Schwarzschild Radius Calculator

Compress any amount of mass into a small enough volume, and it becomes a black hole — the exact radius where that happens is the Schwarzschild radius. It’s one of the simplest formulas in general relativity to write down, and one of the most surprising to actually compute: Earth’s Schwarzschild radius is about the size of a marble.

The formula

That’s it — no exponent, no approximation. Double the mass and the radius exactly doubles; this is a straight proportionality, not merely a power law that happens to look linear. A 1 M☉ object has $r_s$ ≈ 2.95 km, so a 10 M☉ black hole’s event horizon sits at almost exactly 29.5 km — ten times the mass, ten times the radius, precisely.

What it actually means

Every object technically has a Schwarzschild radius — it’s just usually far smaller than the object itself, meaning nothing physical happens at that radius. The Sun’s Schwarzschild radius is about 3 km, deep inside its actual ~696,000 km surface; nothing would need to change for the Sun to “become” a black hole unless it were somehow compressed down to that size, which stellar physics doesn’t permit for a star of its mass. Only when a real, physical event horizon actually forms — in a collapsed massive star’s core, or at a galaxy’s center — does $r_s$ become the size of something real.

Real black holes, real scales

• A 10 M☉ stellar-mass black hole: $r_s$ ≈ 29.5 km — smaller than most cities.
• Sagittarius A* (the Milky Way’s central black hole, ~4.3 million M☉): $r_s$ ≈ 12.7 million km, about a fifth of Mercury’s orbital distance from the Sun.
• M87* (the first black hole ever directly imaged, ~6.5 billion M☉): $r_s$ ≈ 19.2 billion km, or about 128 AU — larger than Pluto’s orbit around the Sun.

An important caveat: this is Schwarzschild, not Kerr

This formula describes an idealized non-rotating, uncharged black hole. Every black hole that actually forms from stellar collapse carries at least some spin, and a spinning (Kerr) black hole’s event horizon follows a different formula entirely: $r_+ = r_g(1 + \sqrt{1-a_*^2})$ , where $r_g = GM/c^2 = r_s/2$ and $a_*$ is the dimensionless spin. At zero spin this reduces to exactly $r_s$ above; at maximal spin, the horizon shrinks to just half of $r_s$ . For the spin- dependent version — and to see how it relates to the innermost stable orbit around a rotating black hole — see this site’s Black Hole ISCO Calculator.

The same equation, solved the other way

Escape velocity is $v_{\rm esc} = \sqrt{2GM/r}$ — the speed needed to leave a mass’s gravity for good. Set $v_{\rm esc} = c$ and solve for $r$ instead of for the speed, and out falls $r_s = 2GM/c^2$ exactly. The Schwarzschild radius isn’t a separate fact about black holes bolted onto escape velocity — it’s the radius at which escape velocity reaches $c$ . This site’s Escape Velocity Calculator is the same formula approached from the other direction, complete with a comparison ladder from a small asteroid up through a neutron star.

Reading the two visuals

• The event-horizon diagram draws the horizon at a fixed size on screen with a scale bar underneath showing what real distance it represents — the only sane way to compare sizes ranging from millimeters (a small asteroid’s horizon) to hundreds of AU (M87*) on one page. The plain-language comparison above it (“about the size of…”) does the same job in words.
• The mass-vs-radius chart plots every preset — from Earth to M87* — on one log-log chart, where the relationship appears as a single straight line. That line’s slope is exactly 1, which is what makes it worth stating plainly: the straightness here reflects genuine direct proportionality, not just “a power law that happens to plot straight.”