====== Hydrostatic equilibrium ======

**Hydrostatic equilibrium** is the condition in which the internal **pressure gradient** within a self-gravitating body exactly balances the inward pull of **gravity** at every layer.  
This balance maintains the structural stability of stars, planets, and other gaseous celestial bodies.

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Consider a thin spherical shell of gas with infinitesimal thickness \( dr \), located at a distance \( r \) from the center of a star.  
At radius \( r \), the **pressure** is \( P(r) \) and the **density** is \( \rho(r) \).

The inner surface of the shell experiences a pressure \( P_1 = P(r) \), while the outer surface experiences \( P_2 = P(r + dr) \).  
The infinitesimal pressure difference across the shell is

$$
dP = P_2 - P_1.
$$

The **net outward force** on the shell due to this pressure difference is

$$
F_P = -A \, dP,
$$

where \( A = 4\pi r^2 \) is the **surface area** of the shell and the outward direction is defined as positive.  
The negative sign ensures that when \( dP < 0 \) (pressure decreases outward), the resulting \( F_P \) is outward.

The **gravitational force** acting inward on this same shell of thickness \( dr \) is

$$
F_G = -\frac{G \, M(r) \, \rho(r) \, A \, dr}{r^2},
$$

where \( G \) is the **gravitational constant** and \( M(r) \) is the **mass enclosed** within radius \( r \).  
The negative sign indicates that the force acts toward the center (decreasing \( r \)).

For **hydrostatic equilibrium**, the sum of the pressure and gravitational forces must vanish:

$$
F_P + F_G = 0.
$$

Substituting the two forces and canceling \( A \), we obtain

$$
\frac{dP}{dr} = -\frac{G \, M(r) \, \rho(r)}{r^2}.
$$

This is the **differential equation of hydrostatic equilibrium**.  
It expresses that, at every radius \( r \), the **pressure gradient** \( \frac{dP}{dr} \) balances the **gravitational pull** per unit volume.

It is often convenient to define the **local gravitational acceleration**

$$
g(r) = \frac{G \, M(r)}{r^2},
$$

so that the equation can be written as

$$
\frac{dP}{dr} = -\rho(r) \, g(r).
$$

===== Insights =====

This equation shows that, at each point within a star, the **outward pressure gradient** balances the **inward gravitational force**.  
If the pressure gradient were smaller in magnitude, the star would collapse under gravity; if larger, it would expand.  
Thus, hydrostatic equilibrium defines the stable structure of stars and planets.

The **mass function** \( M(r) \) is related to the density by

$$
M(r) = \int_0^r 4\pi {r'}^{2} \rho(r') \, dr',
$$

where \( r' \) is a dummy variable of integration.  
Mass outside the radius \( r \) does not contribute to the gravitational force at \( r \), as follows from **Gauss’s law for gravity** (or the shell theorem).  

The hydrostatic equilibrium equation is one of the four **fundamental equations of stellar structure**, along with those of **mass continuity**, **energy conservation**, and **energy transport**.

===== Inquiries =====

  - Explain how the equation of hydrostatic equilibrium encodes the balance between the pressure gradient and gravity inside a star.  
  - Why does only the mass enclosed within \( r \) contribute to the gravitational force at that radius in a spherically symmetric body?  
  - What happens to a star if its internal pressure gradient becomes smaller in magnitude than required for hydrostatic support?