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Astronomy · Black Holes

Schwarzschild Radius Calculator

Find the event-horizon radius of a black hole from its mass with r_s = 2GM ÷ c².

Scientific notation is fine, e.g. 5.972e24.
Mass unitEnter the mass in kilograms or solar masses.
Real objects — tap to load
Schwarzschild radius (r_s)
2.954km

That is 2,954 m (1.000e+0 M☉). r_s = 2GM ÷ c², with G = 6.674×10⁻¹¹ N·m²/kg² and c = 2.99792458×10⁸ m/s.

The Schwarzschild radius is how small a mass must be squeezed to become a black hole: r_s = 2GM ÷ c². Compress the Sun (1 M☉) inside a radius of about 2.95 km and it forms an event horizon; the Earth would need to shrink to roughly 8.9 mm, smaller than a marble.

What the Schwarzschild radius is

The Schwarzschild radius r_s marks the event horizon of a non-rotating black hole — the surface at which the escape velocity equals the speed of light, so nothing, not even light, can get out. Karl Schwarzschild derived it in 1916 from Einstein’s field equations. Any mass compressed within its own Schwarzschild radius collapses into a black hole; a mass spread over a larger radius stays an ordinary star, planet, or lump of matter.

r_s = 2GM ÷ c²

G = 6.674×10⁻¹¹ N·m²/kg², c = 2.99792458×10⁸ m/s, M is the mass in kg, r_s the radius in metres

Worked example

What is the Sun’s Schwarzschild radius? Use M = 1.989×10³⁰ kg (one solar mass).

  1. 1
    Write the formula. r_s = 2GM ÷ c², with G = 6.674×10⁻¹¹ N·m²/kg² and c = 2.99792458×10⁸ m/s.
  2. 2
    Put the mass in kilograms. One solar mass is 1.989×10³⁰ kg, so M = 1.989×10³⁰ kg.
  3. 3
    Substitute the values. r_s = 2 × 6.674×10⁻¹¹ × 1.989×10³⁰ ÷ (2.99792458×10⁸)².
  4. 4
    Evaluate. The numerator is ≈ 2.655×10²⁰ and c² ≈ 8.988×10¹⁶, giving r_s ≈ 2,954 m ≈ 2.95 km.

Schwarzschild radius of some masses

Event-horizon radius computed with r_s = 2GM ÷ c².

ObjectMassSchwarzschild radius
Earth5.972×10²⁴ kg8.9 mm
The Sun1.989×10³⁰ kg (1 M☉)2.95 km
Sagittarius A*≈ 4.3×10⁶ M☉≈ 1.3×10⁷ km

Reading the result

The Schwarzschild radius is directly proportional to mass: double the mass and you double r_s. That is why the answer scales so cleanly — every solar mass adds about 2.95 km to the horizon. Sagittarius A*, the supermassive black hole at the centre of the Milky Way, holds roughly 4.3 million solar masses, so its horizon spans about 1.3×10⁷ km — larger than the orbit of Mercury.

Crossing inside r_s does not mean hitting a solid surface; the event horizon is a boundary in spacetime, not a wall. It simply marks the point of no return, where escaping would require travelling faster than light. Anything — a star, a planet, or you — compressed to a radius smaller than its own r_s becomes a black hole.

What is the event horizon?
The event horizon is the spherical boundary at the Schwarzschild radius where the escape velocity reaches the speed of light. Once anything crosses it — matter or light — it cannot return, because escaping would require moving faster than light. It is a boundary in spacetime, not a physical surface.
Would the Sun ever become a black hole?
No. The Sun lacks the mass to collapse that far. Stars need a remnant core above roughly 3 solar masses to overcome neutron degeneracy pressure; the Sun will end as a white dwarf about the size of Earth, never squeezing inside its 2.95 km Schwarzschild radius.
Why is the Schwarzschild radius proportional to mass?
Because r_s = 2GM ÷ c² is linear in M — G and c are constants. Doubling the mass doubles the radius. Each solar mass contributes about 2.95 km, so you can estimate any horizon size by multiplying the mass in solar masses by ≈ 2.95 km.
How big is the Earth’s Schwarzschild radius?
About 8.9 mm. Using M = 5.972×10²⁴ kg in r_s = 2GM ÷ c² gives ≈ 0.00887 m, roughly the size of a marble. Earth would have to be crushed to that radius to become a black hole, which no natural process can do.
Does this formula work for rotating black holes?
The Schwarzschild radius describes a non-rotating, uncharged (Schwarzschild) black hole. Real black holes usually spin, described by the Kerr solution, where the horizon is smaller than r_s and the geometry is more complex. For a first estimate of horizon scale, r_s = 2GM ÷ c² is still the standard benchmark.
Why does the escaping object’s mass not appear?
The Schwarzschild radius depends only on the central mass M, not on what tries to escape. It comes from setting the escape velocity equal to c, and escape velocity is independent of the escaping object’s mass. So the horizon size is fixed by the black hole alone.