Schwarzschild Radius Calculator: How Small Before It Is a Black Hole
Schwarzschild Radius Calculator: how small would this have to be?
Put in a mass and the radius it actually has, and this works out the Schwarzschild radius — the size it would have to be squeezed to before it became a black hole — and then the thing worth knowing: how many factors of ten of squeezing that is. The Sun needs five. You need about twenty-five.
Any mass, and the radius that would make it a black hole
Work out the Schwarzschild radius of the Sun. Take the IAU nominal solar mass parameter GM = 1.327 1244×1020 m3 s−2 and G = 6.674 30×10−11 m3 kg−1 s−2, and then say how far from being a black hole the Sun is, given a radius of 6.957×108 m.
One line of real general relativity
- rs
- Schwarzschild radius, in metres. The radius of the event horizon of a non-rotating, uncharged black hole of mass M.
- M
- mass, in kilograms. The only thing rs depends on.
- G
- Newtonian constant of gravitation, 6.674 30(15)×10−11 m3 kg−1 s−2 (CODATA 2022). The worst-known constant in the formula: six figures, against nine for everything else.
- c
- speed of light in vacuum, exactly 299 792 458 m s−1 by the SI definition of the metre. It appears squared, so rs carries a factor of about 9×1016 in the denominator — which is the whole reason the answer is always so small.
- 2GM/c²
- the same combination that makes the Newtonian escape speed equal c: set √(2GM/R) = c and solve for R. Getting the right answer that way is a famous coincidence, not a derivation; the Newtonian argument gives the right number for the wrong reasons.
Worked example
Work out the Schwarzschild radius of the Sun. Take the IAU nominal solar mass parameter GM = 1.327 1244×1020 m3 s−2 and G = 6.674 30×10−11 m3 kg−1 s−2, and then say how far from being a black hole the Sun is, given a radius of 6.957×108 m.
Get the mass in kilograms first. M = GM / G = 1.3271244×1020 / 6.67430×10−11 = 1.988 41×1030 kg. This is the step that loses precision: GM is good to ten figures and G to six, so the mass in kilograms is good to six.
Then rs = 2GM/c². The numerator is 2 × 6.67430×10−11 × 1.98841×1030 = 2.654 25×1020. The denominator is c² = 8.987 551 787×1016. Dividing gives rs = 2953.25 m, which is 2.95 km.
Notice the shortcut. You never needed the kilogram at all: 2 × 1.3271244×1020 / 8.987551787×1016 is the same 2953.2501 m, and it uses only GM, which is the quantity actually measured. The kilogram route and the GM route agree here to every figure shown, because the mass in step 1 was itself GM/G — the G cancels back out.
Now the comparison, which is the part worth remembering. R / rs = 6.957×108 / 2953.25 = 235 571. So the Sun is about 236 thousand times wider than its own Schwarzschild radius.
In factors of ten: log10(235 571) = 5.372. You would have to shrink the Sun by five and a bit orders of magnitude — from 696 thousand kilometres down to three — before it collapsed. That is the number the diagram’s bar is showing.
One sanity check and one warning. Check: the mean density inside that horizon would be 1.8×1019 kg m−3, which is roughly ten times nuclear density — plausible for something that has collapsed. Warning: 2953.25 m is quoted to six figures only because GM and c are. Any Schwarzschild radius you compute from a mass someone measured in kilograms is good to no more than that mass is.
How far from being a black hole is it?
| Object | Mass | Schwarzschild radius r_s | Real radius R | Factors of ten to shrink by |
|---|---|---|---|---|
| You, at 70 kg | 70 kg | 1.04×10−25 m | ≈0.3 m | 24.5 |
| Mount Everest, roughly | ≈1015 kg | 1.5×10−12 m | ≈5000 m | 15.5 |
| The Moon | 7.3×1022 kg | 0.11 mm | 1.74×106 m | 10.2 |
| The Earth | 5.9722×1024 kg | 8.870 mm | 6.3781×106 m | 8.86 |
| The Sun | 1.98841×1030 kg | 2953.25 m | 6.957×108 m | 5.37 |
| A white dwarf, 0.6 Msun | 1.2×1030 kg | 1772 m | ≈7×106 m | 3.60 |
| A neutron star, 1.4 Msun | 2.78×1030 kg | 4135 m | ≈11 000 m | 0.42 |
| Sagittarius A*, 4.3×106 Msun | 8.6×1036 kg | 1.27×1010 m | — it is already a black hole | negative |
The same mass, squeezed: why the density is the wrong thing to look at
| Mass | r_s | Mean density inside r_s |
|---|---|---|
| 1 Msun | 2.95 km | 1.8×1019 kg m−3 |
| 10 Msun | 29.5 km | 1.8×1017 kg m−3 |
| 104 Msun | 2.95×104 km | 1.8×1011 kg m−3 |
| 108 Msun | 2.95×108 km | 1.8×103 kg m−3 |
| 1010 Msun | 2.95×1010 km | 0.18 kg m−3 |
What the Schwarzschild radius is, and what it is not
In 1916, within months of Einstein publishing general relativity, Karl Schwarzschild found an exact solution for the gravitational field outside a spherical mass. The solution has a special radius in it, 2GM/c², where the mathematics changes character completely. Squeeze a mass inside that radius and nothing — not light, not a signal, not a rocket — can get back out. That radius is now called the Schwarzschild radius, and the surface at that radius is the event horizon.
It depends on mass and nothing else. This is the first thing to get straight, because it is counter-intuitive. The Schwarzschild radius does not care what the object is made of, how hot it is, how fast it spins or how big it currently is. Every 70 kg object in the universe has the same Schwarzschild radius of 1.04×10−25 m. The radius box on this page exists only so the calculator can tell you the one genuinely useful thing: the ratio R / rs, and how many factors of ten that is.
Why the answer is always so small. The c² in the denominator is 9×1016, and G in the numerator is 6.7×10−11. Together they put a factor of about 1.5×10−27 metres per kilogram in front of the mass. Gravity is extraordinarily weak and light is extraordinarily fast, and the Schwarzschild radius is where those two facts meet. You need roughly the mass of a star before the answer reaches the size of a city.
The Newtonian coincidence. Set the Newtonian escape speed √(2GM/R) equal to c and you get R = 2GM/c² — the right answer. John Michell did essentially this in 1783 and Laplace repeated it in 1796, well over a century before general relativity. It is a coincidence of the Schwarzschild metric and not a proof: the Newtonian calculation treats light as a slow particle that is thrown up, rises and falls back, which is wrong in every detail. The escape velocity page does the Newtonian version properly, and it is worth doing both to see that the agreement really is only in the final number.
Where this goes wrong in exam answers. Almost always the mass unit. A mass given in solar masses or in grams, fed into the formula as though it were kilograms, gives an answer wrong by thirty orders of magnitude or by three — and neither looks obviously absurd, because nobody has an intuition for what a Schwarzschild radius ought to be. Convert to kilograms first, every time, and then check the answer against the Sun’s 2.95 km. The second most common error is dropping the 2, which halves the answer; the third is squaring G instead of c.
What a horizon is not. There is no surface at rs, nothing to touch and nothing locally unusual to measure: an astronaut crossing the horizon of a large hole feels nothing at the moment of crossing. The horizon is defined by where light can get to, not by anything present at that place. And this formula describes only the simplest case — a non-rotating, uncharged hole. Real black holes spin, and for a rapidly spinning one the horizon sits at a smaller radius than 2GM/c², down to half of it in the extreme case.
Frequently asked questions
Could I be turned into a black hole?
Not by anything known. Your Schwarzschild radius is about 10−25 m, which is a hundred billion times smaller than a proton, and there is no process in the universe that compresses ordinary matter to anything like that. Even a neutron star — the densest stable thing there is — is still about two and a half times bigger than its own Schwarzschild radius, and it got that way only by having a star’s worth of gravity crushing it. Gravity is the only force that can do this job, and gravity needs mass. You do not have enough.
Does a black hole suck everything in?
No more than any other object of the same mass. Outside the horizon the field of a Schwarzschild black hole is almost exactly Newtonian: replace the Sun with a 1-solar-mass black hole and the Earth’s orbit would not change. What is different is that you can now get close — within a few kilometres of all that mass instead of 700 000 — and it is being close that makes the field enormous, not the fact that it is a black hole. The inverse-square law page is doing all the work here.
Why is the Sun’s Schwarzschild radius known better than its mass?
Because rs = 2GM/c² needs the product GM, and GM is what orbits actually measure. Watch a planet and you measure GM directly, to about ten significant figures. To get M in kilograms you must then divide by G, which is known to only about six figures — a relative uncertainty of 2.2×10−5, the worst of any constant in this formula. So the Sun’s Schwarzschild radius is good to ten figures and its mass in kilograms to six. The uncertainty you add by converting is pure loss. This is a nice real example for the error propagation page.
What is the photon sphere, and why is it at 1.5 r_s?
At r = 1.5 rs light can orbit the hole in a circle. It is an unstable orbit — nudge the photon either way and it spirals in or escapes — but it is the reason a black hole has a sharp dark disc in an image with a bright ring around it, and the ring’s size is set by this radius rather than by the horizon. The calculator gives it, and the innermost stable circular orbit for matter at 3 rs, because both are more directly observable than the horizon itself.
Is the Schwarzschild radius the same as the event horizon?
For a non-rotating, uncharged black hole, yes — that is the case this page computes. For a rotating (Kerr) hole the outer horizon sits at a smaller radius, shrinking towards rs/2 as the spin approaches its maximum. Real astrophysical holes do spin, sometimes fast, so treat 2GM/c² as the upper bound on the horizon radius for a given mass rather than as an exact figure for a particular object.
What happens to the density of a black hole as it gets bigger?
It falls, as 1/M². Because rs is proportional to M, the volume inside the horizon goes as M³, so the mean density goes as M/M³. A stellar-mass hole has a mean horizon density around 1019 kg m−3; a ten-billion-solar-mass one is thinner than air. The popular picture of a black hole as the densest thing imaginable is only true for small ones. What is always extreme is the ratio of mass to size — the compactness — and that is exactly what this page measures.
Related calculators
References
- NIST, Fundamental Physical Constants (SP 961, 2022 wallet card), and E. Tiesinga et al., CODATA Recommended Values of the Fundamental Physical Constants: 2022. Every constant on this page is taken from here: G = 6.674 30(15)×10−11 m3 kg−1 s−2, ℏ = 1.054 571 817×10−34 J s, k = 1.380 649×10−23 J K−1 and c = 299 792 458 m s−1. The 2022 report states that no new determination of G was available, so its 2022 and 2018 values are identical; c, h and k are exact by the 2019 SI definitions, and ℏ and the Stefan–Boltzmann constant follow from them exactly.
- International Astronomical Union, Resolution B3 on recommended nominal conversion constants for selected solar and planetary properties (XXIX General Assembly, 2015). Fixes the nominal solar radius at 6.957×108 m, the nominal solar mass parameter GM at 1.327 1244×1020 m3 s−2, the nominal solar luminosity at 3.828×1026 W, the nominal total solar irradiance at 1361 W m−2 and the nominal terrestrial mass parameter at 3.986 004×1014 m3 s−2. The resolution is explicit that these are agreed conversion factors chosen close to the best estimates, not measurements.
- K. Schwarzschild, Über das Gravitationsfeld eines Massenpunktes nach der Einsteinschen Theorie, Sitzungsberichte der Königlich Preussischen Akademie der Wissenschaften 7, 189–196 (1916). The original exact solution, and the first appearance of the radius that carries his name. Cited here for provenance; the numerical work on this page comes from the two constants references above.
- B. Schutz, A First Course in General Relativity, 2nd edition, Cambridge University Press (2009), chapters 10–11. A standard undergraduate treatment of the Schwarzschild solution, the horizon, the photon sphere at 1.5 rs and the innermost stable circular orbit at 3 rs. Named here as the level to read next; it is beyond school syllabuses.