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Since by definition we cannot observe black holes directly, how do astronomers determine the mass of a black hole?

What observational techniques are there that would allow us to determine a black hole's mass?

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3 Answers 3

You can work out the mass of a central gravitating body in a manner somewhat similar to Cameron's answer, but it's a little more complicated. For situations where one object is much more massive than the other:

Newton's law of gravitation says that the gravitational force between two objects is

$$F_\text{grav} = \frac{GMm}{r^2}$$

where $G$ is a universal constant, $M$ is the central mass, $m$ is the orbiting mass, and $r$ is the distance between them.

From Newtonian mechanics, we know that the force required to keep an object traveling in a circular path (AKA centripetal force) is

$$F_\text{cent} = \frac{mv^2}{r}$$

where $m$ is the orbiting mass again, $v$ is its velocity, and $r$ is the radius of the circle.

All the letters in the two scenarios match up, and since gravity is the centripetal force, let's just set them equal:

$$ F_\text{grav} = F_\text{cent} \\ \frac{GMm}{r^2} = \frac{mv^2}{r} $$

Rearrange and we get

$$M = \frac{v^2r}{G}$$

Now if we want to compare the Sun and Earth to the black hole and its satellite,

$$ M_\text{sun} = \frac{v_\text{Earth}^2*r_\text{Earth}}{G} M_\text{BH} = \frac{v_\text{sat}^2r_\text{sat}}{G} $$ dividing one by the other, we get

$$\frac{M_\text{BH}}{M_\text{sun}}=\left(\frac{v_\text{sat}}{v_\text{Earth}}\right)^2\left(\frac{r_\text{sat}}{r_\text{Earth}}\right)$$

MORE PRACTICALLY, HOWEVER,

Instead of using Newtonian mechanics, Kepler's third law is more often used to determine the mass of two objects orbiting each other. It uses quantities that are more easily measurable and can be used no matter the ratio of the masses of the two observed objects.

Kepler's Third Law (generalized from his original work) says that the sum of the masses of two objects orbiting each other is proportional to the the cube of the semi-major axis of the orbit (think radius) divided by the square of the period. M+m (is proportional to) SMA^3/P^2 Doing a proportional reasoning problem with the figures for the Earth and Sun (sum of masses = 1 solar mass, period is one Earth-year, SMA is 1 AU) gives the result

M+m (expressed in Solar masses) = (SMA expressed in AU)^3 / (P expressed in E-yr)^2

Let's say that we observe a small object orbiting a black hole at 5 AU, with a period of 2 E-yr. Then M+m = 5*5*5/2*2 = 31.25 Solar masses. If we assume that the orbiting object is practically nothing compared to the black hole, then the black hole is 31.25 Solar masses.

[EDIT: How to actually get that information-

There are lots of analysis techniques to try to extract semi-major axis and period information from observations.

The simplest case is nearby binary stars, which we can literally see orbit each other, so we can measure those things directly.

Exoplanets are mostly invisible against the glare of their parent stars, so they are mostly detected by the wobble they cause in the star's image as they orbit. The details of that wobble signal provide a wealth of information, some of which can be used to extract period and SMA.

For dense objects like neutron stars orbiting each other, Einstein's theory of General Relativity dictates that they will emit gravitational radiation, or ripples in spacetime. This causes them to lose energy, which makes them slow down. Accurate timing of the lighthouse-like pulses of radiation we receive from them can also yield what we need.

Finally, black holes, especially large ones, have a dramatic effect on their environments. Many have a disk of material spiraling into them. The material heats up to extreme temperatures and emits lots of radiation, allowing us to observe it. Models of the material disks' behavior depend on the mass of the central black hole, so matching a model with observations will give an estimate of the mass.]

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The movie at http://www.astro.ucla.edu/~ghezgroup/gc/pictures/orbitsMovie.shtml is a beautiful visualization of data relating to the Milky Way core -- that "something really massive" is there can be clearly seen

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It is possible to measure the mass of a black hole from the speed of other objects. If we measure the speed of a star orbiting a black hole and the distance of its orbiting radius we can tell its mass. For example we would first locate a star that is traveling at 100 km/s and has an orbiting radius of 150 megameters so we could find the black holes mass.

To do this we would first find a mass such as our Sun and the Earth. We know that the Earth is 149,597,900 km, or about 150 megameters from the Sun, (so this is its orbital radius) and so simply by using pi we can work out how fast it is moving around the Sun. To find the orbital circumfrance we just do

pi*149,597,900 = 469,975,664km (470 megameters) (it'll be a bit more because the orbit isn't perfectly round)

We also know we orbit the Sun every 365.25 days (about 31557600 seconds). To find the speed we just do

469,975,664 km / 31557600 s = 14.9 km/s

Because the star around the black hole has the same distance from us to the Sun, we now know how to work out how much mass it has compared to the sun. To begin with, we know that it is orbiting around 6.7 times faster (100/14.9) and it is the same distance away. This means that it must have a mass 6.7 times greater than the mass of the Sun. The Sun's mass is 1.98892 * 10^30 kg so if we just times this by 6.7 we can find the mass of our black hole:

1.98892 * 10^30 * 6 = 1.193352 * 10^31

That's 1,193,352,000,000,000,000,000,000,000,000 kg!

But that doesn't mean the black hole is 6.7 times larger. In fact, if the black hole we are looking at is an "average" black hole it would be 66297333300000000 cm^3.

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If my maths is wrong someone please correct me. –  Cameron Jun 17 '11 at 16:25
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Yes, that math is wrong. The central mass and velocity of an object in a circular orbit are not linearly proportional. See my answer. –  Andrew Jun 17 '11 at 17:33
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