Kvant Physics Problem 154

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Problem

Estimate how many water droplets are contained in $1~\text{м}^3$ of fog if the visibility is 10 m and the fog settles in 2 hours. The height of the fog layer is 200 m.

The air resistance force acting on a water droplet of radius $R~\text{(м)}$ moving with speed $v~\text{(м/с)}$ is equal to $4{,}3Rv~\text{(Н)}$.

P. L. Kapitsa

Setup and Assumptions

The system consists of small spherical water droplets suspended in still air forming a fog layer of uniform thickness $H = 200~\text{m}$. Each droplet has radius $R$ and density $\rho = 1000~\text{kg/m}^3$. The air is treated as a stationary medium, and droplets are assumed to move vertically under gravity with rapid relaxation to terminal velocity.

The drag force acting on a droplet is given by $F_\text{drag} = 4.3 R v$, where $R$ is measured in meters and $v$ in $\text{m/s}$. The buoyant force of air is neglected compared with the droplet weight. Droplets are assumed spherical, identical in size, and uniformly distributed in space.

The visibility distance is $L = 10~\text{m}$, interpreted as the extinction length for light intensity in the fog. The fog clears in a time $t = 2~\text{h} = 7200~\text{s}$, which corresponds to complete sedimentation of droplets through the layer height $H$.

The unknown quantity is the number density of droplets $n$, and the total number of droplets in $1~\text{m}^3$ equals $n$.

Physical Principles

The motion of each droplet at terminal velocity satisfies force balance between gravity and drag,

$$mg = 4.3 R v,$$

where the droplet mass is

$$m = \frac{4}{3}\pi R^3 \rho.$$

Light attenuation in a dilute suspension of particles follows the exponential law

$$I(x) = I_0 e^{-n \sigma x},$$

where $\sigma = \pi R^2$ is the geometric cross section and $n$ is the number density. The visibility length $L$ is defined by the condition $n \sigma L \sim 1$.

The sedimentation time relates to terminal velocity by

$$v = \frac{H}{t}.$$

Derivation

The terminal velocity follows from substituting the droplet mass into the force balance,

$$\frac{4}{3}\pi R^3 \rho g = 4.3 R v.$$

Canceling one factor of $R$ gives

$$\frac{4}{3}\pi R^2 \rho g = 4.3 v,$$

and solving for $R^2$ yields

$$R^2 = \frac{4.3 v}{\frac{4}{3}\pi \rho g}.$$

The settling velocity is obtained from the clearing time of the fog layer,

$$v = \frac{200~\text{m}}{7200~\text{s}} = 2.78 \times 10^{-2}~\text{m/s}.$$

Substituting numerical values with $\rho = 1000~\text{kg/m}^3$ and $g = 9.8~\text{m/s}^2$ gives

$$R^2 = \frac{4.3 \cdot 2.78 \times 10^{-2}}{\frac{4}{3}\pi \cdot 1000 \cdot 9.8}.$$

The numerator equals $1.20 \times 10^{-1}\text{N}\cdot\text{s/m}$, while the denominator equals $4.19 \times 10^{4}\text{kg/(m·s}^2)$. This gives

$$R^2 = 2.9 \times 10^{-6}~\text{m}^2,$$

and therefore

$$R = 1.7 \times 10^{-3}~\text{m}.$$

The optical condition for visibility $L = 10~\text{m}$ gives

$$n \pi R^2 L \sim 1,$$

so

$$n = \frac{1}{\pi R^2 L}.$$

Substituting $R^2 = 2.9 \times 10^{-6}~\text{m}^2$ yields

$$n = \frac{1}{\pi \cdot 2.9 \times 10^{-6} \cdot 10}.$$

Evaluating the denominator gives $9.11 \times 10^{-5}~\text{m}^2$, hence

$$n = 1.1 \times 10^{4}~\text{m}^{-3}.$$

Result

The number of droplets in $1~\text{m}^3$ of fog equals the number density,

$$n = \frac{1}{\pi R^2 L}, \quad R^2 = \frac{4.3 v}{\frac{4}{3}\pi \rho g}, \quad v = \frac{H}{t}.$$

Substitution gives

$$n = \frac{1}{\pi \cdot 2.9 \times 10^{-6}\text{m}^2 \cdot 10\text{m}} = 1.1 \times 10^{4}~\text{m}^{-3}.$$

$$\boxed{n \approx 1 \times 10^{4}~\text{droplets/m}^3}$$

Sanity Checks

The dimension of $R^2$ follows from $\frac{v}{\rho g}$, which gives $\text{m}^2$ since $v$ has units $\text{m/s}$ and $\rho g$ has units $\text{kg/(m}^2\text{s}^2)$, making the expression consistent.

The radius $R \sim 10^{-3}~\text{m}$ corresponds to millimeter-scale droplets, which is large for fog but consistent with rapid sedimentation over a two-hour clearing time of a 200 m layer.

The extinction estimate $n \pi R^2 L \sim 1$ gives an optical depth of order unity over $10~\text{m}$, matching the stated visibility condition.

The dominant sensitivity lies in the drag law coefficient; a factor-of-two change in the constant $4.3$ would change $n$ only linearly, leaving the order of magnitude $10^{4}~\text{m}^{-3}$ stable.