Equate the <accretion luminosity> to the surface luminosity:
$$
\frac{GM\dot M}{R}\propto R^2T_{\rm eff}^4.
$$
Using $R\propto M^3T_{\rm eff}^{-24}$ from part i gives
$$
M^8\propto\dot M T_{\rm eff}^{68}.
$$
Substitution into $L\propto M^6T_{\rm eff}^{-44}$ yields the accreting <Hayashi track>
$$
\boxed{L\propto\dot M^{3/4}T_{\rm eff}^{7}}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-317/3/ii/solution