Upward acceleration of submerged bubble based on radius

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 6K views
pjackson
Messages
2
Reaction score
0
Hi all!
Im a computer program that needs to simulate the motion of some bubbles. Id like to get a physically correct formula for my bubbles upward acceleration.

I have a perfectly spherical bubble of air of known radius. It is submerged in water. Discounting any friction or surface tensiony down force, what is the formula for its upward acceleration?

What is my starting point for working this out? All I've been able to find are a bunch of gas laws which I can't see how they relate.. Would it be possible to point me in the right direction for this?
 
Physics news on Phys.org
I feel a bit silly that I can click on the word bubbles in my post and be taken to a page that gives me a formula that is nearly there-

https://www.physicsforums.com/mgc_gloss/latex_images/mgc_gloss_equation_319-0.png

However, this is giving me a velocity. I am assuming this is a terminal velocity acheived when the drag of the water levels out with the up force. It would be good to have the underlying acceleration.
 
Last edited by a moderator:
As a simplified analysis you could consider the initial upward force on the bubble is just the Archimedes upthrust and is equal to the weight of the volume of water displaced by the bubble.
F= (4/3) π r³ d g
where r is bubble radius and d is density of liquid. g=9.8 m/s/s
Against that you have the very small downwards force of the weight of the air in the bubble.
Same formula but use density of gas.
The initial acceleration is given by F=ma
F is resultant force and m is mass of gas bubble. [=(4/3) π r³ d]
However, the bubble quickly reaches a terminal velocity.
It also changes shape; and there's other factors in there such as viscosity and surface tension.