molecule receive a force Fx function of the local pressure, ok. Imagine at right with 2 molecules, sum of Fx forces must be lower than at left, no ? There is no effect of quantification ?

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7. Mar 13, 2013

### Staff: Mentor

As said, all forces come in equal and opposite pairs, summing to zero.
I don't know what "quantification" means, but if you are asking if the fact that you have two simultaneous collisions matters, it does. The one molecule can gain more momentum than the individual two molecules had prior to the collision because the total momentum is conserved.

Not sure what that has to do with the previous post, though...

8. Mar 14, 2013

### Gh778

ok, thanks

I thinks I understood my error, in the post #4, the drawing is showing gas with a shape like egg but in fact gas (or liquid) have the same height on the black circle, it's that ? Consider circle like solar system (big d) and mass M like mass of solar system but not in the center.

Edit: I'm drawing an example, and trying to show force on gas (or liquid), the problem (for me) is the red forces, the shape for gas is or not a circle ?

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Last edited: Mar 14, 2013