If we assume that M87* is not rotating (although it definitely IS rotating but this assumption gives us a much easier calculation that can be done quickly), what is the size of its event horizon (i. e., its Schwarzschild radius)? The Schwarzschild radius of a black hole is given by LaTeX: r_{s\:}=\frac{\:2GM}{c^2}r s = 2 G M c 2 The mass of the black hole is LaTeX: 6.5\:\times10^{9\:}\:\mathrm{M_{\od ot}}6.5 × 10 9 M ⊙, where the solar mass is LaTeX: M_{\odot}\:=\:2\:\times10^{30}\:kgM ⊙ = 2 × 10 30 k g Give your answer in AU. For reference, LaTeX: 1\:AU\:=\:1.5\:\times10^{11}\:m1 A U = 1.5 × 10 11 m and the gravitational constant is LaTeX: G\:=\:6.67\:\times10^{-11}\:\:m^{3\ :}kg^{-1}s^{-2}G = 6.67 × 10 − 11 m 3 k g − 1 s − 2 You should also read over section 2.1 in the lab script. Be careful with your units! Do not use equation 1 in the lab script; that equation uses an older estimate of the black hole's mass.