Curvature of fluid surface next to a floating sphere

Question

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.

I start with the Young-Laplace equation:

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

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.

Answer 1

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.