Answer to Question #149642 in Astronomy | Astrophysics for Mallory

Question

 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)

Answer 1

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

$$d_s = \dfrac{4GM}{c^2}$$

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

$$M = 2.4\times 10^{12}\cdot 1.99\times 10^{30}kg = 4.78\times 10^{42}kg$$

Thus, obtain:

$$d_s = \dfrac{4\cdot6.67×10^{−11}\cdot 4.78\times 10^{42}}{(3\cdot 10^8)^2} \approx 2.31\times 10^{16}m$$

The angular diameter (in radians) is:

$$\theta = \dfrac{d_s}{R}$$

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

$$R = 55\cdot 10^6\cdot 9.49\times 10^{15} \approx 5.2\times 10^{23}m$$

Thus, the angular diameter is:

$$\theta = \dfrac{2.31\times 10^{16}m}{5.2\times 10^{23}m} \approx 4.4423\times 10^{-8} rad$$

To convert it to arcseconds, calculate:

$$\theta =4.4423\times 10^{-8}\cdot 3600\cdot 180/\pi \approx 0.009''$$

Answer. 0.009''.