# Why there are black holes that have a mass only 6 times the mass of the Sun?

We know that black holes are actually "black" because no light can escape them due to their gravity and that's why they appear black. That means the mass of the black hole most be extremely large even in a cosmological scale.

If light cannot escape black holes due to the their gravity, and the more massive an object the stronger its gravity is, why there are black holes that have a mass only 6 times the mass of the Sun? Light should escape them in that case and make them visible. No?

• – Dvij D.C. Jun 2 at 16:28
• Why two downvotes to the question and 3 upvotes to the answer? This doesn't make much sense – lcv Jun 2 at 19:08
• @lcv Without making any statement about this particular case, it's possible to have a very good answer to a not-so-good question. In particular, elementary misunderstandings about somewhat advanced subjects (like black holes) tend to attract downvotes. – J. Murray Jun 2 at 20:09
• @J.Murray I don't see anything particularly bad with this question, do you? The OP has the (understandable) misconception that black holes are only due to their mass. – lcv Jun 2 at 20:31
• @lcv No, I don't. I didn't downvote the question, for what it's worth. – J. Murray Jun 2 at 20:50

The intuition that a black hole must have a very large mass is not true. The relevant parameter is how much mass is there within a (volume of some characteristic) radius. In the case of simple spherical objects, if a mass $$M$$ is concentrated within a radius $$2GM/c^2$$ then light (or anything else for that matter) cannot escape from the region $$r< 2GM/c^2$$, and the region $$r< 2GM/c^2$$ is called a black hole. Thus, for any small amount of mass, if it is concentrated within a (volume characterized by a) small enough radius, then it is a black hole. In principle, you can have a black hole of the mass of a human being, but of course, its radius would be ridiculously small. This doesn't mean that all astrophysical stars, no matter their mass, would turn into black holes because the non-gravitational forces within the stars can resist the mass of the star from concentrating up to the required small radius $$2GM/c^2$$ if the mass of the star is not large enough. However, if the mass of the star is large enough (as described by the Chandrashekhar Tolman-Oppenheimer-Volkoff limit$$^*$$), the mass of the star would reach a stage where it's confined within the radius $$2GM/c^2$$, and it would become a black hole.

Edit

Notice that the relevant parameter is $$M/r$$, not $$M/r^3$$. The radius of a (non-rotating uncharged) black hole with mass $$M$$ scales as $$r_s\sim M$$. In other words, the density of a black hole with mass $$M$$ scales as $$M/r^3_s\sim 1/M^2$$. Thus, if you have a black hole with small enough mass (which would correspond to a black hole with a small enough radius), you can get as high a density as you want. There is no fundamental restriction on the maximum density as such besides whatever restrictions might exist on how small you can make a black hole in a quantum theory of gravity.

$$^*$$ Thanks to @CharlesFrancis for this correction. The Chandrashekhar limit is the limit on the maximum mass of a stable white dwarf which can either devolve into a neutron star or a black hole if the mass is higher than this limit. However, the Tolman-Oppenheimer-Volkoff limit is the limit on the maximum mass of a neutron star beyond which it would devolve into a black hole.

• So in other words, it is a matter of density and not mass. Right? – mil Jun 2 at 17:05
• you mean Tolman-Oppenheimer-Volkoff limit. – Charles Francis Jun 2 at 17:13
• @CharlesFrancis Thanks for the correction. I have added an addendum, kindly correct me if my understanding is still mistaken. – Dvij D.C. Jun 2 at 17:19
• @mil Not quite. The relevant parameter is $M/r$, not $M/r^3$. The radius of a black hole with mass $M$ scales as $r_S\sim M$. In other words, the density of a black hole with mass $M$ scales as $M/r_s^3\sim 1/M^2$. Thus, if you have a black hole with small enough mass (which would correspond to a black hole with a small enough radius), you can get as high a density as you want. There is no fundamental restriction on the maximum density as such besides whatever restrictions might exist on how small you can make a black hole in a quantum theory of gravity. – Dvij D.C. Jun 2 at 17:28
• Note also that $r>r_s$ at the TOV limit. – Rob Jeffries Jun 2 at 20:17

what makes a black hole is how much mass is squeezed down into how much space, which establishes how strong the force of gravity is at its surface. That in turn determines its escape velocity; once the escape velocity equals the speed of light, a black hole forms.

If you squeezed the earth down to the size of a pea, the surface gravity of that pea would be great enough to form a black hole.

If you squeezed the mass of the entire sun down into a sphere with a diameter of 6 kilometers or less, a black hole would form.

In the universe today, gravity is the only force capable of compressing matter enough to form a black hole all on its own. The minimum size of a black hole thus formed is somewhere between 1.5 and 3 solar masses.

That size of black hole is produced when a star runs out of fuel and thence cools off enough for gravity to overcome the pressure caused by heat inside the star.

(Since the earth contains so little mass compared to 1.5 solar masses, gravity will never be capable of squeezing the earth down to the size of a pea.)

• Minimum mass of black hole is about 3 solar masses (Tolman-Oppenheimer-Volkoff limit) Below that and above 1.39 solar masses (the Chandrasekhar limit), a neutron star forms. – Charles Francis Jun 2 at 17:26
• Current thought is the T-O-V result may have been in error, which if fixed results in the range I quoted. – niels nielsen Jun 2 at 17:34
• TOV themselves got a very low result. Seems like the figure is still being revised, but current estimates about 2.3 solar masses en.wikipedia.org/wiki/… – Charles Francis Jun 2 at 17:46