Define the mean density inside the horizon of a non-rotating black hole of mass as

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Relativistic stars and compact objects · Relativistic stars and compact objects

Relativity and Gravitation, Ch Escape velocity, the Schwarzschild radius and the idea of a black hole

Problem

Define the mean density inside the horizon of a non-rotating black hole of mass MM as ρˉ=M/(43πrs3)\bar\rho = M/\left(\tfrac{4}{3}\pi r_s^{3}\right), with rs=2GM/c2r_s = 2GM/c^{2}. Estimate the properties of the hole whose mean density equals that of air at room temperature and pressure, ρˉ=1.2 kg m−3\bar\rho = 1.2\ \mathrm{kg\,m^{-3}}. ((a)) Estimate its mass, in kg and in solar masses. ((b)) Estimate its Schwarzschild radius, in m and in astronomical units (1 au=1.496×1011 m1\ \mathrm{au} = 1.496 \times 10^{11}\ \mathrm{m}). ((c)) Evaluate the compactness parameter GM/rc2GM/rc^{2} at r=rsr = r_s, and state whether the first-order redshift formula z≈ΔΦ/c2z \approx \Delta\Phi/c^{2} of Chapter 9, with ΔΦ=Φrec−Φem\Delta\Phi = \Phi_{\rm rec} - \Phi_{\rm em}, may be applied to light emitted from just outside this horizon. Use G=6.674×10−11 N m2 kg−2G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}, c=2.998×108 m s−1c = 2.998 \times 10^{8}\ \mathrm{m\,s^{-1}} and M⊙=1.989×1030 kgM_\odot = 1.989 \times 10^{30}\ \mathrm{kg}.

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