# Curvature of fluid surface next to a floating sphere

I want to estimate the curvature $$\kappa$$ of the fluid surface next to a floating sphere. The situation is static and shown here:

The fluid density is $$\rho$$, downward gravity is $$g$$, sphere radius is $$R$$, and solid-liquid-air surface tension is $$\sigma$$.

I know that:

1) The sphere is half submerged, so that the free surface coincides with the sphere center.

2) $$\frac{\sigma}{\rho g R^2} << 1$$, which implies that surface tension effects are negligible compared to gravitational effects.

$$P_i - P_o = \sigma (\frac{1}{R} + \frac{1}{R_2})$$

where one radius of curvature is the sphere radius $$R$$, and $$\kappa = \frac{1}{R_2}$$ is the curvature of the free surface.

Now, use a simple hydrostatic force balance to find $$P_i$$. Note that the pressures at points 1 and 2 below are the same.

Since we have $$P_1 = P_2$$, this implies that

$$P_i = P_{atm} - \rho g h$$

The outer pressure $$P_o = P_{atm}$$.

Substituting into Young-Laplace,

$$-\rho g h = \sigma (\frac{1}{R} + \kappa)$$

Now, since surface tension effects are negligible, $$\rho g h \sim 0$$ and the final result is:

$$\kappa \sim - \frac{\sigma}{R}$$

I'm confused by the negative sign, and I'm not sure if my $$\rho g h \sim 0$$ assumption is valid.

• Near the surface of the sphere you cannot ignore surface tension effects. In your analysis, you use $R$ as the characteristic length scale for the Bond number however the region which you are considering is not characterized $R$. A more appropriate length scale may be the capillary length. – Ragnar Dec 31 '18 at 17:50
• @PapaZulu, what is the "capillary length" here? I thought it would be the sphere circumference $2 \pi R$. Note that capillary length and capillary rise are not necessarily the same thing, correct? – Drew Dec 31 '18 at 18:49

I found this figure online from a book titled "Capillarity and wetting phenomena : drops, bubbles, pearls, waves" by de Gennes et al. According to the figure an estimate of the curvature near a floating sphere is given by $$\kappa = 1/\ell_c$$ where $$\ell_c = \sqrt{\sigma/\rho g}$$ is the capillary length. The capillary length is the length scale where surface tension and gravity forces are of the same order of magnitude.
• That's a helpful source you cited, and I think you're right about the capillary length. However, in my problem I'm given $\frac{\sigma}{\rho g R^2} << 1$, which may not be the case in your example. I'm not sure how to use this or how to interpret it. – Drew Dec 31 '18 at 22:23