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. The Newtonian tidal field of Chapter 7 gives, for two freely falling particles separated radially by ξ\xi at radius rr from a point mass MM, a relative acceleration of magnitude 2GMξ/r32GM\xi/r^{3} (a stretch). ((a)) Show that the tidal stretch across a radial separation ξ\xi evaluated at r=rsr = r_s can be written entirely in terms of the mean density, Δa=83πGρˉ ξ\Delta a = \tfrac{8}{3}\pi G\bar\rho\,\xi, and hence also as Δa=c6ξ/4G2M2\Delta a = c^{6}\xi/4G^{2}M^{2}. ((b)) Evaluate Δa\Delta a across ξ=2.0 m\xi = 2.0\ \mathrm{m} for the hole whose mean density inside the horizon equals that of water, ρˉ=1.00×103 kg m−3\bar\rho = 1.00 \times 10^{3}\ \mathrm{kg\,m^{-3}}. ((c)) Find the mass of that hole, in kg and in solar masses. 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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