B Coalescence of Drops: Exploring Kinetic Energy & Star Formation

[gianeshwar]

Hello friends

Coalesced drops form a bigger drop and there is extra energy spare.Now this energy is said to give kinetic energy to bigger drop.Which factor decides the velocity of coalesced drop?
Is it just any accidental push after coalescence?
I think also that the process is just feasible with delta G of system negative .
Can I compare it with p-p cycle of star formation as well.
Question is scattered but I want to see the underlying processes and compare them.
Thanks!

 

[Charles Link]

For coalesced drops, the energy that is gained is from surface tension, because the surface area of the combined drops is less than the sum of the parts. The extra energy can go into heating of the drops rather than into any translational motion. For stars, I think the attraction can be a combination of gravitational and electrostatic, but I have little expertise in that area.

 

[jbriggs444]

Clearly, the added energy cannot go into bulk translational motion. Conservation of momentum dictates the final velocity of the merged drop.

The energy can go at least in part to oscillations in the droplet, for instance, an oscillation between an oblate spheroid and a prolate spheroid. It could also lead to the ejection of smaller droplets.

 

[gianeshwar]

Thank you friends!
Anyway why not a translational motion possible .Is it possible to consume extra energy in translational motion or even spin motion if there is any accidental negligibly small foce or torque to start.

 

[jbriggs444]

Anyway why not a translational motion possible .Is it possible to consume extra energy in translational motion or even spin motion if there is any accidental negligibly small foce or torque to start.

Conservation of momentum and conservation of angular momentum will tell you how much linear velocity and how much spin results.

 

[sophiecentaur]

Anyway why not a translational motion possible

The answer to this question is basically why no one has come up with a 'reactionless drive', despite many attempts and fake solutions. Sit in a box, in space and try to move (or change the velocity of) your (you + box) centre of mass by moving about inside the box. Whatever you do - running against the side and smashing into it etc. - will not change the total Momentum.
Of all the principles ('laws') of Science, the Conservation of Momentum Law seems to be the most robust. We have not yet found a single exception.

 

[gianeshwar]

Thank you friends ! I understand that since coalesced drops had total linear momentum and total angular momentum zero so after coalescing there is neither any linear momentum nor any angular momentum of big drop.
Only an external force or torque or both could do so.
Similar things we can expect in star formation I think.
Delta G(Gibbs free energy) of this isolated system is negative I think .Bigger drop must be more stable.
Is my argument completely justified?
Thanks!