Escape Velocity Calculator: From Leaving Earth to Falling Into a Black Hole

Fire a cannonball fast enough off a mountaintop and it doesn't just fly farther — past a certain speed, it stops coming down at all. Isaac Newton drew exactly this scene in A Treatise of the System of the World to explain orbits, and the same simple idea, pushed a little further, explains something Newton never could have imagined: black holes. Ask the same question — "how fast do you have to go to leave, for good?" — of an object so dense that even light can't go fast enough, and you've arrived at the Schwarzschild radius, the boundary of a black hole.

This page is an escape velocity calculator you fire rather than read. Pick a body — the Moon, Mars, Earth, Jupiter or the Sun — dial in a launch speed and fire. The ball crashes back down, loops forever without landing, or leaves and never returns, and every shot's trail stays on screen until you have rebuilt Newton's own diagram out of your own attempts. Below that sits a press. It crushes a body at constant mass while you watch what can still get off a shrinking surface, each speed class failing in turn until even light can't manage it. When light gives up, you've made a black hole, and the radius where that happens is the Schwarzschild radius.

What Is Escape Velocity and How Do You Calculate It?

Escape velocity is the minimum speed an object needs — launched once, with no further thrust — to break free of a body's gravity forever. Give it exactly that speed and it just barely makes it to infinity, arriving there with essentially nothing left. Give it more, and it leaves with speed to spare.

The formula falls out of a straightforward energy argument. An object of mass $m$ has kinetic energy $\frac{1}{2}mv^2$ and gravitational potential energy $-\frac{GMm}{r}$ (negative, because gravity is attractive — it takes work to pull the object away, not push it). "Just barely escaping" means the total mechanical energy is exactly zero: kinetic energy exactly cancels the gravitational binding energy at the surface.

$\frac{1}{2}mv_{esc}^2 = \frac{GMm}{R}$

The object's own mass $m$ cancels out completely. Solve for $v$ and you get:

$v_{esc} = \sqrt{\frac{2GM}{R}}$

where $M$ is the mass of the body you're escaping, $R$ its radius, and $G$ the gravitational constant, $6.674 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2$. Notice what's missing from that formula: the mass of whatever's escaping. A feather and a rocket need exactly the same escape velocity — the difference is entirely in how much energy it takes to get each of them up to that speed, not in the speed itself.

Earth's escape velocity, 11.19 km/s, is worth sitting with for a second — that's just over 40,000 km/h, or about 33 times the speed of sound. No rocket actually needs to hit that speed while still down at the surface (they build up speed gradually, fighting less gravity the higher they climb), but 11.19 km/s is what a single unpowered shot would need.

Does Escape Velocity Depend on Direction or Mass?

Not on the mass of the escaping object, as the derivation above shows. Direction matters less than you'd think, too: fired straight up, an object needs exactly $v_{esc}$. Fired at an angle, or even sideways, it still escapes at exactly the same speed — what changes is the shape of the path getting there, not the speed threshold itself. That's a direct consequence of energy conservation: "just enough energy to reach infinity" doesn't care which way the kinetic energy happens to point.

Newton's Cannon: Why Speed Alone Decides Orbit or Escape

Here's the thought experiment Newton actually used to argue that orbits and falling apples are the same physics. Imagine a cannon on a mountain so tall its tip pokes above the atmosphere, firing cannonballs perfectly horizontally.

Fire slowly, and the cannonball arcs out a little and lands — ordinary projectile motion, gravity pulling it down while it drifts forward. Fire faster and it goes farther before landing, the arc flattening out. Keep increasing the speed, and at some point the ball is falling around the curve of the planet exactly as fast as the ground curves away beneath it — it never lands. That's an orbit. Push the speed higher still, past $\sqrt{2}$ times that orbital speed, and gravity can no longer bend the path into a closed loop at all: the cannonball leaves and keeps leaving, forever.

Newton could only draw that figure. Below, you get to fire it — one shot at a time. The strip under the picture names both boundaries from the start; what it cannot tell you is which side of them any particular shot falls on, and that is what firing is for.

Fire a dozen shots and you will have redrawn, by hand, the figure Newton put in his System of the World: a fan of arcs from one mountaintop, each reaching further round the world than the last, until one of them stops coming down. The tick strip underneath is the same discovery written as data: red gives way to amber at one dashed line, amber to green at the other.

One honest detail about the numbers. The cannon stands on Newton's mountain rather than on the ground: $r_0 = 1.05R$, five per cent up. That isn't decoration — a horizontal shot fired from exactly $R$ at less than orbital speed starts at the top of its ellipse and dips below the surface immediately, so every slow shot would bury itself before you saw an arc. Every speed the widget quotes is therefore the value at that altitude, about 2.4% below the textbook surface figure in the table above: 10.92 km/s for Earth rather than 11.19. The physics is identical. And because every speed on the dial is measured against v_esc at that same launch altitude, the 1.00 × mark sits exactly on the escape boundary you can watch happen — the altitude moves the number, not the threshold.

What Is the Difference Between Orbital Velocity and Escape Velocity?

Both numbers come from the exact same balance of forces, just answering a different question. Orbital velocity asks "how fast to fall in a circle around this radius?" — set gravitational force equal to the centripetal force requirement (see our centripetal force calculator for that derivation on its own) and you get $v_{orbit} = \sqrt{GM/R}$. Escape velocity asks a bigger question — "how fast to leave the gravity well altogether?" — and the answer is always exactly $\sqrt{2}$ times larger:

$v_{esc} = \sqrt{2}\, v_{orbit}$

That factor of $\sqrt{2} \approx 1.414$ isn't a coincidence or a rough approximation — it drops straight out of comparing the two energy equations, and it's exactly why the amber ticks on the strip above give way to green where they do. It has a bit of history behind it, too: Soviet engineers labelled these the "cosmic velocities" — orbital speed is the first cosmic velocity, escape speed the second. For a full treatment of what a stable orbit actually looks like over time — ellipses, periods, Kepler's laws — see our orbital mechanics post; this page is deliberately about the boundary case, not the stable orbits sitting on either side of it.

How Do Black Holes Form?

A star spends most of its life in a standoff: gravity crushing inward, the outward pressure from nuclear fusion pushing back. Fusion wins that argument for billions of years — until the fuel runs out. For an ordinary star like the Sun, gravity wins gently: the core settles into a white dwarf, held up by quantum pressure between crowded electrons, and that's the end of the story.

For a star roughly 20 times the Sun's mass or more, gravity wins violently. The core collapses in on itself in under a second, the outer layers detonate outward as a supernova, and whatever's left of the core keeps falling inward with nothing left to stop it. If that remaining core is heavier than roughly 3 solar masses, not even the exotic pressure that supports a neutron star can hold the line. The collapse doesn't stop. A black hole is what "doesn't stop" looks like: matter compressed past the point where its own escape velocity would exceed the speed of light.

That last clause is the entire definition — and it connects straight back to the formula from the first section.

What Is the Schwarzschild Radius?

Take the escape velocity formula and ask a slightly unsettling question: what radius would a given mass $M$ need to be squeezed into for its escape velocity to equal $c$, the speed of light — the fastest anything can ever travel?

$c = \sqrt{\frac{2GM}{R_s}}$

Square both sides and solve for $R$:

$R_s = \frac{2GM}{c^2}$

This is the Schwarzschild radius, named for Karl Schwarzschild, who derived it properly using Einstein's general relativity in 1916 — remarkably, working out an exact solution to Einstein's field equations for a spherical mass while serving on the Russian front in the First World War, only months before his death. The Newtonian derivation above isn't technically rigorous (photons don't have mass, and gravity this close to a black hole isn't Newtonian at all), but it lands on precisely the right answer anyway — one of those odd coincidences in physics that turns out not to be a coincidence once you understand the deeper theory underneath it.

Compress any mass down inside its own Schwarzschild radius and the result is a black hole: a region where the escape velocity genuinely exceeds the speed of light, so nothing — not a rocket, not a radio signal, not light itself — can climb back out. The boundary itself, at radius $R_s$, is called the event horizon.

That is easy enough to write down and very hard to feel. The press below is an attempt at the feeling: one slider squeezing one body at a mass that never changes, its surface firing test launches the entire time you crush it.

Two things in that sequence are worth pulling apart. That the classes fail in speed order is obvious the moment you see it. What isn't obvious is that the last class to fail is special: light's threshold isn't one milestone among five, it is the definition. Setting $v_{esc} = c$ is the equation $R_s = 2GM/c^2$ solves, so when the light streaks curl back the surface has reached the ring, and nothing below it will ever be heard from outside again. (The press is Newtonian throughout, which is John Michell's "dark star" of 1783 — an idea Laplace arrived at independently in 1796 — rather than Einstein's black hole. Real light doesn't slow down. General relativity arrives at the same $R_s$ by an entirely different route, which is the coincidence-that-isn't from a moment ago.)

How Small Would Earth Have to Be to Become a Black Hole?

Earth's entire mass, squeezed into a sphere just 8.87 mm in radius — about the size of a marble — would become a black hole. The Sun would need to shrink to a radius of about 2.95 km: a sphere 5.9 km across, a few minutes' drive at highway speed, still holding onto its full mass.

Now put a neutron star next to that column. It packs about 1.4 solar masses — $2.78 \times 10^{30}$ kg — into a ball roughly 12 km in radius, and its Schwarzschild radius is 4.14 km. The compression it still needs is a factor of 2.9. Earth is nearly nine orders of magnitude from its own horizon; a dead stellar core has covered all but the last factor of three of that same distance. That gap, not the mass, is the whole reason black holes come from collapsed stars and never from planets. Nothing in the ordinary universe squeezes a planet at all, while a star's own collapse routinely does almost the entire job — and dropping a little more mass onto a neutron star finishes it without anyone's help.

Real Black Holes: From Stellar Remnants to Sagittarius A*

The smallest black holes that can form from stellar collapse sit right around 3 solar masses — the theoretical floor below which the leftover pressure in a collapsing core can still hold the line as a neutron star instead. A black hole that size has a Schwarzschild radius of about 8.9 km — an event horizon roughly 18 km across, about the span of a city.

Far larger black holes sit at the centers of galaxies. Sagittarius A*, the supermassive black hole anchoring the Milky Way, weighs in at about 4.3 million solar masses — its Schwarzschild radius works out to roughly 12.7 million km, comfortably smaller than Mercury's 0.39 AU orbit around the Sun. M87*, the first black hole ever directly imaged — by the Event Horizon Telescope in 2019 — is far bigger still: about 6.5 billion solar masses, giving it a Schwarzschild radius of roughly 128 AU, bigger than Neptune's entire 30 AU orbit. Swap M87* in for our Sun and its event horizon alone would swallow every planet in the solar system. Working out distances on this scale — to the galactic center, to a galaxy 55 million light-years away — leans on the same cosmic distance ladder covered in our stellar distances post, just stretched to its most extreme rungs.

Worked Examples for Physics Exams