- #36

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Yes, this seems right (as a check for myself ;) ). So this is one example of a general relativistic problem in which a naive intuition gives the wrong answer (the most massive black hole pulls the least), while a Newtonian intuition gives the right answer (the Schwarzschild radius combined with the fact that gravity falls of as one over r squared gives a surface gravity inversely proportional to mass, so the lightest black hole will pull the most). This is a really nice scenario to show different aspects of general relativity (surface gravity as measured at "infinity", surface gravity as measured near the horizon, etc).PeterDonis said:In other words, the areal radius ##r## in both cases is ##2M## + 1 meter. Ok.10 solar masses is not "supermassive", but basically you're asking how things vary with the mass of the hole, so ok.

I'll use SI units throughout.

Some useful constants:

One solar mass: ##1.9885 \times 10^{30}##

Newton's gravitational constant ##G##: ##6.6732 \times 10^{-11}##

Speed of light ##c##: ##299792458##

1 Year: ##31558118.4##

##r = 2GM/c^2## plus 1 meter for

One solar mass hole: ##2953.897##

Ten solar mass hole: ##29529.97##

Redshift factor at ##r## for

One solar mass hole: ##0.01839761##

Ten solar mass hole: ##0.005813797##

Proper acceleration at 1 m above horizon for

One solar mass hole: ##8.266231 \times 10^{15}##

Ten solar mass hole: ##2.617423 \times 10^{15}##

##2GM / c^2 r## at 500 ly for

One solar mass hole: ##6.242327 \times 10^{-16}##

Ten solar mass hole: ##6.242327 \times 10^{-15}##

So we can treat this as "infinity" for both holes. That means the redshift factor above is the factor by which we multiply the proper acceleration at 1 meter above the horizon, to get the force exerted at infinity. So...

Force exerted at infinity for

One solar mass hole: ##1.520789 \times 10^{14}##

Ten solar mass hole: ##1.521716 \times 10^{13}##

Both of these are quite close to the surface gravity, which is the limit of the redshifted proper acceleration at the horizon, and which in SI units is ##c^4 / 4 G M##. So...

Surface gravity for

One solar mass hole: ##1.521819 \times 10^{14}##

Ten solar mass hole: ##1.521819 \times 10^{13}##

Notice how the ten solar mass value 1 meter above the horizon is closer to the limiting surface gravity; that is because 1 meter is a smaller increment in terms of the mass of the hole.

Thanks (and sorry for the late reply)!