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I was looking At a couple different replies to some questions on here and I kept seeing people saying that larger mass black holes have a lower gravitaional strength on their surface than lower mass black holes. I am no expert by any stretch of the imagination but I would of assumed that the larger the mass the more powerful the gravity. Does anyone know why this is

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    $\begingroup$ You are thinking of surface gravity $$ \kappa =\frac {c^{4}}{4GM}$$ which is a little more subtle. See this. $\endgroup$
    – joseph h
    May 10, 2023 at 7:18
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    $\begingroup$ Basically, the more massive the black hole, the higher its gravity, but also the larger its radius. Both the gravity force and the radius increase linearly with mass. However, the "surface gravity" is (Newtonian approximation!) proportional to $M / R^2$, which means it decreases with an increase of mass. An even more directly measurable quantity is the tidal force which goes as $M/R^3$, meaning it decreases even more rapidly with increasing mass. $\endgroup$ May 10, 2023 at 7:55
  • $\begingroup$ Possible duplicate/closely related: Do supermassive black holes contain a singularity? "...a person on the surface of the Earth and one at the event horizon of a 10 million M☉ black hole experience about the same tidal force". See also the discussion in: Spaghettification inside a black hole? $\endgroup$
    – Quillo
    May 10, 2023 at 13:34

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I kept seeing people saying that larger mass black holes have a lower gravitational strength on their surface

A black hole has no physical surface. If somebody said "surface," they may have been talking about the event horizon of the black hole. That's a "surface" only in a mathematical sense. If you fell into a super-massive black hole, you would not see or feel any thing change when you passed through the point of no return.

...gravitational strength...

The event horizon actually is defined by gravitational strength. The strength of gravity at the event horizon is exactly the strength that prevents light from escaping. It's the same no matter the size of the black hole.

Whoever said gravitational strength, they probably were talking about the gravitational gradient.

Maybe you have heard of spaghettification. That's the idea that, as you fall in to a black hole, gravity tries to pull your head away from your feet while simultaneously squeezing your sides. It happens because (let's say, you are falling head-first) your head is closer to the center of the black hole than are your feet. Your head feels a stronger gravitational attraction.

Now, if your head is at the event horizon of a supermassive black hole with a Swarzschild radius of a billion kilometers, then your head is experiencing a mind-bogglingly strong gravitational field, but your feet, which are only about two meters further away are feeling almost exactly the same thing.

On the other hand, if your head is at the event horizon of a stellar mass black hole with a radius of only a few tens of kilometers, then your head is feeling exactly the same gravity as in the supermassive case, but the ratio of the lesser gravity that your feet are feeling to what your head is feeling is much greater. It's that difference between what your head feels and what your feet feel—It's the gradient—that will pull you apart.

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  • $\begingroup$ The gravitational length contraction near the horizon approximately is $\sqrt{r_s/(r-r_s)}$. With $r_s=20\,km$ and person’s height of $2\,m$, the contraction is $100$. Thus the coordinate distance from the feet to the horizon is approximately $2\,cm$, not $2\,m$ when the head is close to the horizon. $\endgroup$
    – safesphere
    May 10, 2023 at 18:24
  • $\begingroup$ I think the first part of this answer is incorrect. An event horizon is defined as the boundary at which the escape velocity exceeds the speed of light. It's not defined by a particular strength of the gravitational force - differently sized black holes have different "surface gravity" at their event horizons. physics.stackexchange.com/questions/636832/…). $\endgroup$ May 10, 2023 at 18:27
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    $\begingroup$ @NuclearHoagie it’s not incorrect. The diverging gravitational acceleration is equivalent to the luminal escape velocity. The “surface gravity” has nothing to do with this plus nothing “exceeds the speed of light”, so your comment is incorrect on all accounts. $\endgroup$
    – safesphere
    May 10, 2023 at 18:33
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It's possible that they were talking about the tidal force at the event horizon (see Solomon Slow's answer), but also possible that they were talking about the so-called surface gravity of a black hole.

Despite its name, the surface gravity isn't the acceleration you'd feel if stationary near the event horizon (standing on a thin shell around the black hole, say, so that there is a surface). That goes to infinity at the event horizon. Rather, it's that acceleration divided by the gravitational time dilation factor, which also goes to infinity at the event horizon. The ratio of the two ends up approaching a finite limit.

That limit has some physical significance: it's proportional to the temperature of the Hawking radiation, and I believe it's also the force per unit mass you'd have to exert on a fishing line to keep an object suspended just above the event horizon from a great height. (At least these lecture notes by Thomas Hartman, p.23, agree with me.) But it's not the acceleration you'd feel if stationary near the horizon.

That the surface gravity decreases with increasing mass is somewhat intuitive for Schwarzschild black holes, where it works out to the Newtonian-looking $GM/r^2$, and $r$ is the Schwarzschild radius, which is proportional to $M$. It's not so intuitive in other cases. In particular, the surface gravity of an extremal black hole of any size is zero.

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