Black holes and the Schwarzschild radius

Builds on: Sections S4, 3.1 and 3.2
Reading time: 40 min
Exercises: 6
Olympiad problems: 2

In infrared images of the centre of the Milky Way, stars can be followed for years as they turn round a small region that gives out radio waves, called Sagittarius A*. Their orbits weigh what is inside, just as the orbit of a moon weighs a planet. They show that an enormous mass occupies a very small region. The mass alone does not make this a black hole. What matters is how tightly it is packed. In this section we find the radius that separates an ordinary gravitating body from a black hole, and what that radius tells us about black holes of different masses.

10.1.1Mass and size

Before you readMake a guess first

Suppose the Sun were replaced by a black hole of exactly the same mass, without disturbing Earth. What would happen to Earth's orbit?

Far outside a spherical body, gravity depends on its mass and our distance from its centre. Shrinking the body while keeping its mass does not change that distant pull. It does, however, let us approach closer to the centre, where gravity is stronger and escaping takes more energy.

A compact object has a large mass in a small volume. White dwarfs and neutron stars are the dense remains of stars and still have material surfaces. A black hole has a region from which no light signal can reach a distant observer. We need a measure of size as well as mass to tell these objects apart. The full story of how stars leave these remains belongs to Chapter 7, but we can understand their gravity without it.

The escape speed is the speed a projectile needs to reach a very large distance with no speed left, without further propulsion after launch. At a distance $r$ from a spherical body of mass $M$ , (S.21) gives $v_{esc} = \sqrt{2 G M / r}$ , where $G$ is the gravitational constant.

At the surface of the Sun, whose radius is $6.957 \times 10^{8}$ m, the escape speed is about 618 km/s. Compressing the Sun while keeping its mass would let a projectile start closer to the centre, so it would need a larger escape speed.

10.1.2The Schwarzschild radius

The speed of light in a vacuum is $c$ . As we reduce the radius at fixed mass, the Newtonian escape-speed estimate eventually reaches $c$ . The radius where this happens is the Schwarzschild radius. General relativity gives it a physical meaning as the horizon radius of a black hole.

ResultThe Schwarzschild radius(10.1)

R_{s} = \frac{2 G M}{c^{2}}

$R_{s}$	the Schwarzschild radius
$M$	the mass of the object
$G, c$	the gravitational constant and the speed of light in a vacuum

For an isolated, non-rotating, uncharged black hole, this is the radius of its event horizon. The radius is proportional to the mass, with 2.95 km for each solar mass.

Quick checkA black hole of ten solar masses

What is the horizon radius of a non-rotating, uncharged black hole of ten solar masses?

Show the derivation

Setting the Newtonian escape-speed estimate equal to $c$ and squaring gives $c^{2} = 2 G M / R_{s}$ . Multiplying by $R_{s}$ and dividing by $c^{2}$ gives (10.1).

The constants are $G = 6.674 \times 10^{- 11}$ m³ kg⁻¹ s⁻² and $c = 2.998 \times 10^{8}$ m/s. For the Sun's mass, $M_{⊙} = 1.988 \times 10^{30}$ kg,

\begin{aligned} R_{s} & = \frac{2 G M_{⊙}}{c^{2}} \\ = 2952 m = 2.95 km . \end{aligned}

The same formula gives a radius twice as large for twice the mass. This is why the solar-mass conversion works for any mass.

10.1.3What the horizon means

The escape-speed argument finds the correct length, but it does not prove that light is trapped. Newton's expression for the kinetic energy of a projectile cannot be applied to light. The explanation comes from Einstein's general relativity, the theory in which mass changes the geometry of space and time and therefore the possible paths of light.

The event horizon is the boundary of the region from which a signal can never reach a distant observer. A signal sent outwards from just outside the horizon can escape. From inside it, even an outward-directed light signal cannot get out. Light still travels locally at $c$ . The horizon has no solid surface that light strikes or that a falling astronaut lands on.

We can also calculate a for Earth or the Sun. They are much larger than their own Schwarzschild radii and have no event horizons. The Sun would have to fit inside a radius of about 2.95 km to form a black hole with its present mass. This is a hypothetical compression. The Sun will end its life as a white dwarf.

Figure 10.1Distance from an event horizon

MassRadius of X

Horizon radius
Radius of X
Newtonian escape estimate at X

The figure shows the horizon of a non-rotating, uncharged black hole, with a fixed kilometre scale. O marks the centre of the drawing, H is on the horizon, and X is outside it at radius

r

. Increasing the mass enlarges the horizon. Changing the radius of X moves X along OX. The scale compares the radii in (10.1), which differ from distances measured by a radial ruler near the hole. The escape estimate in the table comes from Newtonian gravity. The figure can be set to , , or .

The useful comparison is $R_{s} / r$ . Replacing $2 G M$ in the escape-speed expression by $c^{2} R_{s}$ gives

\frac{v_{esc}}{c} = \sqrt{\frac{R_{s}}{r}} .

At four Schwarzschild radii the Newtonian estimate is $c / 2$ , whatever the black hole's mass. Doubling the mass doubles both the horizon radius and the distance of a point at four horizon radii. It leaves this ratio unchanged. When $R_{s} / r$ is small, Newtonian gravity is a good approximation. Near the horizon we need general relativity to describe motion and light.

Example 10.1The size of the Galactic centre's black hole

For a size estimate, take the mass of Sagittarius A* as $4.0 \times 10^{6} M_{⊙}$ and treat it as non-rotating and uncharged. What is its Schwarzschild radius, in kilometres and astronomical units? Take 1 au to be $1.496 \times 10^{8}$ km.

Solution

By (10.1), using the constants before rounding,

\begin{aligned} R_{s} & = \frac{2 G M}{c^{2}} \\ = 1.18 \times 10^{7} km . \end{aligned}

In astronomical units,

R_{s} = \frac{1.18 \times 10^{7}}{1.496 \times 10^{8}} = 0.079 au .

AnswerThe radius is about 12 million km, or 0.079 au, smaller than Earth's distance from the Sun. A black hole can be much larger than a star in kilometres. Its size must be judged alongside its mass.

Example 10.2Earth's Schwarzschild radiusfinish it

Earth's mass is $5.972 \times 10^{24}$ kg. What is its Schwarzschild radius in millimetres?

The start

By (10.1), $R_{s} = 2 G M / c^{2}$ . Substituting the mass in kilograms with the SI values of $G$ and $c$ gives a radius in metres.

Your stepNow calculate the radius and convert it to millimetres.

10.1.4A larger black hole has a lower mean density

Before you readMake a guess first

One non-rotating black hole has ten times the mass of another. How do the mean densities calculated inside their Schwarzschild radii compare?

In an olympiad problem, the mean density of a black hole usually means its mass divided by the ordinary spherical volume calculated from $R_{s}$ . Since the radius grows in proportion to mass, that volume grows as the cube of the mass. Dividing the mass by the volume therefore gives a density that falls as the inverse square of the mass.

ResultThe conventional mean density inside the Schwarzschild radius(10.2)

\begin{aligned} \bar{ρ} & = \frac{M}{\frac{4}{3} π R_{s}^{3}} \\ = \frac{3 c^{6}}{32 π G^{3} M^{2}} \end{aligned}

$\bar{ρ}$	the mass divided by the spherical volume assigned to the horizon radius
$M, R_{s}$	the mass and its Schwarzschild radius

This is a useful comparison of mass and size. It does not say that a black hole is a uniform ball of matter, or describe the density or the true spatial volume inside its horizon.

For a solar mass the value is $1.84 \times 10^{19}$ kg/m³. Increasing the mass by a factor of $10^{8}$ lowers that value by a factor of $10^{16}$ , to about 1840 kg/m³. Thus a very massive black hole can have a conventional mean density comparable to familiar materials. This does not allow light to escape. The horizon condition is set by mass and radius together.

Quick checkDoubling the mass

A non-rotating black hole doubles its mass. What is its new conventional mean density divided by its old one?

Common errorsSection 10.1

The Schwarzschild radius is a radius. An angular diameter or a physical diameter needs twice that value.
The escape-speed calculation identifies the correct radius, but light does not obey the Newtonian projectile-energy equation. An event horizon is a boundary for possible signals, and light still travels locally at $c$ .
A distant orbit depends on the central mass. Making a body smaller at fixed mass does not give it a larger pull at the same distant point.
Calculating a Schwarzschild radius for an ordinary star does not give it an event horizon. Its matter lies far outside that radius.
The conventional mean density falls as $M^{- 2}$ . It is not a measurement of matter spread uniformly inside the horizon.

ExercisesSection 10.16 questions

10.1.1
What is the Schwarzschild radius of a black hole of eight solar masses?
km
10.1.2
A non-rotating, uncharged black hole has a horizon radius of 60.0 km. What is its mass in solar masses?
solar masses
10.1.3
A spherical neutron star has mass $1.4 M_{⊙}$ and a material radius of 12 km. Its Schwarzschild radius is 4.13 km. What does this comparison tell us?
10.1.4
At nine Schwarzschild radii from a black hole, what is the Newtonian escape-speed estimate divided by $c$ ?
10.1.5
Black hole A has a mass of $10 M_{⊙}$ and black hole B has a mass of $100 M_{⊙}$ . What is the conventional mean density of A divided by that of B?
times
10.1.6
What happens to a light signal sent locally outwards from inside an event horizon?

Olympiad problemsPast IOAA problems, restated2 problems

2007The Sun compressed into a black hole2 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 1.14. We have restated it in our own words and kept the original data. The solution was last checked on 22 October 2026.

Suppose the Sun collapsed into a non-rotating black hole, keeping its mass. What would be the radius of its event horizon?

I solved it without the solution

AnswerThe horizon radius would be about 2.95 km. The question is hypothetical. The Sun is not massive enough to end its life as a black hole.

2009The mean density of a supermassive black hole10 points

Adapted from IOAA 2009 (Tehran), theory problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 22 October 2026.

A black hole has mass $1 \times 10^{8} M_{⊙}$ . Calculate its mean mass density inside the Schwarzschild radius, using the spherical volume $4 π R_{s}^{3} / 3$ .

I solved it without the solution

AnswerThe conventional mean density is about 1840 kg/m³. This is a comparison based on mass and horizon radius, rather than a statement that matter fills the interior uniformly.