Hydrostatic equilibrium plus the ideal gas law is enough to estimate what the center of a star is like, without solving a single differential equation. Astrophysicists call this an order-of-magnitude estimate: replace every derivative by a ratio of typical sizes and keep only the physics.
Why it works
In $dP/dr = -Gm\rho/r^2$, replace $dP/dr$ by $P_c/R$, $m$ by $M$, $r$ by $R$, and $\rho$ by $M/R^3$. Then $P_c/R \sim GM(M/R^3)/R^2$, so $P_c \sim GM^2/R^4$. The gas is ideal, so $P = nkT$ with particle density $n = \rho/(\mu m_p)$, where $\mu$ is the mean mass of a particle in units of $m_p$ (about 0.6 for ionized solar gas). Then $$kT_c = \frac{P_c\,\mu m_p}{\rho} \sim \frac{GM^2/R^4}{M/R^3}\,\mu m_p = \frac{GM\mu m_p}{R}.$$ It says something simple: a particle at the center has thermal energy comparable to the gravitational energy per particle, $GMm_p/R$. That is what it takes to hold the star up, which is why the core must be hot (see temperature and molecular energy).
Key idea
$$P_c \sim \frac{GM^2}{R^4}, \qquad T_c \sim \frac{GM\mu m_p}{kR}.$$ For the Sun these give about $10^{15}$ Pa and $10^{7}$ K, close to detailed models (about $2.5 \times 10^{16}$ Pa and $1.57 \times 10^{7}$ K).
Worked examples
Worked example: The Sun's core temperature
Estimate $T_c$ for the Sun from $kT_c \sim GMm_p/R$.
- Multiply $G$, $M$ and $m_p$: $GMm_p = 2.22 \times 10^{-7}$ J·m.
- $kR = (1.38 \times 10^{-23})(6.96 \times 10^{8}) = 9.60 \times 10^{-15}$ J/K.
- $T_c \sim 2.3 \times 10^{7}$ K. With $\mu = 0.62$, $1.4 \times 10^{7}$ K.
Answer: Tens of millions of kelvins; models give 15.7 million K.
Worked example: Temperature from pressure and density
Solar models give the Sun's center $P = 2.5 \times 10^{16}$ Pa and $\rho = 1.5 \times 10^{5}$ kg/m³, with $\mu \approx 0.85$ (the core is now rich in helium). What temperature does the ideal gas law give?
- $n = \rho/(\mu m_p) = 1.5 \times 10^{5}/(0.85 \times 1.67 \times 10^{-27}) = 1.06 \times 10^{32}$ m⁻³.
- $T = P/(nk) = 2.5 \times 10^{16}/(1.06 \times 10^{32} \times 1.38 \times 10^{-23})$.
Answer: About $1.7 \times 10^{7}$ K, consistent with the models' 15.7 million K.
Worked example: Bigger stars are only a little hotter inside
On the main sequence $R \propto M^{0.8}$ roughly. How does $T_c$ scale with mass?
- $T_c \propto M/R \propto M/M^{0.8} = M^{0.2}$.
- A 10 $M_\odot$ star: $10^{0.2} = 1.6$.
Answer: Only about 1.6 times the Sun's central temperature, yet, as later topics show, thousands of times its luminosity.
Common mistake
Thinking the Sun's center is about as hot as its surface, as if the Sun were the same all through. Gravity squeezes the inner layers under the weight of everything above, and gas under that pressure must be hot to push back: about 15.7 million K at the center against 5,772 K at the surface. Check with the estimate: $GMm_p/(kR)$ is millions of kelvins, so a cool interior could not hold the Sun up.