Re: Measuring the mass of a black hole

On 17 Feb, 16:50, "PD" <TheDraperFam...@xxxxxxxxx> wrote:
On Feb 17, 9:42 am, "Paul W" <1337...@xxxxxxxxx> wrote:



On 15 Feb, 16:59, "PD" <TheDraperFam...@xxxxxxxxx> wrote:

On Feb 15, 4:06 am, "Paul W" <1337...@xxxxxxxxx> wrote:

Hello,

When a black hole candidate is discovered, how do you actually measure
it's mass? Are there different techniques for doing it?

Thanks,
Paul W.

Typically by watching stuff orbit around it.
Watching the Moon's orbit is how we know the mass of the Earth, and
watching the Earth's orbit is how we know the mass of the Sun.

PD

Are there any formulae to use for this?

For the moon orbit and the earth orbit, the Newtonian approximation
works very well. I won't do the same for black holes, because the
relativistic notation is dense.

For the Newtonian case, the force between two objects of mass M and m
is
GMm/r^2
and Newton's law of motion is F=ma, where m is the mass of the
orbiting object
and the left hand side is what we just wrote down
GMm/r^2 = ma
and if we approximate the orbit as a circle, then we also know the
acceleration
4*pi^2*r/T^2, where r is the radius of the orbit and T is the time for
one orbit.
Putting all this together we have
GMm/r^2 = 4m*pi^2*r/T^2
The mass of the orbiting object is m and it cancels out, leaving
M = 4*pi^2*r^3/(G*T^2)
So if we know the distance from the Earth to the Moon (by parallax,
say), and we measure T, then we've measured the mass of the Earth M.

The constant ratio between r^3 and T^2 is fairly famous, being one of
Kepler's laws.

I've made a number of approximations in the above -- the moon doesn't
really travel in a perfect circle, the Earth also orbits a center of
gravity, etc -- but this is good enough for a 1% accurate answer.

PD

What if there is a gas cloud or an accretion disc around the measured
object? Will this still apply?

[Thanks for all your help, and sorry for asking so many questions =)]

.

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