Past exam of the astrophysics course of the University of Oxford / 2021 / 6

For a spherical mass distribution, the shell theorem says the circular gravitational acceleration is $GM(<R)/R^2$. Balancing it against the centripetal acceleration $V_c^2/R$ givesAt $R=16\mathrm{kpc}=4r_0$, the galaxy rotation curve gives $V_c=200(1-e^{-4})=196.337\mathrm{km}{\mathrm{s}}^{-1}$. With $R=4.93708\times 10^{20}\mathrm{m}$,This is the total dynamically inferred enclosed mass, including any dark matter; the galaxy rotation curve alone does not separate its components.

Put $V_0=200\mathrm{km}{\mathrm{s}}^{-1}$. A Taylor expansion near the centre givesTherefore the circular angular frequency isThus the central motion approaches solid-body rotation. Numerically ${\Omega }_0=1.6204\times 10^{-15}{\mathrm{s}}^{-1}$, and the central orbital period tends toThe finite limit of $\Omega$ accompanies $V_c\to 0$ at $r\to 0$; it does not imply a finite central linear speed.

At large radius $V_c\to V_0$, so the spherical enclosed mass is $M(<r)\simeq V_0^{2}r/G$. A spherical mass density satisfies $dM/dr=4\pi r^2\rho (r)$, and henceEquivalently, a local power law $\rho \propto r^{-p}$ with enclosed mass dominated by that extended profile has $M\propto r^{3-p}$ and $V_c^2\propto r^{2-p}$. A flat galaxy rotation curve requires $p=2$.

Differentiating the exact mass profile verifies the limit and the finite central density:As $x\to 0$, the bracket is $3x^2+O(x^3)$, so $\rho (0)=3V_0^2/(4\pi G{r}_0^2)$. As $x\to \infty$, it tends to one, recovering the $r^{-2}$ tail. Extending the flat galaxy rotation curve to infinity makes total mass diverge linearly; a real finite halo must eventually depart from this idealization.

Let the uniform thin galactic disk have outer radius $R_d$ and surface mass density $\Sigma =M/(\pi R_d^2)$, where $M$ is the mass calculated in a. If the redistribution retains the stated outer radius, set $R_d=16\mathrm{kpc}$; otherwise $M$ alone does not determine both $R_d$ and $\Sigma$.

A ring of radius $s$ and width $ds$ has mass $dM=2\pi \Sigma sds$. By axial symmetry its transverse gravitational force components cancel. At signed height $z$, its vertical force on a star of mass $m$ isIntegrating the rings gives, for $z\ne 0$,ThusFor a positive distance above the plane, this is $\mathbf{F}=-2\pi Gm\Sigma [1-z/\sqrt{z^2+R_d^2}]\hat{\mathbf{z}}$, directed toward the disk. With $R_d=16\mathrm{kpc}$, the inferred $\Sigma \simeq 178.3M_{\odot }\mathrm{p}{\mathrm{c}}^{-2}$ sets its normalization.

The far-field expansion gives $F_z\simeq -GMm\mathrm{sgn}(z)/z^2$, as required for a point-like total mass. Close to either face, $F_z\to \mp 2\pi Gm\Sigma$, the infinite-sheet limit. The ideal zero-thickness model has a jump in gravitational field at $z=0$; the formula is understood through its one-sided limits. Its symmetric mean is zero on the plane, and a disk of finite thickness resolves the discontinuity.