AstrophysicsGalaxies and Cosmology

Supermassive Black Holes

Stars orbiting the center of the Milky Way weigh its black hole at about 4 million suns.

Not started Jump to practice

At the center of the Milky Way, in the constellation Sagittarius, sits a compact radio source called Sagittarius A*. For about 30 years, two teams led by Andrea Ghez and Reinhard Genzel followed individual stars as they whipped around it, using infrared light that passes through the dust. Their orbits are Kepler ellipses, and they weigh the central object exactly as visual binaries weigh stars. Its size is set by the Schwarzschild radius.

Why it works

The star S2 has an orbital period of 16.0 years and an angular semi-major axis of $0.1255''$ at a distance of about 8.18 kpc. In AU, $a = 0.1255 \times 8180 \approx 1030$ AU. With Kepler's third law in solar units, $M = a^3/P^2 = 1030^3/16.0^2 \approx 4.2 \times 10^{6}\,M_\odot$. That much mass inside S2's closest approach, about 120 AU, can only be a black hole: no cluster of stars could be packed that tightly and survive.

Key idea

The Milky Way's central black hole has about $4 \times 10^{6}$ solar masses. Most large galaxies have a central supermassive black hole, with masses up to billions of $M_\odot$, typically a few tenths of a percent of the mass of the galaxy's central bulge, and a tiny fraction of the whole galaxy.
S2's orbit around Sagittarius A*, to scale: a long ellipse with the black hole at one focus. The speeds come from the vis-viva equation for $4.2 \times 10^{6}\,M_\odot$.

Worked examples

Worked example: Weighing Sagittarius A*

S2 has $a = 0.1255''$, $P = 16.0$ yr, and the galactic center is 8.18 kpc away. Find the black hole's mass.
  1. $a = 0.1255 \times 8180 = 1030$ AU.
  2. $M = 1030^3/16.0^2 = 1.09 \times 10^{9}/256$.

Answer: About $4.2 \times 10^{6}\,M_\odot$. (Ghez and Genzel shared the 2020 Nobel Prize in Physics for this work, with Roger Penrose.)

Worked example: How fast at closest approach?

S2's orbit has $a = 1030$ AU and $e = 0.886$. Find its speed at closest approach.
  1. $r_p = a(1 - e) = 117$ AU; angular momentum per mass $h = \sqrt{GMa(1 - e^2)}$, and $v_p = h/r_p$.
  2. With $GM = 5.6 \times 10^{26}$ m³/s² (for $4.2 \times 10^{6}\,M_\odot$): $v_p \approx 7.7 \times 10^{6}$ m/s.

Answer: About 7,700 km/s, 2.6% of the speed of light. In 2018 astronomers measured S2's light reddened by gravity at that moment, as general relativity predicts.

Worked example: How big does it look?

The black hole's shadow has a diameter of about $\sqrt{27}\,r_s$. For $4.3 \times 10^{6}\,M_\odot$ at 8.2 kpc, what angle does it span?
  1. $r_s = 2GM/c^2 = 1.27 \times 10^{10}$ m, so the shadow is $6.6 \times 10^{10}$ m across.
  2. Angle: $6.6 \times 10^{10}/2.53 \times 10^{20} = 2.6 \times 10^{-10}$ rad $= 54$ microarcseconds.

Answer: About 54 μas, like a doughnut on the Moon seen from Earth. The Event Horizon Telescope imaged a ring of about this size in 2022, after imaging M87's much larger black hole in 2019.

Common mistake

Thinking the central black hole holds most of a galaxy's mass. Four million solar masses is enormous for one object, but the Milky Way's stars alone total tens of billions of solar masses, and with dark matter the galaxy is near $10^{12}\,M_\odot$. The black hole is about 0.0004% of that. Its gravity dominates only within a few parsecs of the center. Check with the rotation curve: $v^2r/G$ at 8 kpc is $10^{11}\,M_\odot$, 25,000 times the black hole.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

Loading practice…

Builds on

Leads to

This is one of the top rungs of Astrophysics.

Sources and further reading

  • OpenStax Astronomy 2e, section 25.4 (The Center of the Galaxy) and section 27.2 (Supermassive Black Holes: What Quasars Really Are)
  • The Nobel Prize in Physics 2020, press release (nobelprize.org)
@teddyrookbook GitHub