In short

The escape velocity from a distance r from a body of mass M is the minimum speed needed to reach infinity: v_{\text{esc}} = \sqrt{2GM/r}. From Earth's surface this is about 11.2 km/s. The orbital velocity for a circular orbit at radius r is the speed that makes gravity supply exactly the right centripetal pull: v_{\text{orb}} = \sqrt{GM/r}. At the surface of Earth this is about 7.9 km/s. The two are connected by a clean factor: v_{\text{esc}} = \sqrt{2}\,v_{\text{orb}} — so escape speed is about 41% faster than circular-orbit speed at the same radius.

In 1687, Newton drew a thought experiment that haunts every physics textbook. Climb an impossibly tall mountain. Stand at the top and fire a cannon horizontally. A modest cannonball drops to the ground a few kilometres downrange — the familiar parabola. Now fire a more powerful cannon. The ball flies farther before hitting the ground. Keep increasing the firing speed, and at some magic value, the ball's curved trajectory matches the curvature of the Earth itself. The ball falls forever and never hits the ground. It has become a satellite.

Fire even harder, and the ball's trajectory is no longer a closed circle — it is an ellipse, swinging outward and returning. Fire harder still, and the trajectory opens entirely — the ball flies off forever, leaving Earth behind. Somewhere between "keeps falling around Earth" and "leaves forever" there is a threshold: a specific speed. Below it, you have a bound satellite. Above it, you have a probe bound for Mars, for the outer planets, or for deep space.

This article derives both magic numbers — the speed that puts you in orbit and the speed that sets you free — from a single principle that works for every planet, moon, star, and black hole: the conservation of energy. ISRO's engineers use the same formulas to plan Chandrayaan's escape from low Earth orbit to a lunar trajectory, and every astrophysicist uses them to estimate how fast gas must be moving to escape a star's gravity.

Two distinct questions

The article tackles two related but different problems.

Problem 1: Orbital velocity. You want to keep going around a body at constant radius r. The orbit is a circle. What speed do you need?

Problem 2: Escape velocity. You want to leave the body entirely — get so far away that gravity is negligible and coast off forever. You fire straight up (or in any direction with the right energy). What is the minimum speed that does this?

Both answers fall out of the same two ingredients — Newton's law of universal gravitation and the gravitational potential energy of a two-body system. But the derivations feel different: orbital velocity comes from balancing forces (gravity must provide the centripetal pull), while escape velocity comes from accounting for energy (kinetic energy at launch must equal gravitational binding energy). Work through each in turn, then stitch them together.

Orbital velocity — a balance of forces

A satellite in a circular orbit of radius r around a planet of mass M moves at constant speed v. Its direction is always tangent to the circle — purely sideways, never toward the planet. Yet its velocity vector is constantly changing direction, which means the satellite is accelerating. The acceleration points toward the planet's centre (the definition of circular motion) and has magnitude:

a_{\text{centripetal}} = \frac{v^2}{r}

Something must supply this acceleration. The only force on the satellite is gravity, pulling it toward the planet's centre with magnitude:

F_{\text{grav}} = \frac{GMm}{r^2}

For the satellite to stay in a circular orbit, this force must provide exactly the centripetal pull — no more, no less:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Why: Newton's second law says F = ma. The force on the satellite is gravity GMm/r^2. The acceleration for circular motion at radius r with speed v is v^2/r. Equating the two gives the condition for stable circular orbit — the gravitational pull matches exactly what is needed to bend the satellite into a circle.

The m (satellite mass) cancels. That is worth pausing on — the orbital speed does not depend on the mass of the satellite. A 1 kg balloon and a 400 000 kg space station, at the same altitude, orbit at the same speed. Solve for v:

v^2 = \frac{GM}{r}
\boxed{v_{\text{orb}} = \sqrt{\frac{GM}{r}}}
Satellite in a circular orbit around Earth A satellite in a circular orbit around Earth. Arrows show the gravitational force toward Earth's centre and the tangential velocity perpendicular to the radius. M satellite (m) v F = GMm/r² r
A satellite in a circular orbit at radius $r$ around a body of mass $M$. Gravity pulls the satellite toward the centre with force $GMm/r^2$. The satellite's velocity $v$ is tangent to the circle (perpendicular to the radius). The orbital speed is set by the condition that gravity supplies exactly the centripetal force: $v = \sqrt{GM/r}$.

This is the orbital velocity. Its consequences are broad:

v_{\text{orb}}(R_E) = \sqrt{\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{6.371 \times 10^6}} = \sqrt{6.255 \times 10^7} \approx 7910 \text{ m/s}

About 7.91 km/s, or around 28 500 km/h — fast enough that a satellite at that speed orbits the Earth every 85 minutes.

A useful rewrite using surface gravity

At Earth's surface, g_0 = GM_E/R_E^2, so GM_E = g_0 R_E^2. Substituting into the orbital velocity formula at the surface:

v_{\text{orb}}(R_E) = \sqrt{\frac{GM_E}{R_E}} = \sqrt{\frac{g_0 R_E^2}{R_E}} = \sqrt{g_0 R_E}

Numerically: \sqrt{9.81 \times 6.371 \times 10^6} = \sqrt{6.250 \times 10^7} \approx 7906 m/s. Same answer in a form that uses only g_0 and R_E — two numbers you can memorise easily.

Escape velocity — an energy argument

Now the second question: how fast must you throw something straight up for it to leave Earth and never come back?

At first this feels like a harder problem. The object is not in a steady orbit; it is in a violent struggle with gravity, decelerating the whole way out. Do you need to integrate the equation of motion to answer this? No — energy conservation cuts through the whole calculation in a couple of lines.

Setting up the energy equation

When the object is at distance r from the centre, its kinetic and gravitational potential energies are:

\text{KE} = \tfrac{1}{2}mv^2
U = -\frac{GMm}{r}

Why: the gravitational potential energy of a two-body system, with the zero of potential taken at infinity, is -GMm/r. It is negative because the system is bound — you have to do work on it to separate the bodies. At infinity, U = 0; as r shrinks, U becomes more and more negative (the bodies are more tightly bound).

The total mechanical energy is:

E = \text{KE} + U = \tfrac{1}{2}mv^2 - \frac{GMm}{r}

With no non-conservative forces (ignore air drag and radiation losses), energy is conserved. Whatever E is at launch, it has the same value everywhere along the trajectory.

The condition for escape

Think about what "escape" means physically. You want the object to reach infinity with some velocity (possibly zero) and stop being "trapped" by Earth's gravity. At infinity, U = 0. The kinetic energy at infinity is non-negative: \text{KE}_\infty \ge 0.

Therefore the total energy at infinity is:

E_\infty = \text{KE}_\infty + 0 = \text{KE}_\infty \ge 0

Since E is the same at launch and at infinity:

E_{\text{launch}} \ge 0
\tfrac{1}{2}mv^2 - \frac{GMm}{r} \ge 0

The minimum launch speed that just barely reaches infinity (with \text{KE}_\infty = 0) is obtained by setting E_{\text{launch}} = 0:

\tfrac{1}{2}mv_{\text{esc}}^2 = \frac{GMm}{r}

The m cancels:

v_{\text{esc}}^2 = \frac{2GM}{r}
\boxed{v_{\text{esc}} = \sqrt{\frac{2GM}{r}}}

Why: E = 0 is the threshold between bound and unbound orbits. E < 0 means the object is trapped — as r grows, KE shrinks and hits zero before r reaches infinity. E > 0 means the object arrives at infinity still moving. E = 0 is the exact boundary: just enough energy to coast to infinity and arrive there at rest.

A profound point: like orbital velocity, escape velocity does not depend on the launched object's mass. A 1 kg rock and a 300 000 kg Saturn V rocket both need 11.2 km/s to escape Earth. What differs is the energy required (scales with mass), not the speed.

Earth's escape velocity — the 11.2 km/s

Plug in Earth's numbers:

v_{\text{esc}}(R_E) = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}} = \sqrt{1.251 \times 10^8} \approx 11\,186 \text{ m/s}

About 11.2 km/s, or 40 270 km/h. Throw something straight up at this speed from Earth's surface, and it just barely reaches infinity. Fire it faster, and it arrives at infinity with some leftover speed (and continues sailing through interplanetary space). Fire it slower, and it slows to zero before reaching infinity and falls back — maybe on a highly eccentric ellipse, but eventually back to Earth.

Like orbital velocity, you can rewrite escape velocity using g_0 and R_E alone:

v_{\text{esc}} = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2g_0 R_E}

At Earth's surface: \sqrt{2 \times 9.81 \times 6.371 \times 10^6} \approx 11\,180 m/s.

The clean relationship between the two

Stand back and compare:

v_{\text{orb}}(r) = \sqrt{\frac{GM}{r}}, \qquad v_{\text{esc}}(r) = \sqrt{\frac{2GM}{r}}

Divide one by the other:

\frac{v_{\text{esc}}}{v_{\text{orb}}} = \sqrt{\frac{2GM/r}{GM/r}} = \sqrt{2}
\boxed{v_{\text{esc}} = \sqrt{2}\,v_{\text{orb}}}

At the same radius, escape velocity is exactly \sqrt{2} \approx 1.414 times the orbital velocity. So a satellite already in a circular orbit needs only to boost its speed by about 41% to escape — a much smaller boost than you might intuitively guess.

Why the factor of √2? An energy view

Here is the intuitive picture. At radius r:

The ratio of these two energies is 2:

\frac{\text{KE}_{\text{esc}}}{\text{KE}_{\text{orb}}} = \frac{GMm/r}{GMm/(2r)} = 2

Escape demands twice the kinetic energy of an orbit at the same radius. Since kinetic energy is proportional to v^2, the speed ratio is \sqrt{2}. The factor of \sqrt{2} is not arbitrary — it encodes the fact that a circular orbit has exactly half the kinetic energy of the full binding energy, a deep property of inverse-square gravitational systems (it generalises to the virial theorem, treated in Satellites — Energy and Binding).

Explore the energy diagram

The interactive figure below shows the total mechanical energy as a function of launch speed at Earth's surface. Drag the red point to change the launch speed. Below the dashed line (E < 0) the trajectory is bound — the object comes back. At the dashed line (E = 0) the object just barely escapes. Above the line (E > 0) the object escapes with speed to spare.

Interactive: total energy vs launch speed A graph of total mechanical energy E per unit mass versus launch speed from Earth's surface. A parabola rises from the bottom-left at zero speed to zero at 11.2 km per second and positive above. A draggable point explores the relationship. 0 4 8 11.2 16 launch speed v (km/s) 0 +40 +80 −40 −80 E/m (MJ/kg) E = 0 (escape threshold) v_esc v_orb drag red point to change launch speed
Total mechanical energy per unit mass, $E/m = \tfrac{1}{2}v^2 - GM_E/R_E$, as a function of launch speed from Earth's surface. Below 11.2 km/s, $E < 0$ and the object is bound (it comes back). At 11.2 km/s exactly, $E = 0$ and the object just barely escapes. Above 11.2 km/s, $E > 0$ and the object has leftover energy at infinity. The dashed vertical lines mark the orbital velocity (7.91 km/s) and escape velocity (11.2 km/s) — the first is the speed that circles Earth, the second is the speed that leaves.

Why direction does not matter for escape

A subtle point: the escape velocity derivation used only the magnitude of the launch velocity. It did not care whether the object was fired straight up, or sideways, or at 45°.

This makes sense physically because gravitational potential energy depends only on distance from the source, not on direction. Energy is conservative: the amount of kinetic energy you need to reach infinity is fixed by the potential energy well you start in, regardless of which way you climb out.

There is one caveat. If you fire horizontally at speed v_{\text{esc}}, you do reach infinity — but along an open parabolic trajectory, not straight up. The parabolic path goes off to infinity but never back; kinetic energy decays to zero as r \to \infty. If you fire at 45°, you get a different parabolic path, also reaching infinity. If you fire straight up, you get a radial trajectory. All paths reach infinity; none loops back.

The catch is what happens on Earth: if you fire sideways, the trajectory would collide with the atmosphere and the ground well before reaching infinity. So in practice, rockets are launched vertically (or nearly so) until they are above the atmosphere, then turned sideways to build up horizontal speed. The rocket equations handle the staged burn, but the energy budget is still set by the escape condition.

Worked examples

Example 1: The ISS orbital speed

The International Space Station orbits Earth at an average altitude of 408 km. Taking R_E = 6371 km, M_E = 5.972 \times 10^{24} kg, and G = 6.674 \times 10^{-11} N·m²/kg², find (a) the orbital radius, (b) the orbital speed, (c) the orbital period, and (d) the escape speed from that altitude.

The ISS in circular orbit 408 km above Earth Earth with a dashed circular orbit passing 408 km above the surface. A small rectangle represents the ISS. Arrows show its velocity tangent to the orbit and gravity toward Earth's centre. Earth ISS v g h = 408 km surface
The ISS orbits at altitude 408 km. At that height, gravity still has 89% of its surface strength, but the station is moving sideways fast enough to keep falling past Earth forever.

Step 1. Orbital radius.

r = R_E + h = 6371 + 408 = 6779 \text{ km} = 6.779 \times 10^6 \text{ m}

Why: the "radius" of the orbit is measured from Earth's centre, not from the surface. Altitude + Earth radius = orbital radius.

Step 2. Orbital speed.

v_{\text{orb}} = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{6.779 \times 10^6}}
v_{\text{orb}} = \sqrt{\frac{3.986 \times 10^{14}}{6.779 \times 10^6}} = \sqrt{5.881 \times 10^7} \approx 7669 \text{ m/s}

Why: GM_E = 3.986 \times 10^{14} m³/s² is a convenient grouping to memorise — it is the "gravitational parameter" of Earth. Dividing by r and taking the square root gives the orbital speed. Note the units work: m³/s² divided by m gives m²/s², whose square root is m/s.

So the ISS cruises at about 7.67 km/s — or 27 600 km/h.

Step 3. Orbital period.

T = \frac{2\pi r}{v_{\text{orb}}} = \frac{2\pi \times 6.779 \times 10^6}{7669} = \frac{4.259 \times 10^7}{7669} \approx 5553 \text{ s}

Converting: 5553/60 = 92.6 minutes.

Why: period = circumference divided by speed. A full orbit is 2\pi r long, traversed at constant speed v_{\text{orb}}.

The ISS circles Earth in about 93 minutes — sunrise and sunset every 45 minutes for the astronauts.

Step 4. Escape speed from the ISS altitude.

v_{\text{esc}} = \sqrt{2}\,v_{\text{orb}} = \sqrt{2} \times 7669 \approx 10\,846 \text{ m/s}

Or compute directly:

v_{\text{esc}} = \sqrt{\frac{2GM_E}{r}} = \sqrt{2 \times 5.881 \times 10^7} = \sqrt{1.176 \times 10^8} \approx 10\,845 \text{ m/s}

Why: at any radius r, escape velocity is \sqrt{2} times the orbital velocity at that same radius. The two derivations agree — a useful check.

Result: ISS orbital radius 6779 km, orbital speed 7.67 km/s, period 93 minutes. Escape speed from ISS altitude is 10.85 km/s — slightly less than 11.2 km/s (the surface value), because the station is already partway out of Earth's gravity well.

What this shows: to escape Earth from the ISS altitude, you would need to add only 10.85 - 7.67 = 3.18 km/s of speed. That is the kind of velocity increment (Δv) mission planners tally when designing interplanetary trajectories — lunar and Martian missions often launch first into low Earth orbit, then boost by a few km/s to enter a transfer orbit. The ISS is already halfway out of the well.

Example 2: Escape velocity from the Moon

The Moon has mass M_M = 7.35 \times 10^{22} kg and radius R_M = 1.737 \times 10^6 m. Find (a) the escape velocity from the Moon's surface and (b) the ratio of the Moon's escape velocity to Earth's. (c) If a cricket bowler could project a ball at Earth's Mach-1 speed (~340 m/s) on the Moon, could the ball escape?

Comparison of Earth and Moon escape velocities Earth on the left with a tall arrow labelled 11.2 km/s. Moon on the right with a shorter arrow labelled 2.38 km/s. Earth 11.2 km/s Moon 2.38 km/s Arrows scaled proportionally: Earth escape is ~4.7× Moon escape
Escape velocity from Earth (11.2 km/s) and from the Moon (2.38 km/s). The Moon is much less massive and smaller, so its gravity is much weaker — a ball thrown from the Moon's surface at less than 2.38 km/s always comes back.

Step 1. Moon's escape velocity.

v_{\text{esc}}(\text{Moon}) = \sqrt{\frac{2GM_M}{R_M}} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 7.35 \times 10^{22}}{1.737 \times 10^6}}
= \sqrt{\frac{9.810 \times 10^{12}}{1.737 \times 10^6}} = \sqrt{5.647 \times 10^6} \approx 2377 \text{ m/s}

Why: same formula as for Earth, with Moon's mass and radius. The Moon is about 81 times less massive than Earth and about 3.7 times smaller in radius, so the product M/R shrinks by a factor of about 22, and the square root gives a factor of about 4.7 between the escape velocities.

Step 2. Ratio to Earth's escape velocity.

\frac{v_{\text{esc}}(\text{Moon})}{v_{\text{esc}}(\text{Earth})} = \frac{2377}{11\,186} \approx 0.213

The Moon's escape velocity is about 21% of Earth's — or Earth's is about 4.7 times the Moon's.

Step 3. Can a 340 m/s throw escape the Moon?

No — 340 m/s is about 1/7 of the Moon's escape velocity of 2377 m/s. A ball thrown at 340 m/s from the Moon's surface would reach some altitude, stop, and fall back.

How high would it reach? Set \tfrac{1}{2}mv^2 = U(R_M + h) - U(R_M) at the highest point:

\tfrac{1}{2}v^2 = \frac{GM_M}{R_M} - \frac{GM_M}{R_M + h}
\tfrac{1}{2}(340)^2 = \frac{GM_M \cdot h}{R_M(R_M + h)}
5.78 \times 10^4 = \frac{(6.674 \times 10^{-11})(7.35 \times 10^{22}) \cdot h}{R_M (R_M + h)}

For small h \ll R_M, replace R_M + h \approx R_M in the denominator:

5.78 \times 10^4 \approx \frac{4.905 \times 10^{12} \cdot h}{(1.737 \times 10^6)^2} = \frac{4.905 \times 10^{12}}{3.017 \times 10^{12}} \cdot h \approx 1.626\,h
h \approx \frac{5.78 \times 10^4}{1.626} \approx 3.55 \times 10^4 \text{ m} \approx 35.5 \text{ km}

Why: at 340 m/s launch speed, the KE converts to PE. Using the small-h approximation \Delta U \approx mg_M h, where g_M = GM_M/R_M^2 \approx 1.625 m/s² is the Moon's surface gravity, we get h = v^2/(2g_M) = (340)^2/(2 \times 1.625) \approx 35\,600 m. The two approaches agree.

So the 340 m/s ball rises about 35 km, then falls back — high but not escaping. To actually escape the Moon, you need roughly 7 times that speed, or 2377 m/s.

Result: Moon's escape velocity is 2.38 km/s, about 21% of Earth's. A 340 m/s throw on the Moon falls back after rising about 35 km.

What this shows: smaller, lighter worlds have much lower escape velocities. The Moon's low escape velocity is why it has no atmosphere — molecules of a would-be atmosphere are moving fast enough, at lunar surface temperatures, that a significant fraction escape to space over geological timescales. By contrast, a massive body like Jupiter has escape velocity 59.5 km/s at its cloud tops, enough to hold on to the lightest gas (hydrogen) forever. The same formula that tells you how to launch a rocket also tells you whether a planet can hold its air.

Common confusions

You now have the two velocities and understand their \sqrt{2} ratio. What follows is for readers who want the full classification of conic-section trajectories by energy, a derivation of escape velocity for non-radial launches, and the introduction to black holes — the objects whose escape velocity is so large it reaches the speed of light.

Trajectories classified by total energy

The nature of an object's trajectory in an inverse-square gravity field is determined entirely by its total energy E = \tfrac{1}{2}mv^2 - GMm/r. This is a deep result known to Johannes Kepler empirically and to Newton analytically:

  • E < 0: The orbit is a closed ellipse (or a circle as a special case). Bound to the central body. Returns to its starting configuration periodically.
  • E = 0: The trajectory is a parabola. Barely escapes — reaches infinity with zero residual speed.
  • E > 0: The trajectory is a hyperbola. Escapes with leftover speed; at infinity, moves along a straight asymptote at a non-zero speed equal to \sqrt{2E/m}.

The threshold E = 0 — the parabolic case — is exactly the escape-velocity condition. You can say: escape velocity is the launch speed at which the trajectory transitions from an ellipse to a hyperbola, passing through a parabola at the exact threshold.

The speed at infinity for a hyperbolic trajectory is called the excess hyperbolic velocity and is given by:

v_\infty^2 = v_{\text{launch}}^2 - v_{\text{esc}}^2

For a probe launched at 13 km/s from Earth's surface: v_\infty = \sqrt{13^2 - 11.2^2} = \sqrt{44.56} \approx 6.67 km/s. That is the speed with which the probe would arrive at the edge of interplanetary space, far from Earth but before the Sun's gravity has much to do with its motion.

A more careful look at "escape": escape requires infinite time

In the exact E = 0 trajectory, the object reaches infinity — but it takes an infinite time to do so. The object's velocity asymptotes to zero as r \to \infty. Mathematically, r(t) \sim (t)^{2/3} for a radial parabolic trajectory. So strictly, "escape" is the statement that the trajectory is unbound (reaches arbitrarily large r), not that it reaches a specific distance in a specific time.

In practice, long before an object gets "to infinity" (a physical impossibility), it gets far enough from Earth that other gravitational sources dominate — the Sun, the Moon, Jupiter. So the relevant comparison is: did the object get far enough from Earth to be captured by one of these other bodies, or by the outward Hubble expansion? Interplanetary trajectories use escape velocity only as a first approximation; the full calculation involves patched conics or n-body simulation.

Escape velocity for a rotating planet

Launching eastward from the equator, a rocket starts with a speed of about 465 m/s relative to Earth's centre (this is the equatorial rotation speed). Subtracting that from the 11.2 km/s escape velocity, the rocket only needs to supply 11.2 − 0.465 = 10.74 km/s relative to the launch pad. Launching from the equator eastward saves about 4% of the energy budget — which is why the Indian launch site in Sriharikota was chosen near the equator and why most rockets launch eastward.

More generally, if the launch velocity is \vec{v}_{\text{launch}} in the ground frame, and the ground itself is moving at \vec{v}_{\text{ground}} relative to Earth's centre (because of rotation), then the relevant speed for escape-velocity calculation is:

\vec{v}_{\text{net}} = \vec{v}_{\text{launch}} + \vec{v}_{\text{ground}}

and escape requires |\vec{v}_{\text{net}}| \ge 11.2 km/s. At Sriharikota (13.7° N), |\vec{v}_{\text{ground}}| \approx 451 m/s eastward, which is almost all captured when launching eastward.

Black holes — the extreme case

The escape velocity formula v_{\text{esc}} = \sqrt{2GM/r} has a fascinating property. For a fixed mass M, the escape velocity grows as r shrinks. What if you packed M into a small enough r that v_{\text{esc}} \ge c, the speed of light?

Setting v_{\text{esc}} = c and solving for r:

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

This radius r_s is the Schwarzschild radius — the size below which an object of mass M becomes a black hole, because even light cannot escape. For the Sun (M = 1.989 \times 10^{30} kg): r_s \approx 2.95 km — if you compressed the Sun into a sphere 3 km across, it would be a black hole. For Earth: r_s \approx 8.9 mm — compress Earth into a marble, and nothing could leave it.

The full general-relativistic treatment of black holes is more subtle (escape velocity as derived from Newtonian energy conservation is not strictly valid for relativistic objects), but the Schwarzschild radius happens to come out to the right answer, and that is a pleasant coincidence — Newtonian gravity pointing, correctly, at the existence of objects whose escape velocity equals c.

Why black holes really exist

Notice that the expression r_s = 2GM/c^2 is linear in M. Double the mass, double the Schwarzschild radius. This means that more massive black holes are larger, not smaller. A supermassive black hole with mass 10^9 M_\odot (like the one at the centre of M87) has a Schwarzschild radius of roughly 3 \times 10^{12} m — bigger than the orbit of Pluto. Such an object is easier to form than a stellar-mass black hole because the density required is lower (density scales as M/r_s^3 \propto 1/M^2).

This relationship between mass, radius, and density explains why we see black holes of every scale in the universe — from stellar-mass black holes formed when massive stars collapse, to supermassive ones at galactic centres. The escape velocity formula, this article's subject, is secretly a formula about when gravity wins.

Where this leads next