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Ellie Snyder
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Homework Statement
Soon astronomers will be imaging the “shadow” of light from the event horizons of black holes. Since black holes are very small, this achievement seems impossible. Nevertheless, it is possible for three reasons:
1. Supermassive black holes have large event horizons.
2. Radio wavelengths of light are not affected by interstellar dust and can explore regions close to the black holes.
3. By building a radio telescope effectively as large as the Earth, astronomers can overcome the relatively poor resolution associated with long wavelengths and achieve high angular resolution.
a. Calculate the Schwarzschild radii (in kilometers) of the supermassive black holes associated with our Milky Way galaxy (4x10^8 M_Sun) and the giant elliptical galaxy M87 (3.5x10^9 M_Sun).
b. The distances of these black holes are 8.4 kpc and 16.4 mega-parsecs, respectively. What are the expected angular diameters of the innermost circular orbits, assuming a non-rotating black hole?
c. Show that radio astronomers can resolve (1.22*(wavelength/D)) these objects at =0.7 mm (c=wavelength*f; f=450 GHz) if they combine separate radio telescopes over the diameter of the Earth (12,742 km).
Homework Equations
a. R_Sch=(2GM)/c^2=(2.95 km/M_Sun)*M
b. Diam.=2arctan(d/2D)
c. Angular resolution=1.22*(wavelength/D)
The Attempt at a Solution
a. Using the eqn with 2.95/solar mass I calculated Schwarzschild Radii of 1.18*10^9 km for our galaxy's black hole, and 1.0325*10^10 km for M87.
b. Would I plug in the calculated radii for the variable d and the given distances for D to calculate the angular diameter?
c. Just convert the 0.7 mm and 12,742 km into similar units, then plug them into the angular resolution eqn?