3.2. Escape velocity¶
In most potentials, bodies that move fast enough can “escape” the gravitational field of a mass distribution. This means that such bodies do not orbit the center-of-mass of this mass distribution—because gravity is a long-range force, the gravitational field is technically felt even at extremely large distances. That the energy is conserved allows us to compute the escape velocity \(v_\mathrm{esc}\): the minimum velocity necessary to “escape” the gravitational field, starting from position \(\vec{x}\) to a resting position at \(r=\infty\). The minimum velocity would cause the body to arrive at infinity with zero velocity, so energy conservation gives \begin{equation}\label{eq-vesc} \frac{v_\mathrm{esc}^2}{2} + \Phi(\vec{x}) = \frac{0^2}{2} + \Phi_\infty\,,\quad \mathrm{or}\quad v_\mathrm{esc} = \sqrt{2[\Phi_\infty-\Phi(\vec{x})]}\,. \end{equation} Because we normally try to define the potential such that \(\Phi_\infty=0\), this typically simplifies to \(v_\mathrm{esc} = \sqrt{-2\Phi(\vec{x})}\). For a spherical mass distribution, we have \(v_\mathrm{esc} = \sqrt{-2\Phi(r)}\). Note that we have simply written \(\Phi_\infty\) rather than specifying how we approach infinity; for a potential for which it is possible to escape, the potential at infinity cannot depend on the direction of infinity, so this is well defined.
Thus, the escape velocity measures the depth of the potential well. This means that the escape velocity contains information about the mass distribution outside of the radius at which it is measured, unlike the circular velocity, which only depends on the mass contained within the radius at which it is measured (at least for a spherical mass distribution, but approximately so for all mass distributions). This should be clear from the meaning of the “escape” velocity: the speed necessary for a body to escape must depend on the cumulative amount of gravitational attraction the body needs to overcome out to \(\infty\). It is also clear from the expression for the gravitational potential in a spherical potential in Equation (2.21): the second term depends on the density outside of \(r\).
In the Milky Way, it is possible to measure the escape velocity near the Sun by looking at the distribution of stellar velocities and attempting to identify a high-velocity cut-off. The cut-off arises because stars at the escape velocity and beyond should be very rare, as they will escape the Galaxy and never return. This has been done with ever-increasing data from large-scale surveys measuring velocities for solar neighborhood stars (e.g., Leonard & Tremaine 1990; Smith et al. 2007; Piffl et al. 2014; Deason et al. 2019) with a local escape speed of \(v_\mathrm{esc} \approx 550\,\mathrm{km\,s}^{-1}\) (we discuss this measurement in more detail in Chapter 6.2). We can get a rough estimate of the total mass of the Milky Way by comparing this measured escape velocity to the escape velocity that the Milky Way would have if all of its mass were within the solar orbit. We can use Newton’s second shell theorem to get the escape velocity in this case \begin{equation} v_\mathrm{esc}(r=8\,\mathrm{kpc}) = \sqrt{2\frac{GM(r<8\,\mathrm{kpc})}{8\,\mathrm{kpc}}}\,, \end{equation} and substitute the estimate of \(M(r<8\,\mathrm{kpc})\) from the simple estimate of the mass within the solar circle in Equation (2.25). This gives \(v_\mathrm{esc} = 311\,\mathrm{km\,s}^{-1}\), well below the observed value of \(550\,\mathrm{km\,s}^{-1}\). The expression for the gravitational potential in a spherical potential in Equation (2.21) demonstrates that we need to assume a density law outside of the solar orbit if we want to match the observed escape velocity (the potential outside of \(r\) depends on \(\int_r^\infty\mathrm{d}r'\,\rho(r')\,r'\), not on the mass directly). Let’s use the mass–radius relation that we derived from the fact that rotation curves are flat in Equation (2.26): \(\rho \propto r^{-2}\). For this particular density law, the mass and potential do not converge as we go to \(r\rightarrow \infty\), so let’s define the edge of the Milky Way as being at \(100\,\mathrm{kpc}\) outside of which the density is zero (this is just a rough guess, given that we need to add lots of mass to explain the measured escape velocity; in Chapter 2.4.6, we saw that the standard definition of “edge” that is the virial radius gives a radius around \(250\,\mathrm{kpc}\)). Then we can show that the potential difference between \(r_\infty = 100\,\mathrm{kpc}\) and \(r_0 = 8\,\mathrm{kpc}\) is given in terms of the mass \(M(r<8\,\mathrm{kpc})\) and the mass \(\Delta M\) between \(r_0\) and \(r_\infty\) \begin{align} \Delta \Phi & = \Phi(r_\infty)-\Phi(r_0)\\ & = -G\,\left[\frac{M(r<8\,\mathrm{kpc})+\Delta M}{r_\infty}-\frac{M(r<8\,\mathrm{kpc})}{r_0}-\frac{\Delta M}{r_\infty-r_0}\,\ln (r_\infty/r_0)\right]\nonumber\,. \end{align} The potential differs from the standard logarithmic potential from Chapter 2.4.5, because of the hard edge at \(r_\infty\), which gives rise to the first two terms. Plugging this into \(v_\mathrm{esc} = \sqrt{2\Delta\Phi}\) and using the enclosed mass from the simple estimate of the mass within the solar circle in Equation (2.25), we find that (this calculation can be simplified by assuming that \(M(r<8\,\mathrm{kpc}) \ll \Delta M\)) \begin{equation} \Delta M \approx 1.4\times 10^{12}\,M_\odot\,! \end{equation} That the escape velocity near the Sun is approximately \(550\,\mathrm{km\,s}^{-1}\) therefore implies that the total mass of the Milky Way must be \(\approx 10^{12}\,M_\odot\), more than an order of magnitude larger than the mass we estimate to lie within the solar orbit. This simple estimate is remarkably accurate: a variety of sophisticated methods show that the Milky Way’s total mass is \(\approx 0.8\) to \(1.6 \times 10^{12}\,M_\odot\), in good agreement with our estimate. This mass is much larger than the total mass that we see in stars and gas, again directly implying the presence of a large amount of dark matter in the Milky Way.
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