In the vastness of space, there is a sphere made of oil floating. At what diameter will the total pressure at the center of the sphere, due to gravity and surface tension, be the smallest? (The density of oil is $0.8 \mathrm{~g} / \mathrm{cm}^{3}$, surface tension $\alpha=30$ dyn $/ \mathrm{cm}$.
In the center of a sphere, the pressure is composed of the curvature pressure due to surface tension and the hydrostatic pressure.

The curvature pressure in a fluid sphere with radius $R$ and surface tension $\alpha$ (for example, the solution to problem 1248)

$$
p_{g}=2 \alpha / R
$$

The hydrostatic pressure in a fluid layer with thickness $h$ and density $\varrho$ under gravitational acceleration $g$ is

$$
p=\varrho \cdot g \cdot h
$$

In our problem, the issue is that the gravitational acceleration decreases as we move towards the center of the fluid sphere. At a distance $r$ from the center, a force

$$
g(r)=f \frac{(4 / 3) r^{3} \pi \varrho}{r^{2}}
$$

acts on a unit mass, since the mass of the sphere with radius $r$ is $(4 / 3) r^{3} \pi \varrho$, and the gravitational field strength of the parts located beyond $r$ is zero, similar to the electric field of a charged metal sphere.

Let's divide the sphere into thin spherical layers such that the variation of $g(r)$ within a layer can be neglected. Then the hydrostatic pressure in the layers with thickness $\Delta r$ is approximately

$$
p_{h} \approx \Sigma \varrho \cdot g(r) \cdot \Delta r
$$

which, as the division becomes finer, approaches the integral

$$
p_{h}=\int_{0}^{R} \varrho g(r) \mathrm{d} r
$$

Using (1),

$$
p_{h}(R)=(4 / 3) \pi \cdot f \cdot \varrho^{2} \cdot \int_{0}^{R} r \mathrm{~d} r=(2 / 3) \pi \cdot f \cdot \varrho^{2} \cdot R^{2}
$$

The total pressure is the sum of $p_{g}$ and $p_{h}$

$$
p(R)=2 \alpha / R+(2 / 3) \pi f \varrho^{2} R^{2}
$$

We need to find the minimum of this function. The necessary condition for a minimum is that the first derivative

$$
p^{\prime}(R)=-2 \alpha / R^{2}+(4 / 3) \pi f \varrho^{2} R=\left(2 / 3 R^{2}\right)\left(2 \pi / \varrho^{2} R^{3}-3 \alpha\right)
$$

is zero, from which

$$
R=R_{0}=\sqrt[3]{\frac{3 \alpha}{2 f \pi \varrho^{2}}}
$$

From (2), it is clear that for $R>R_{0}$, $p^{\prime}(R)>0$, so the function $p(R)$ is increasing; for $R<R_{0}$, $p^{\prime}(R)<0$, so the function $p(R)$ is decreasing. Therefore, the pressure indeed has a minimum at $R=R_{0}$.

With the given data, $R_{0}=6.95 \mathrm{~m} \approx 7 \mathrm{~m}$, so the diameter of the sphere is $14 \mathrm{~m}$. The pressure at the center of the oil sphere is $1.4 \cdot 10^{-6} \mathrm{kp} / \mathrm{cm}^{2}$.

László Sparing (Szombathely, Nagy Lajos Gymnasium, 4th grade)