Question #149642

 At a distance of 55 million light years from Earth, and with a mass equal to 2.4e12 solar masses, calculate the angular diameter of that black hole's event horizon as viewed from the Earth. (express your answer in arcseconds)

1
Expert's answer
2020-12-10T11:07:23-0500

The diameter of that black hole's event horizon (doubled Schwarzschild radius, see https://en.wikipedia.org/wiki/Schwarzschild_radius) is given as


ds=4GMc2d_s = \dfrac{4GM}{c^2}

where G=6.67×1011m3kg1s2G = 6.67×10^{−11} m^3⋅kg^{−1}⋅s^{−2} is the gravitational constant, c=3108m/sc = 3\cdot 10^8m/s is the speed of light, and MM is the mass of the black hole. Since it is equal to 2.4e12 solar masses, obtain:


M=2.4×10121.99×1030kg=4.78×1042kgM = 2.4\times 10^{12}\cdot 1.99\times 10^{30}kg = 4.78\times 10^{42}kg

Thus, obtain:

ds=46.67×10114.78×1042(3108)22.31×1016md_s = \dfrac{4\cdot6.67×10^{−11}\cdot 4.78\times 10^{42}}{(3\cdot 10^8)^2} \approx 2.31\times 10^{16}m

The angular diameter (in radians) is:


θ=dsR\theta = \dfrac{d_s}{R}

where RR is the distance from the Earth to the black hole. Since this distance is 55 million light years, and one light year is approximately 9.46×1015m9.46\times 10^{15}m, the distance is:


R=551069.49×10155.2×1023mR = 55\cdot 10^6\cdot 9.49\times 10^{15} \approx 5.2\times 10^{23}m

Thus, the angular diameter is:


θ=2.31×1016m5.2×1023m4.4423×108rad\theta = \dfrac{2.31\times 10^{16}m}{5.2\times 10^{23}m} \approx 4.4423\times 10^{-8} rad

To convert it to arcseconds, calculate:

θ=4.4423×1083600180/π0.009\theta =4.4423\times 10^{-8}\cdot 3600\cdot 180/\pi \approx 0.009''

Answer. 0.009''.


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