What holds a star up against its own gravity?
March 10, 2026
Without support, the Sun should collapse in about 50 minutes.
So what has kept it stable for 4.6 billion years?
The outward force comes from a pressure gradient: the gas below a layer must push harder than the gas above it.
By the end of class, you should be able to:
Support comes from a difference across a layer, not from an absolute number.
Consider a thin shell at radius \(r\) with thickness \(dr\).
Gravity
Pressure difference
\[ \frac{dP}{dr} = -\frac{G\,M(r)\,\rho(r)}{r^2} \tag{1}\]
Read it like a sentence
If \(dP/dr = 0\) everywhere inside a star, the star would:
\[ P_c \sim \frac{GM^2}{R^4} \tag{2}\]
Scaling story
\(P_c \sim \frac{GM_\odot^2}{R_\odot^4} \approx 10^{16}~\text{dyn}/\text{cm}^2\)
\[ 2 K_\text{th} + U_\text{grav} = 0 \tag{3}\]
What makes this deep
Self-gravitating systems do the opposite of a coffee cup on your desk.
\[ T_c \sim \frac{\mu G M m_p}{k_B R} \tag{4}\]
Why this matters
More massive stars are hotter inside, but not by orders of magnitude.
What we used
What we have not used yet
That is the setup for the next lecture: once the core is hot enough, how does fusion actually happen?
| Step | This lecture |
|---|---|
| Observable | A star stays roughly steady for billions of years |
| Model | Pressure gradient balances gravity |
| Inference | The core must be dense, high-pressure, and hot |
We never “see” the solar core directly.
We infer it from the physics required to keep the Sun standing.
Two stars have the same mass, but Star B has half the radius of Star A.
Target idea: the more compact star needs much larger support and therefore a hotter, denser interior.
Gravity can make the core hot enough.
Next question: how do protons actually fuse when they repel each other?
Optional Expansion
Use these if we want the slower derivation or a fuller pressure story.
For a shell of area \(A\) and thickness \(dr\):
\[ F_\text{pressure} = [P(r) - P(r+dr)]A \approx -\frac{dP}{dr}A\,dr \]
\[ F_\text{grav} = -\frac{G M(r)\rho A\,dr}{r^2} \]
Set \(F_\text{pressure} + F_\text{grav} = 0\) to recover hydrostatic equilibrium.
Keep the same mass, shrink the radius by a huge factor, and the required central pressure skyrockets.
That is why later in Module 3 ordinary gas pressure stops being enough, and degeneracy pressure has to take over.

ASTR 201 • Module 3, Lecture 2