How this instrument works
The Schwarzschild radius is the size a given mass would have to be compressed to before its own gravity stops light from leaving. It falls out of setting the classical escape velocity, v = √(2GM ⁄ r), equal to c and solving for r, which gives r = 2GM ⁄ c². Every mass has one — Earth's is about 8.9 millimetres, an ordinary car's is far smaller than a proton — but almost nothing ever gets squeezed anywhere near its own Schwarzschild radius, because ordinary matter resists that kind of collapse long before gravity could win.
That the escape-velocity shortcut gives the right number is closer to a coincidence than a proof. Karl Schwarzschild found the same formula properly in December 1916, solving Einstein's field equations exactly for a non-rotating, uncharged, spherical mass, while serving as an artillery officer on the Russian front; he mailed the solution to Einstein weeks before he died of a skin disease contracted there. Birkhoff's theorem later confirmed that exact exterior solution holds for any spherical mass, whether it sits still or pulses in size, which is why one formula covers a static star and a collapsing one alike.
The radius marks a boundary, not a place where the mass sits: past it, no outgoing light or signal can climb clear, which is the working definition of an event horizon. The formula also carries a real limit — it assumes zero spin and zero charge. A rotating mass needs the Kerr solution instead, whose horizon sits closer to the centre than this formula predicts, shrinking toward half the Schwarzschild value for a black hole spinning as fast as physics allows.
- Enter the object's mass in the Mass field — kilograms by default, with solar masses and Earth masses on the unit menu for astronomical bodies.
- Leave the constants alone: G and c are built into the formula and are not fields you set.
- Read the answer in the Schwarzschild radius field, switching to kilometres or astronomical units once the mass gets stellar.
- To judge whether a body is actually a black hole, compare this radius against its real, observed size — a mass confined inside its own Schwarzschild radius is one.
Worked example — the Sun's Schwarzschild radius
Take the Sun's mass, M = 1.989×10³⁰ kg. First 2GM: 2 × 6.674 30×10⁻¹¹ × 1.989×10³⁰ = 2.655037×10²⁰ m³ ⁄ s². Then c²: 299,792,458² = 8.987552×10¹⁶ m² ⁄ s². Dividing gives R_s = 2954.12655506 m — call it 2954.13 m, or 2.954 km, the radius the entire Sun would have to be compressed inside for its escape velocity to equal the speed of light.
That is roughly the size of a small town, not a star — which is the point of running the number at all. The Sun cannot collapse to 2.95 kilometres, because nothing in its physics compresses it that far; its core will end as a white dwarf roughly Earth-sized, held up by electron degeneracy pressure long before gravity gets anywhere close. The Schwarzschild radius is still a real, calculable figure for the Sun's mass regardless of whether the Sun could ever reach it — that is exactly what makes it useful, since it separates what gravity alone would demand from what a given star actually does.
Questions
What does the Schwarzschild radius actually represent?
It is the radius a given mass would need to be compressed inside for its escape velocity to equal the speed of light — the point past which nothing, not even light, can climb back out. That boundary is called the event horizon. Every mass, from a coin to a galaxy, has a Schwarzschild radius; almost none of them are ever small enough to reach it.
Why compute this for the Sun if the Sun isn't a black hole?
Because the formula only needs a mass, not a fate. The Sun's entire 1.989×10³⁰ kg, compressed to 2.95 kilometres, would satisfy the escape-velocity condition — but stars this size end as white dwarfs, not black holes, since their mass is too low for the core collapse that would actually get them there. The radius is a real, checkable number regardless of which outcome the star faces.
Does this formula work for a spinning black hole?
Not exactly. R_s = 2GM ⁄ c² assumes zero rotation and zero charge. A rotating mass is described by the Kerr metric, whose event horizon sits closer to the centre than the Schwarzschild value — down to half of it for a hole spinning at the theoretical maximum. Treat this instrument's output as the non-rotating limit, useful for scale but not exact for a known spinning hole.
How is this different from just calculating an escape velocity?
Setting Newtonian escape velocity equal to c and solving for r happens to land on the correct radius, but it is a coincidence of algebra, not a derivation that holds up physically — Newtonian mechanics does not actually apply once gravity is that strong. The same number falls out properly from Karl Schwarzschild's 1916 exact solution to Einstein's field equations, which is where the formula gets its name and its validity.
What is Earth's Schwarzschild radius?
About 8.9 millimetres — roughly a marble. Earth's actual radius is around 6,371 kilometres, so the planet sits nowhere near collapsing into a black hole; its mass would need to be squeezed down by a factor of more than 700 million to reach that size, far beyond anything Earth's gravity or composition could ever produce on its own.
Can the Schwarzschild radius be larger than the object itself?
For everyday objects, never — a car's Schwarzschild radius is trillions of times smaller than the car. For a large enough concentration of mass at low enough density, though, the Schwarzschild radius can exceed the object's physical size before any exotic collapse happens; this is the reasoning behind estimates that a sufficiently big, sufficiently diffuse cloud of gas could in principle already sit inside its own event horizon.