Escape Velocity Calculator: Compare Planets, Moons, Custom Bodies

Newtonian gravity holds up beautifully for the escape-velocity question across the whole solar system, with one important caveat. The formula v_esc = √(2GM/r) drops out of basic energy conservation, and it gives 11.186 km/s for Earth's surface, 2.376 km/s for the Moon, 5.027 km/s for Mars, 59.5 km/s for Jupiter, 617.5 km/s for the Sun's photosphere. JPL's ephemerides agree with these numbers to better than 1 part in 10⁶, which is plenty for any mission you'd realistically plan with this calculator. Newton works.

The breakdown happens at the extremes. Push the radius down toward r_s = 2GM/c² (the Schwarzschild radius), and v_esc reaches the speed of light. At that point Newtonian energy conservation has predicted something the framework can't describe, because the geometry of spacetime is no longer flat. For Earth, r_s ≈ 8.87 mm, which is about 7×10⁸ times smaller than Earth's actual radius, so we're fine. For the Sun, r_s ≈ 2.95 km vs. its actual 696,000 km radius, also fine. For the supermassive black hole Sagittarius A* at the galactic center (4.3×10⁶ solar masses), r_s ≈ 12.7×10⁶ km. Inside that you can't escape no matter how fast you go, and the formula v_esc = c is the boundary, not a velocity you can actually achieve.

The replacement at the strong-field end is general relativity, where escape becomes a question about geodesics and event horizons rather than algebraic energy balance. For everything you'd realistically launch a rocket from, GR corrections are buried below 10⁻⁸ of the Newtonian answer, and we ignore them. For mission planning around a stellar-mass or supermassive black hole, you'd need the Schwarzschild metric. The other place classical physics needs help is in the rocket equation, where the propellant exhaust velocity is comparable to the spacecraft's target velocity. That's where you bring in the Tsiolkovsky equation and the staging argument, both of which we'll work through below.

The escape-velocity formula falls out of one bookkeeping choice: take the gravitational potential energy at infinity to be zero. Then for a mass m at distance r from a body of mass M, U(r) = −GMm/r. Negative. The minus sign is what makes the math work. Total energy E = ½mv² + U. Bound orbits have E < 0, parabolic escape has E = 0, hyperbolic escape has E > 0. Setting E = 0 at the surface and solving for v gives the escape velocity.

The mass m of the escaping object cancels. A 1 kg probe and a 1,000 kg satellite need the same v_esc. What matters is the central body's gravitational parameter μ = GM and the starting distance r from its center. Two technical choices to flag.

Why infinity, not the surface, for the zero of U. You could pick any reference, and the physics would still work as long as you're consistent. But choosing U(∞) = 0 makes "is the orbit bound?" answer with the sign of E. Other choices (U at the surface, U at the center) would put the threshold for escape at some non-zero energy you'd have to remember separately. Pick the convention that makes the inequality clean.

μ = GM is known better than G or M separately. The gravitational constant G is one of the worst-measured fundamental constants in physics: CODATA 2018 gives 6.674 30(15)×10⁻¹¹ m³ kg⁻¹ s⁻², a relative uncertainty of ~2×10⁻⁵. Earth's mass M and the gravitational constant G are individually known only to that level. But the product μ = GM is measured directly from spacecraft tracking and orbital periods, and we know μ_Earth = 398,600.4418 ± 0.0008 km³/s² (about 2×10⁻⁹ relative uncertainty, ten thousand times better than G alone). This is why astrodynamics codes always use μ rather than computing G × M.

Units note. μ in km³/s² is standard in astrodynamics. To go to SI base units (m³/s²), multiply by 10⁹.

Escape velocity is one specific case of a family of orbits, and Kepler's laws set the geometry. First law: orbits are conic sections with the central body at one focus. Bound orbits are ellipses (circle is the special case e = 0). The marginal case is a parabola (e = 1, exactly v_esc). Beyond that, hyperbolas (e > 1, faster than escape, with leftover speed v_∞ at infinity).

Second law: a line from the central body to the orbiting object sweeps equal areas in equal times. That's angular momentum conservation in disguise. For escape problems it implies that radial speed is fastest at periapsis, which is why launches that benefit from extra altitude (going through a parking orbit before the trans-Mars injection burn, for example) make engineering sense.

Third law: T² = (4π²/μ) a³, where a is the semi-major axis. This relates orbital period directly to size. For Earth-orbit examples, a circular orbit at 6,771 km (ISS altitude) has period T = 92.7 min. A circular orbit at the Moon's distance (a = 384,400 km) has T = 27.3 days, which is the sidereal lunar month.

The energy connection is what ties Kepler back to escape. For any conic-section orbit around a body with parameter μ, the specific orbital energy is ε = −μ/(2a) for an ellipse. Period and energy are two views of the same number: a is the semi-major axis, T is the period, ε is the specific energy, and they're bound together by μ. Setting ε = 0 (a → ∞, T → ∞) gives the parabolic escape case, where the orbit is "an ellipse with infinite semi-major axis" and v at the starting radius equals v_esc. Hyperbolic escape has ε > 0 and a < 0 by convention, with v_∞ = √(2ε) the speed at infinity.

The relation between escape and circular speed at the same altitude is exactly √2. Kinetic energy at circular speed equals half the magnitude of the potential energy (the virial theorem for bound orbits). Kinetic energy at escape speed equals the full magnitude. Twice the energy means √2 times the velocity. So if you're already in a 7.67 km/s circular orbit at 400 km altitude, you only need an extra 0.414 × 7.67 ≈ 3.18 km/s of delta-v to escape, not the full 10.85 km/s. That's why mission designs almost always go to a parking orbit first.

The differences between escape velocities for solar-system bodies are dramatic, and they show up squared in energy budgets. Run the numbers before talking strategy.

The KE/kg column is what really matters for fuel budgets. Jupiter's escape velocity is only 5.3× Earth's, but the energy per kilogram is 5.3² ≈ 28× higher. That's why Galileo, Cassini, and Juno never planned on returning a sample from Europa or Io: any ascent stage from those moons has to fight Jupiter's gravity well too. Sample-return missions from inner-solar-system targets (asteroids, Mars) are barely tractable. A return from Titan (Saturn's moon) or Europa is an order-of-magnitude harder problem because of the planet's well, even though local v_esc from those moons is small.

v_esc falls as 1/√r, not linearly with r. By the time you're at GEO altitude, you've cut escape velocity by 61%. By the Moon's distance, by 87%. But you have to expend the energy to get to that altitude in the first place, and the total delta-v from Earth's surface is still about 11.2 km/s no matter what altitude you're aiming for as the "launch point". Going via orbit lets you use staging, gravity assists, and aerobraking, all of which reduce propellant mass.

Light-speed limit. Set v_esc = c in the Newtonian formula and solve for the radius: r_s = 2GM/c². That's the Schwarzschild radius. For Earth, r_s ≈ 8.87 mm. For the Sun, r_s ≈ 2.95 km. For our galaxy's central black hole, ~12.7×10⁶ km. At r < r_s, escape is impossible no matter what speed you achieve, but the Newtonian formula has stopped being trustworthy long before that. Inside the strong-field regime you need general relativity: the "escape" question becomes a question about light cones and timelike geodesics, not algebraic energy balance. LIGO's GW150914 detection in 2015 saw two black holes of about 36 and 29 solar masses spiraling and merging, and the event-horizon radii of those (around 100 km) are where the Newtonian picture is unusable.

Bound vs. unbound trajectories. Total specific energy ε = v²/2 − μ/r is the discriminator. ε < 0: bound, the object will return. ε = 0: marginally unbound, parabolic, v at the starting point equals v_esc exactly. ε > 0: hyperbolic escape with leftover speed v_∞ = √(2ε) at infinity. For interplanetary missions, "C₃" is the term-of-art for 2ε (units km²/s²). Mars Curiosity's C₃ at trans-Mars injection was about 8.4 km²/s², meaning it left Earth's sphere of influence with v_∞ ≈ 2.9 km/s relative to Earth. Voyager 1, on a Solar System escape trajectory, left with C₃ around 102 km²/s², meaning v_∞ ≈ 10.1 km/s relative to Earth. New Horizons launched with C₃ ≈ 159, the highest of any spacecraft ever, to reach Pluto in under a decade.

Multi-body subtleties. The single-body formula assumes only one gravitating mass matters. Reality is messier. Inside the Hill sphere (the region where a body's gravity dominates over its parent's), the local v_esc applies. Outside the Hill sphere, the parent body takes over. Earth's Hill sphere extends to about 1.5×10⁶ km (about 4 lunar distances). A rocket that escapes Earth at 11.2 km/s is still bound to the Sun unless it then changes its heliocentric velocity. Voyager 1's post-Earth-escape velocity in heliocentric coordinates is what determined whether it was ultimately Solar System-bound, and Jupiter's 1979 gravity assist was what put it onto an unbound trajectory.

Earth's escape velocity from the surface is 11.186 km/s, computed from μ_Earth = 398,600 km³/s² and R_Earth = 6,371 km: v_esc = √(2 × 398,600 / 6,371) = √125.10 ≈ 11.19 km/s. The arithmetic isn't the hard part. The hard part is whether you can build a rocket that delivers that delta-v, and if so, how big it has to be. The Tsiolkovsky rocket equation gives the answer.

Apollo's lunar return is the easy direction. From the lunar surface, v_esc is 2.38 km/s. The Apollo Lunar Module's ascent stage carried about 2.4 tonnes of propellant for a 4.7-tonne ascent stage, giving a mass ratio of around 2.5 and easily reaching the 1.85 km/s needed to rendezvous in lunar orbit (not full escape, just orbit). With a single hypergolic engine and no atmosphere to fight, the lunar departure was tractable from the start. Coming home from Earth is the punishment direction.