Some planet consists entirely of an incompressible liquid of density $\rho$. The temperature in the depth of the planet is constant. Find the dependence of pressure on depth. The radius of the planet $R$.
Decision:
The distance from the given point to the center of the planet will be characterized by the radius $r$. It is required to find the dependence of the pressure $p$ inside the planet on $r$, i.e. the function $p(r)$.
Consider a narrow column of liquid located along the radius of the planet. One base of the column is at a distance r from the center of the planet, and the other is at the surface. Let us denote the cross-sectional area of the column by $S$. The resultant force acting on this column is the sum of the pressure force
$F_{1}=p(r)S$, (1)
directed along the radius toward the surface, and of the gravitational force $F_{2}$ directed in the opposite direction. Let us find the force of gravity $F_{2}$. The difficulty of its calculation consists in taking into account that different small elements of the column under consideration are subject to different gravity forces.
Let us select a small element of the column, the lower base of which is located at a distance $r$ from the center of the planet, and the upper one at a distance $r + dr$. The volume $dV$ of this element is given by the formula $dV = S dr$, and the mass $dm = \rho dV$. This element of the column is acted upon by the force of gravity from the fluid enclosed in a ball of radius r with mass $M(r) = \rho \frac{4}{3} \pi r^{2}$. The resultant action on mass $dm$ from the rest of the planet is zero. The action of a ball of mass $M(r)$ on mass $dm$ is equivalent to the action of a point mass $M(r)$ located at the center of the ball. Given this observation, for the force $dF$ acting from the side of the ball of mass $M(r)$ on the mass $dm$, we can write:
$dF=G \frac{M(r)dm}{r^{2}}= \frac{4}{3} \pi \rho^{2} Srdr$.
Dividing both parts of this equality by $dr$, we obtain
$f(r)=\frac{dF}{dr}= G \frac{4}{3} \pi \rho^{2} Srdr $,
where $f(r)$ is the force of gravity acting on a column element of unit length located at a distance r from the center of the planet; $f(r)$ is a linear function of r. The resulting gravity force $F_{2}$ acting on the entire column under consideration is numerically equal to the area of the figure bounded by the graph of the function $f(r)$ from above, by the abscissa axis from below, and by the initial and final ordinates from the sides:
$F_{2}=\frac{Rf(R)-rf(r)}{2}=G \frac{2}{3} \pi \rho^{2} S (R^{2}-r^{2})$. (2)
Since the fluid column under consideration is in equilibrium, $F_{1}=F_{2}$. Equating the right-hand sides of Equations (1) and (2), we find:
$p(r)= G \frac{2}{3} \pi \rho^{2} (R^{2}-r^{2})$
Passing from the variable $r$ to the "depth" $h=R-r$ in this expression, we obtain
$p(r)= G \frac{4}{3} \pi R \rho^{2} h \left ( 1-\frac{h}{2R} \right )= \rho gh \left ( 1-\frac{h}{2R} \right )$.
Here $g=G \frac{4}{3} \pi R \rho$ is the free-fall acceleration at the surface of the planet. At shallow depths ($h \||| R$), the pressure is described by the well-known expression
$p(h) \approx \rho gh$.
The pressure at the center of the planet (at $h=R$) is equal to
$p(0)= \frac{\rho gR}{2}$.
Decision:
The distance from the given point to the center of the planet will be characterized by the radius $r$. It is required to find the dependence of the pressure $p$ inside the planet on $r$, i.e. the function $p(r)$.
Consider a narrow column of liquid located along the radius of the planet. One base of the column is at a distance r from the center of the planet, and the other is at the surface. Let us denote the cross-sectional area of the column by $S$. The resultant force acting on this column is the sum of the pressure force
$F_{1}=p(r)S$, (1)
directed along the radius toward the surface, and of the gravitational force $F_{2}$ directed in the opposite direction. Let us find the force of gravity $F_{2}$. The difficulty of its calculation consists in taking into account that different small elements of the column under consideration are subject to different gravity forces.
Let us select a small element of the column, the lower base of which is located at a distance $r$ from the center of the planet, and the upper one at a distance $r + dr$. The volume $dV$ of this element is given by the formula $dV = S dr$, and the mass $dm = \rho dV$. This element of the column is acted upon by the force of gravity from the fluid enclosed in a ball of radius r with mass $M(r) = \rho \frac{4}{3} \pi r^{2}$. The resultant action on mass $dm$ from the rest of the planet is zero. The action of a ball of mass $M(r)$ on mass $dm$ is equivalent to the action of a point mass $M(r)$ located at the center of the ball. Given this observation, for the force $dF$ acting from the side of the ball of mass $M(r)$ on the mass $dm$, we can write:
$dF=G \frac{M(r)dm}{r^{2}}= \frac{4}{3} \pi \rho^{2} Srdr$.
Dividing both parts of this equality by $dr$, we obtain
$f(r)=\frac{dF}{dr}= G \frac{4}{3} \pi \rho^{2} Srdr $,
where $f(r)$ is the force of gravity acting on a column element of unit length located at a distance r from the center of the planet; $f(r)$ is a linear function of r. The resulting gravity force $F_{2}$ acting on the entire column under consideration is numerically equal to the area of the figure bounded by the graph of the function $f(r)$ from above, by the abscissa axis from below, and by the initial and final ordinates from the sides:
$F_{2}=\frac{Rf(R)-rf(r)}{2}=G \frac{2}{3} \pi \rho^{2} S (R^{2}-r^{2})$. (2)
Since the fluid column under consideration is in equilibrium, $F_{1}=F_{2}$. Equating the right-hand sides of Equations (1) and (2), we find:
$p(r)= G \frac{2}{3} \pi \rho^{2} (R^{2}-r^{2})$
Passing from the variable $r$ to the "depth" $h=R-r$ in this expression, we obtain
$p(r)= G \frac{4}{3} \pi R \rho^{2} h \left ( 1-\frac{h}{2R} \right )= \rho gh \left ( 1-\frac{h}{2R} \right )$.
Here $g=G \frac{4}{3} \pi R \rho$ is the free-fall acceleration at the surface of the planet. At shallow depths ($h \||| R$), the pressure is described by the well-known expression
$p(h) \approx \rho gh$.
The pressure at the center of the planet (at $h=R$) is equal to
$p(0)= \frac{\rho gR}{2}$.