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un:hydrostatic-equilibrium [2025/10/25 13:38] – [Questions] asadun:hydrostatic-equilibrium [2025/10/26 07:50] (current) asad
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 ====== Hydrostatic equilibrium ====== ====== Hydrostatic equilibrium ======
  
-Hydrostatic equilibrium is the condition under 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.+**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.
  
 {{:courses:ast301:hydrostatic.webp?nolink&650|}} {{:courses:ast301:hydrostatic.webp?nolink&650|}}
  
-Consider a thin spherical shell of gas with thickness \( dr \) located at a distance \( r \) from the center of a star.   +Consider a thin spherical shell of gas with infinitesimal thickness \( dr \)located at a distance \( r \) from the center of a star.   
-At this radius, the gas has a density \( \rho(r) \)and the pressure is \( P(r) \).  +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 inner surface of the shell experiences a pressure \( P_1 = P(r) \), while the outer surface experiences \( P_2 = P(r + dr) \).  
-The differential pressure is therefore:+The infinitesimal pressure difference across the shell is
  
 $$ $$
-dP = P_2 - P_1+dP = P_2 - P_1.
 $$ $$
  
-The net **upward (outward)** force on the element due to pressure difference is:+The **net outward force** on the shell due to this pressure difference is
  
 $$ $$
-F_P = -A \, dP+F_P = -A \, dP,
 $$ $$
  
-where \( A \) is the area of the shell and the positive direction is defined as outward from the center +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 element is:+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}+F_G = -\frac{G \, M(r) \, \rho(r) \, A \, dr}{r^2},
 $$ $$
  
-Here, \( M(r) \) represents the total mass enclosed within radius \( r \), and \( \) is the gravitational constant.+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 \( \)).
  
-For the element to remain in **hydrostatic equilibrium**, the net force must vanish:+For **hydrostatic equilibrium**, the sum of the pressure and gravitational forces must vanish:
  
 $$ $$
-F_G + F_P = 0+F_P + F_G = 0.
 $$ $$
  
-Substituting the two forces gives:+Substituting the two forces and canceling \( A \), we obtain
  
 $$ $$
-\dP = -\frac{G M(r) \, \rho(r) \, A \, dr}{r^2}+\frac{dP}{dr} = -\frac{G \, M(r) \, \rho(r)}{r^2}.
 $$ $$
  
-Simplifying, we obtain the **differential equation of hydrostatic equilibrium**:+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.
  
-$$ +It is often convenient to define the **local gravitational acceleration**
-\frac{dP}{dr} = -\frac{G M(r) \, \rho(r)}{r^2} +
-$$ +
- +
-or equivalently,+
  
 $$ $$
-\frac{dP}{dr} = -\rho(r) \, g(r)+g(r) = \frac{G \, M(r)}{r^2},
 $$ $$
  
-where the local **gravitational acceleration** is:+so that the equation can be written as
  
 $$ $$
-g(r) = \frac{G M(r)}{r^2}+\frac{dP}{dr} = -\rho(r) \, g(r).
 $$ $$
  
-===== Notes =====+===== Insights =====
  
-This equation states that, at every point within a star, the outward pressure gradient balances the inward gravitational pull.   +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, the gas would collapse under gravity; if it were larger, the star would expand.   +If the pressure gradient were smaller in magnitude, the star would collapse under gravity; if larger, it would expand.   
-Thus, hydrostatic equilibrium defines the fundamental balance governing the structure of stars and planets.+Thus, hydrostatic equilibrium defines the stable structure of stars and planets.
  
-The mass function \( M(r) \) is found by integrating the density:+The **mass function** \( M(r) \) is related to the density by
  
 $$ $$
-M(r) = \int_0^r 4\pi r'^2 \rho(r') \, dr'+M(r) = \int_0^r 4\pi {r'}^{2\rho(r') \, dr',
 $$ $$
  
-Mass outside the radius \( r \) does not contribute to the gravitational force at \( r \), in accordance with **Gauss’s law for gravity**.   +where \( r' \) is a dummy variable of integration.   
-The hydrostatic equilibrium equation is one of the core equations in the theory of stellar structure, along with the equations of mass continuity, energy conservation, and energy transport.+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 =====
  
-===== Questions ===== +  - Explain how the equation of hydrostatic equilibrium encodes the balance between the pressure gradient and gravity inside a star.   
-1. Explain how the equation of hydrostatic equilibrium represents the balance between gravity and pressure inside a star.   +  Why does only the mass enclosed within \( r \) contribute to the gravitational force at that radius in a spherically symmetric body  
-2. Why does only the mass enclosed within radius \( r \) contribute to the gravitational force at that point+  What happens to a star if its internal pressure gradient becomes smaller in magnitude than required for hydrostatic support?
-3. What happens to a star if its internal pressure gradient becomes smaller than required for hydrostatic equilibrium?+
  
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