# Gravity of black holes

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

• You are thinking of surface gravity $$\kappa =\frac {c^{4}}{4GM}$$ which is a little more subtle. See this. May 10, 2023 at 7:18
• 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. May 10, 2023 at 7:55
• 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? May 10, 2023 at 13:34

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.

• 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. May 10, 2023 at 18:24
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.