Schwarzschild Radius

Equivalent forms

r_s = \frac{2GM}{c^2}

A single invariant built from G, M, and c that defines the boundary of a black hole.

Symbol	Name	SI unit	Dimension	Range
$r_s$	Schwarzschild Radiusoutput Event horizon radius	m	L	0 – 10000000000
$M$	Mass Total mass of the object	kg	M	1 – 1e+40
$G$	Gravitational Constant Newton's constant (fixed)	m^3 kg^-1 s^-2	M^-1 L^3 T^-2	6.674e-11 – 6.674e-11
$c$	Speed of Light Speed of light in vacuum (fixed)	m/s	LT^-1	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Imperial
$r_s$	m	cm	ft
$M$	kg	g	slug
$G$	m^3 kg^-1 s^-2	cm^3 g^-1 s^-2	ft^3 slug^-1 s^-2
$c$	m/s	cm/s	ft/s

Where it holds

Valid for a non-rotating, uncharged spherically symmetric mass (Schwarzschild metric). Rotating masses use Kerr radius.

Dimensional analysis

r_s \to [L]

2GM/c^2 \to [M^-1\,\mathrm{L}^3\,\mathrm{T}^-2]\cdot [M]/[L^2\,\mathrm{T}^-2] = [L]

Both sides length.

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Discovery

Karl Schwarzschild · 1916

Schwarzschild derived the first exact solution of Einstein's field equations while serving on the Russian front during WWI.

Try this

What radius would Earth need to be crushed to, to become a black hole?

With M = 5.972e24 kg, compute r_s = 2GM/c^2 ≈ 8.87 mm.

Research status: stable

Real-world applications

Black hole mass estimation from EHT images
LIGO merger waveform templates
Accretion disk physics
Primordial black hole searches

Common misconceptions

r_s is not a physical surface — free-falling observers feel nothing crossing it
Newtonian derivation gives the correct answer only by coincidence
Schwarzschild radius grows linearly with M, not with sqrt(M)

Experimental verification

Event Horizon Telescope imaging of M87* and Sgr A* (2019, 2022) confirmed shadow sizes consistent with 2GM/c^2 to within 10%.

Derivation

Set Newtonian escape speed

v_esc = \sqrt{2GM/r}

equal to c:

c^2 = 2GM/r_s \to r_s = 2GM/c^2

.

The same radius drops out of the metric component

g_tt = 0

in the Schwarzschild solution of general relativity.

Limiting cases

M \to 0

⟶

r_s \to 0

No mass, no horizon.

M \to M_sun

⟶

r_s \approx 2.95\,\mathrm{km}

The Sun would fit inside 3 km to become a black hole.

M \to 10^9\,\mathrm{M}_sun

⟶

r_s \approx 3e12\,\mathrm{m}

Supermassive black hole horizon

\approx 20\,\mathrm{AU}

.

What if…

What if the mass rotates?

Use the Kerr metric: horizon shrinks as spin increases, down to GM/c^2 at maximal spin.

What if you fall through r_s?

Tidal effects dominate locally; an outside observer sees you frozen at the horizon due to infinite redshift.

What if G were

10\times

larger?

Every black hole's horizon would be $10\times$ bigger, and stars would collapse far more easily.

1

Earth as a black hole

Given ·

M:: 5.972e+24
G:: 6.674e-11
c:: 299792458

Find · r_s

Steps

Numerator: $2GM = 2 \times 6.674e-11 \times 5.972e24 \approx 7.97e14$
Denominator: $c^2 \approx 8.988e16$
$r_s \approx 7.97e14 / 8.988e16 \approx 8.87e-3\,\mathrm{m}$

Result ·

r_s = 2 \times 6.674e-11 \times 5.972e24 / (299792458)^2 \approx 8.87e-3\,\mathrm{m} \approx 8.87\,\mathrm{mm}

.

2

Sgr A* supermassive

Given ·

M:: 8.54e+36
G:: 6.674e-11
c:: 299792458

Find · r_s

Steps

$2GM \approx 1.14e27$
$c^2 \approx 8.988e16$
$r_s \approx 1.27e10\,\mathrm{m}$

Result ·

M \approx 4.3e_{6} M_sun \to r_s \approx 1.27e10\,\mathrm{m} \approx 0.085\,\mathrm{AU}

.

Lorentz Factor→Energy–Momentum Relation→Time Dilation→