When an object enters water with high velocity, (like in Why is jumping into water from high altitude fatal?), most of it's kinetic energy will be converted, eg to accelerate water, deform the object etc. -
What is the relevance of the surface tension to this?

Are the effects related to surface tension just a small part, or even the dominant part regarding the forces.

  • $\begingroup$ practice pools use bubblers to break the surface tension, so they at least contribute $\endgroup$ – ratchet freak Apr 4 '14 at 16:21
  • 2
    $\begingroup$ @ratchetfreak: The bubblers used in diving practice operate by reducing the density of the water by interspersing it with a large volume of air, not by eliminating surface tension. $\endgroup$ – DumpsterDoofus Apr 4 '14 at 17:04
  • $\begingroup$ @DumpsterDoofus actually thinking about it the bubbler would provide a compressible substance which cushions better then "solid" water $\endgroup$ – ratchet freak Apr 4 '14 at 17:18

Unless I have made a conceptual mistake (which is very possible), surface tension plays essentially no role in the damping of the impact of a fast-moving object with a liquid surface.

To see this, a simple way to model it is to pretend that the water isn't there, but only its surface is, and see what happens when an object deforms this surface. Let there be a sphere of density $\rho=1.0\text{g/cm}^3$ and radius $r=1\text{ft}$ with velocity $v=200\text{mph}$, and let it collide with the interface and sink in halfways, stretching the interface over the surface of the sphere.

Before the collision, the surface energy of the patch of interface that the sphere collides with is $$E_i=\gamma A_1=\gamma\pi r^2$$ and after collision, the stretched surface has a surface energy of $$E_f=\gamma A_2=2\gamma\pi r^2$$ and so the energy loss by the sphere becomes $$\Delta E=E_f-E_i=\gamma\pi r^2$$ which in the case of water becomes (in Mathematica):

<< PhysicalConstants`
r = 1 Foot;
\[Gamma] = 72.8 Dyne/(Centi Meter);
Convert[\[Pi] r^2 \[Gamma], Joule]

0.0212477 Joule

Meanwhile, the kinetic energy of the ball is $$E_k=\frac{1}{2}\left(\frac{4}{3}\pi r^3\rho\right)v^2$$ which is:

\[Rho] = 1.0 Gram/(Centi Meter)^3;
v = 200 Mile/Hour;
Convert[1/2 (4/3 \[Pi] r^3 \[Rho]) v^2, Joule]

474085 Joule

and hence the surface tension provides less than one millionth of the slowdown associated with the collision of the sphere with the liquid surface. Thus the surface tension component is negligible.

I'd suspect that kinematic drag provides most of the actual energy loss (you're basically slamming into 200 pounds of water and shoving it out of the way when you collide), but I've never taken fluid dynamics so I'll await explanations from people with more experience.


It all depends on the length scale. At human scale, surface tension is sure not to contribute. At the one of an insect, it will.

The Weber number allows one to compare surface tension to inertial effects. (http://en.wikipedia.org/wiki/Weber_number)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.