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If round ball $A$ and $B$ have weight 55 kg and 25 kg respectively and there distance is 20 meters then what would be the attraction force if we dive these two objects in the water?

I am assuming there is a downwards gravity $g$

There are two components here: downwards gravitational force and horizontal

Downwards Gravitational Force: This would be the weight of the object subtracted from the buoyant force $$\sum F_y=mg-\rho Vg$$ Horizontal Gravitational Force: This is simply our known equation $\sum F_x=\frac{GMm}{r^2}$

Now we simply take the vector sum of these two forces to get the resultant force. Or $$\sum F=\sqrt{F_x^2+F_y^2}=\sqrt{(mg-\rho V g)^2+\left(\frac{GMm}{r^2}\right)^2}.$$

Not sure if this helps get an answer.

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    $\begingroup$ While the horizontal force is negligible to begin with, note that you also have a horizontal buoyancy effect (for the same reason you have it vertically, i.e., because the water is also attracted horizontally) $\endgroup$
    – Hagen von Eitzen
    Commented Oct 27, 2019 at 21:03
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    $\begingroup$ You should check out the physics stack exchange instead $\endgroup$
    – gen-ℤ ready to perish
    Commented Oct 27, 2019 at 22:11

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I believe the intention of the question is to adjust the weight of the two balls due to the existing water medium, i.e. $m’= m -\rho V$. Then, their gravitational attraction force becomes,

$$F = \frac{G(m-\rho V)^2}{r^2}$$

instead of $\frac{Gm^2}{r^2}$ without water.

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  • $\begingroup$ I might be wrong, but the gravitational attraction is independent of the medium. While there is a buoyancy force on the balls, their masses do not change. $\endgroup$
    – Andrei
    Commented Oct 28, 2019 at 4:11
  • $\begingroup$ Note that the two masses are not points. They are round balls with volumes, which remove two balls of water and effectively reduce the masses. $\endgroup$
    – Quanto
    Commented Oct 28, 2019 at 4:26
  • $\begingroup$ That should be compensated by hydrostatic pressure $\endgroup$
    – Andrei
    Commented Oct 28, 2019 at 4:40

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