The <line mass> is
$$
\lambda=2\pi\int_0^{R_f}\rho(R)R\,dR.
$$
Using $d[xJ_1(x)]/dx=xJ_0(x)$ and $J_1(x)=-J_0'(x)$ gives
$$
\lambda=2\pi\rho_1a^2x_1J_1(x_1)
=-\frac{\rho_1K_{\rm eff}}Gx_1J_0'(x_1).
$$
Since $x_1\simeq2.4$ and $J_0'(x_1)\simeq-0.52$,
$$
\boxed{
\lambda\simeq1.25\frac{\rho_1}{G}
\left(K+\frac{B_0^2}{8\pi\rho_0^2}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 2 / a / iii / Solution

{title2=$\lambda$}

The line mass is mass per unit length along a filament. For an axisymmetric density profile $\rho(R)$,
$$
\lambda=2\pi\int_0^\infty\rho(R)R\,dR,
$$
with the upper limit replaced by the filament radius when the density has compact support.


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