Schwarzschild Radius Calculator
Calculate the Schwarzschild radius (event horizon) of a black hole from its mass. Find density, Hawking temperature, evaporation time, and tidal forces.
Schwarzschild Radius Calculator
Calculator
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Formula: r_s = 2GM / c^2
Worked example โ Schwarzschild radius: 29.54 km | Density: 1.84 x 10^14 kg/m^3 | Temp: 6.17 nK
Formula
r_s = 2GM / c^2
Where r_s = Schwarzschild radius, G = gravitational constant (6.674 x 10^-11 m^3 kg^-1 s^-2), M = mass of the object, c = speed of light (2.998 x 10^8 m/s). This gives the radius at which escape velocity equals the speed of light.
Worked Examples
Example 1: Stellar Black Hole (10 Solar Masses)
Problem:Calculate the Schwarzschild radius, density, and Hawking temperature of a black hole with 10 times the mass of our Sun.
Solution:Mass = 10 x 1.989 x 10^30 kg = 1.989 x 10^31 kg r_s = 2 x 6.674e-11 x 1.989e31 / (2.998e8)^2 r_s = 2.954e20 / 8.988e16 = 29,543 meters = 29.54 km Density = M / (4/3 x pi x r_s^3) = 1.84 x 10^14 kg/m^3 Hawking Temp = hbar x c^3 / (8pi x G x M x k_B) = 6.17 x 10^-9 K
Result:Schwarzschild radius: 29.54 km | Density: 1.84 x 10^14 kg/m^3 | Temp: 6.17 nK
Example 2: Sagittarius A* (4 Million Solar Masses)
Problem:Calculate the event horizon size of the Milky Way's central supermassive black hole at approximately 4 million solar masses.
Solution:Mass = 4 x 10^6 x 1.989e30 = 7.956 x 10^36 kg r_s = 2 x 6.674e-11 x 7.956e36 / (2.998e8)^2 r_s = 1.062e27 / 8.988e16 = 1.181 x 10^10 m = 1.181 x 10^7 km In AU = 1.181e7 / 1.496e8 = 0.0789 AU Density = 1.14 x 10^6 kg/m^3 (about the density of gold)
Result:Schwarzschild radius: 11.81 million km (0.079 AU) | Density: ~gold
Frequently Asked Questions
What is the Schwarzschild radius and what does it represent?
The Schwarzschild radius defines the size of the event horizon of a non-rotating, uncharged black hole. Named after Karl Schwarzschild, who derived this solution to Einstein's field equations in 1916, it represents the critical radius at which the escape velocity equals the speed of light. Any object compressed within its own Schwarzschild radius becomes a black hole from which nothing, not even light, can escape. The formula is remarkably simple: r_s = 2GM/c^2, where G is the gravitational constant, M is the mass, and c is the speed of light. For our Sun, the Schwarzschild radius is approximately 2.95 kilometers, meaning if you compressed the entire mass of the Sun into a sphere less than 3 km in radius, it would become a black hole. For Earth, this critical radius is only about 8.87 millimeters.
How does black hole mass relate to its size and density?
The relationship between a black hole's mass and its Schwarzschild radius is perfectly linear: doubling the mass doubles the radius. However, this creates a counterintuitive result regarding density. Since volume scales as the cube of the radius while mass scales linearly with radius, the average density within the event horizon actually decreases as mass increases. A stellar black hole of 10 solar masses has a density of about 2 x 10^14 g/cm^3, far exceeding nuclear density. However, supermassive black holes like Sagittarius A* (4 million solar masses) have an average density comparable to water, and the largest known supermassive black holes (billions of solar masses) can have average densities lower than air. This means you could theoretically cross the event horizon of a sufficiently large black hole without experiencing extreme tidal forces.
What is Hawking radiation and how long do black holes last?
Hawking radiation is a theoretical process by which black holes slowly lose mass and energy through quantum mechanical effects near the event horizon. Proposed by Stephen Hawking in 1974, it arises from virtual particle-antiparticle pairs that form near the event horizon, where one particle falls in while the other escapes as thermal radiation. The temperature of this radiation is inversely proportional to the black hole's mass: smaller black holes are hotter and radiate faster. A stellar-mass black hole of 10 solar masses has a Hawking temperature of only about 6 x 10^-9 Kelvin, far colder than the cosmic microwave background, meaning it currently absorbs more energy than it emits. The evaporation time for such a black hole is approximately 2 x 10^67 years, vastly longer than the current age of the universe at 13.8 billion years.
What happens at the event horizon of a black hole?
The event horizon is not a physical surface but a mathematical boundary in spacetime. An astronaut crossing the event horizon of a sufficiently large black hole would not notice anything unusual at the moment of crossing, as local spacetime appears normal. However, to a distant observer, the astronaut would appear to slow down, redshift, and eventually fade from view, never quite reaching the horizon due to extreme gravitational time dilation. The tidal forces experienced depend on the black hole's mass: for stellar-mass black holes, tidal forces at the horizon are extremely intense, enough to stretch and compress matter in a process called spaghettification. For supermassive black holes, tidal forces at the horizon can be quite mild because the radius is so large. Once past the event horizon, all worldlines point toward the singularity, making escape physically impossible regardless of speed or thrust.
How are real black holes different from the Schwarzschild solution?
The Schwarzschild solution describes an idealized, non-rotating, electrically neutral black hole in a vacuum. Real astrophysical black holes almost certainly rotate, making the Kerr metric a more accurate description. Rotating black holes have two important differences: they possess an ergosphere outside the event horizon where spacetime itself is dragged along with the rotation, and their singularity is ring-shaped rather than a point. The event horizon of a Kerr black hole is smaller than the Schwarzschild radius for the same mass, with maximum rotation reducing it by half. Charged black holes are described by the Reissner-Nordstrom metric (non-rotating) or the Kerr-Newman metric (rotating and charged), though astrophysical black holes likely carry negligible net charge. Despite these differences, the Schwarzschild radius remains a useful first approximation and upper bound on the event horizon size.
What are the photon sphere and the innermost stable circular orbit?
Both are fixed multiples of the Schwarzschild radius, which makes them easy to read off any mass. The photon sphere sits at 1.5 times the Schwarzschild radius, the unique distance where light can circle the hole on a closed but unstable path. The innermost stable circular orbit, or ISCO, lies at 3 times the Schwarzschild radius; inside it no massive particle can hold a stable circular orbit, so accretion discs around non-rotating black holes truncate there and the ISCO sets the inner edge of the X-ray emitting region.
How much does time slow down for an observer hovering near the event horizon?
In the Schwarzschild metric a stationary clock at radius r ticks slower than a distant clock by the factor equal to the square root of one minus the Schwarzschild radius divided by r. At twice the Schwarzschild radius that factor is about 0.707, at ten times it is about 0.949, and it falls to zero exactly at the horizon. The same factor gravitationally redshifts emitted light, which is why an infalling signal appears to a distant observer to dim and stretch toward infinite wavelength rather than ever arriving from the horizon.
Why are tidal forces at the horizon gentler for a more massive black hole?
Tidal acceleration across a body of length L scales as GM times L divided by the cube of the distance. Evaluated at the horizon, the distance is itself proportional to M, so the cube in the denominator wins and the tidal stretch falls off as one over the mass squared. A ten solar mass hole would tear a human apart well outside its 30 kilometre horizon, whereas the horizon of a several million solar mass hole is so wide that the tidal gradient there is unremarkable. Spaghettification is therefore a small-black-hole phenomenon.
Why is a black hole's shadow larger than its Schwarzschild radius?
Gravitational lensing bends light around the hole, so the dark patch seen by a distant telescope is the capture cross-section for photons, not the horizon itself. For a non-rotating hole that shadow has a radius of the square root of 27 times GM over c squared, roughly 2.6 Schwarzschild radii, giving an apparent diameter about 5.2 times the Schwarzschild radius. This is why the ring imaged by the Event Horizon Telescope around M87* is several times wider than the naive horizon size computed from the mass alone.
Why does a Newtonian escape-velocity argument give the same formula as general relativity?
Setting the Newtonian escape speed equal to c gives r = 2GM over c squared, exactly the relativistic result, but the agreement is a numerical coincidence of the Schwarzschild solution rather than evidence that the Newtonian reasoning is valid. In the eighteenth century Michell and Laplace used that argument to imagine dark stars from which light could not depart, yet Newtonian gravity has no event horizon, no light-speed limit and no one-way surface. The relativistic derivation instead identifies where the metric coefficient vanishes and outward light cones tip inward.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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