At fixed radiative efficiency, the <Eddington ratio> implies $\dot m\propto f_{\rm Edd}M$. Far outside the <innermost stable circular orbit>, part (a) gives $\Sigma\propto\dot m/\nu$. The supplied viscosity law therefore gives, in physical radius,
$$
\Sigma
\propto
f_{\rm Edd}^{7/10}
\left(\frac\alpha{0.1}\right)^{-4/5}
M^{19/20}R^{-3/4}.
$$
The disk mass follows by radial integration:
$$
M_d(R)=2\pi\int^{R}\Sigma(R')R'\,dR'
\propto
f_{\rm Edd}^{7/10}
\left(\frac\alpha{0.1}\right)^{-4/5}
M^{19/20}R^{5/4}.
$$
Since the <Schwarzschild radius> satisfies $R_S\propto M$, replacing $R$ by $(R/R_S)R_S$ contributes another factor $M^{5/4}$. Thus
$$
\boxed{
M_d(R)=C_1
f_{\rm Edd}^{7/10}
\left(\frac\alpha{0.1}\right)^{-4/5}
\left(\frac{M}{10^6M_\odot}\right)^{11/5}
\left(\frac R{R_S}\right)^{5/4}
},
$$
so
$$
(k_1,k_2,k_3,k_4)=
\left(\frac7{10},-\frac45,\frac{11}5,\frac54\right).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 347 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-347/2/d/solution