Given a star's mass, radius and average composition (e.g. 90% H, 10% He), is there a formula to estimate the core temperature of that star?

I only found one for a lower bound but that wasn't very accurate.

  • 2
    $\begingroup$ There isn't any simple formula. You need a computer code that calculates the rates of the relevant nuclear reactions along with transport of the heat via both radiation and convection. Opacities and nuclear reaction rates are functions of temperature, so it's complicated. $\endgroup$
    – John Doty
    Mar 21, 2022 at 21:05
  • $\begingroup$ @JohnDoty Ok, I found the formula $T = \frac{GMm_{p}}{rk_{B}}$ for a rough estimate (but still very inaccurate). Do you by chance know a better approximation that at least works for main sequence stars? $\endgroup$
    – 299792458
    Mar 21, 2022 at 21:12
  • $\begingroup$ $M$ being the total mass, $m_{p}$ a proton's mass, $r$ the star's radius and $k_{B}$ Boltzmann's constant $\endgroup$
    – 299792458
    Mar 21, 2022 at 21:13
  • $\begingroup$ I guess you could probably also use the average particle mass in the star instead of a proton's mass for $m_{p}$ for a better approximation... $\endgroup$
    – 299792458
    Mar 21, 2022 at 21:16
  • $\begingroup$ Your relation $kT \approx GM_\text{sun}m_\text{particle} / r$ appears to be an application of the virial theorem. $\endgroup$
    – rob
    Mar 21, 2022 at 21:42

2 Answers 2


There are approximate formulas for this, but to do this right you really need an equation of state for matter in the stellar core. Writing that down is really, really hard, which is why predicting the temperature and density profiles for stellar cores is really, really hard.


Yes, the formula you quoted in comments is an application of the viral theorem, that says for a star in equilibrium that twice its internal kinetic energy plus it's potential energy is zero. This can be written $$\Omega = - 3 \int P\ dV = -3\int P\ \frac{dm}{\rho}$$ where $\Omega$ is gravitational potential, $P$ is pressure, $\rho$ is density and $dm$ is a mass element.

A rough solution is to assume the star is a uniform sphere with an average pressure, density and temperature. In which case $$-\frac{3GM^2}{5R} = -3\frac{PM}{\rho}$$ where $M$ is the stellar mass and $R$ the radius. Writing $\rho = 3M/4\pi R^3$ and assuming an ideal, perfect gas with a mass per particle of $\mu$ and temperature $T$, then $$\frac{GM}{5R} = \frac{\rho k_B T}{\mu \rho}$$ $$T = \frac{G\mu}{5k_B}\frac{M}{R}$$

This gives the right proportionality but the numerical coefficient of 0.2 is not accurate because the star is not uniform; the gravitational potential is not that of a sphere and the density, temperature and pressure vary with radius. A 1 solar mass, 1 solar radius star, with the composition you mention ($\mu = 0.58\times 1.67\times 10^{-27}$ kg) we get $T = 0.2 \times 1.3\times 10^7$ K. This is not so bad for an average temperature, but not close to the core temperature.

A more accurate approximation comes from assuming a polytropic equation of state with $P \propto \rho^{\alpha}$. For a star like the Sun (or of higher mass), where the energy is largely transported radiatively, it turns out $\alpha \simeq 4/3$. Solving the Lane-Emden equation and assuming an ideal gas then gives a new numerical coefficient of 1.17 (for a largely convective star with $\alpha =5/3$, the dtar is more centrally condensed and the coefficient would be 1.86).

Thus your answer for a sun-like star is $$ T_c \simeq 1.17 \frac{GM\mu}{R}\ .$$ and for a lower mass main sequence star, where convection is dominant $$T_c \simeq 1.86 \frac{GM\mu}{R}\ .$$

If you want something more accurate than this then a precise stellar evolution model is required to solve for energy transport at each radius, to account for the non-perfect nature of the gas, radiation pressure and accurately describe the pressure and density profile.


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