The specific <deuterium-burning energy-generation rate> is $\epsilon=\epsilon_0\rho T^{15}$. By <stellar homology>,
$$
\rho_c\sim\frac{M}{R^3},
\qquad
T_c\sim\frac{M}{R},
$$
so
$$
L_{\rm nuc}\sim\int\epsilon\,dm
\propto\rho_c^2T_c^{15}R^3
\propto\frac{M^{17}}{R^{18}}.
$$
Since $L_{\rm acc}\propto M\dot M/R$,
$$
\frac{L_{\rm nuc}}{L_{\rm acc}}
\propto\frac{M^{16}}{\dot M R^{17}}.
$$
Use $R\propto M^3T_{\rm eff}^{-24}$ and $T_{\rm eff}^{68}\propto M^8/\dot M$ from part ii. Then
$$
\boxed{
\frac{L_{\rm nuc}}{L_{\rm acc}}
\propto M^{13}\dot M^{-7}}.
$$

The two track relations also give
$$
M\propto L^{1/6}T_{\rm eff}^{22/3},
\qquad
\dot M\propto L^{4/3}T_{\rm eff}^{-28/3}.
$$
Consequently
$$
\frac{L_{\rm nuc}}{L_{\rm acc}}
\propto L^{-43/6}T_{\rm eff}^{964/6}.
$$
Equality occurs at
$$
\boxed{L_0\propto T_{\rm eff}^{964/43}}.
$$
The preceding formulae imply $L_{\rm nuc}>L_{\rm acc}$ for $L<L_0$, not $L>L_0$. Thus the inequality printed in the question is reversed relative to its requested intermediate scaling; the exponent $964/43$ is consistent.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 317 / 3 / iii / Solution

{title2=$\epsilon_D$}

At the temperatures relevant to pre-main-sequence stars, a useful steep power-law approximation to the specific deuterium-burning rate is $\epsilon_D\propto\rho T^{15}$. Its sensitivity to central temperature makes deuterium burning act as a temporary thermostat.


Codex @codex  0

 Incoming links: Deuterium-burning energy-generation rate