Local shell force balance
Match the cutaway shell to the inset patch: equal pressure cancels, but a pressure difference can balance local weight.
Match the cutaway shell to the inset patch: equal pressure cancels, but a pressure difference can balance local weight.
The shell guide marks the same radius shown in the cutaway, so one local observation can be placed in the whole star.
Separate the exact local law from the global scale estimate, then connect the support requirement to gas temperature.
Use one local shell observation, one global scaling statement, and one thermal bridge statement.
The demo’s full logic chain lives here once you have seen the shell and the profile together.
In spherical symmetry, the local gravitational field at radius $r$ depends only on the mass enclosed within $r$.
Start from spherical symmetry: only the enclosed mass contributes to the local gravitational field at radius $r$.
A shell is supported only if the inner side pushes harder than the outer side.
For a thin local patch of thickness $dr$, outward pressure difference is $(dP/dr)dr$ times area, and it must balance the weight $\rho g\,dr$ times area.
This is an order-of-magnitude support requirement, not the exact central pressure for every stellar structure.
Combine the hydrostatic scale $\Delta P / R \sim \rho G M / R^2$ with $\rho \sim M / R^3$ to get $P_c \sim G M^2 / R^4$.
If thermal gas pressure provides the support, this gives the rough temperature scale required in the deep interior.
Pair the pressure scale with the ideal-gas relation $P\sim \rho k_B T / (\mu m_p)$ and use $\rho \sim M / R^3$.
Pressure on both sides of a shell can cancel. Support requires the lower side to push harder than the upper side. That is why stars need a pressure gradient, not just pressure.
Hydrostatic equilibrium is a local force-balance law. At every radius, the pressure gradient must exactly balance the local weight of the overlying gas.
If the same mass is squeezed into a smaller radius, gravity becomes harder to resist. At fixed mass, the required central pressure scales as $P_c \propto R^{-4}$.
Pressure support in an ordinary star is mainly thermal gas pressure, but maintaining the temperature needed for that pressure requires an energy source. For main-sequence stars, that source is nuclear fusion.
This demo uses toy models to teach the logic of hydrostatic equilibrium. The goal is conceptual understanding and scaling intuition, not a full stellar evolution calculation.
A star can be in hydrostatic equilibrium and still evolve because force balance can hold while temperature, luminosity, and composition change on longer timescales.
Answer before checking. Use one local observation and one global scaling idea as evidence.
Thermal pressure can support a star against gravity, but hydrostatic equilibrium by itself does not tell you where the heat came from. Main-sequence stars stay hot because nuclear fusion replenishes the energy that the star radiates away.
That is why hydrostatic equilibrium does not mean the star is inert. It means the inward and outward forces nearly balance at each radius while the star evolves more slowly through changes in composition and energy flow.
Begin with the local force-balance law:
$$\frac{dP}{dr}=-\rho g=-\frac{G M(r)\rho(r)}{r^2}$$
Then estimate the order of magnitude by taking $dP/dr\sim P_c/R$ and $\rho\sim M/R^3$:
$$\frac{P_c}{R}\sim \frac{G M}{R^2}\frac{M}{R^3} \quad \Rightarrow \quad P_c\sim\frac{G M^2}{R^4}$$
If gas pressure supplies that support, use $P\sim \rho k_B T/(\mu m_p)$ with $\rho\sim M/R^3$:
$$T_c\sim\frac{\mu G M m_p}{k_B R}$$