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Gravitation

Escape Velocity Calculator

Calculate escape velocity for Earth, Moon, Mars, Jupiter, the Sun and any custom body. Also shows orbital velocity and surface gravity.

v_esc = √(2GM/r)
Select a body to calculate escape velocity.

Escape velocity

Escape velocity is the minimum speed needed to escape a gravitational field without further propulsion. The formula v_esc = √(2GM/R) shows it depends only on the body's mass and radius — not on the escaping object's mass. At Earth's surface, this is about 11.2 km/s (≈ 40,000 km/h).

Escape velocity is always exactly √2 times the circular orbital velocity at the same radius. This comes from the ratio of total energy: an orbit has E = −GMm/(2r), escape requires E ≥ 0.

Escape Velocity — The Complete Physics Guide

Escape velocity is the minimum speed an object must reach to escape a planet's (or other body's) gravitational field without further propulsion. At escape velocity, the object's kinetic energy exactly equals the magnitude of its gravitational potential energy — giving it enough energy to reach infinity with zero remaining kinetic energy. Understanding escape velocity is fundamental to rocket science, space mission planning, and understanding why some planets retain atmospheres while others do not.

Deriving Escape Velocity

The derivation uses energy conservation. At the surface of a planet of mass M and radius R, an object of mass m has: total energy = KE + GPE = ½mv² − GMm/R. To just escape (reaching r = ∞ with v = 0), total energy must equal zero (since both KE and GPE → 0 at infinity):

½mv_esc² − GMm/R = 0
v_esc = √(2GM/R)

The mass m of the escaping object cancels entirely — escape velocity is independent of the mass of what is escaping. A cannonball and a spacecraft require the same launch speed to escape from the same point on Earth's surface. This is the same mass-independence as free fall — a consequence of the equivalence of gravitational and inertial mass.

Escape Velocities in the Solar System

Earth's escape velocity: v_esc = √(2 × 6.674×10⁻¹¹ × 5.972×10²⁴ / 6.371×10⁶) = √(125,270,000) = 11.19 km/s (about 40,270 km/h or Mach 33).

BodyMass (kg)Radius (km)Escape Velocity
Moon7.35 × 10²²1,7372.38 km/s
Mars6.39 × 10²³3,3905.03 km/s
Earth5.97 × 10²⁴6,37111.19 km/s
Jupiter1.90 × 10²⁷69,91159.5 km/s
Sun1.99 × 10³⁰695,700617.5 km/s

The Moon's low escape velocity (2.38 km/s) explains why it has no atmosphere — gas molecules at room temperature have average speeds comparable to this, and atmospheric gases gradually escape over geological time. Earth retains its atmosphere because 11.19 km/s is much higher than molecular speeds at our temperatures. Jupiter, with v_esc = 59.5 km/s, easily retains even hydrogen and helium.

Real Rocket Launches — Why Not Just Hit Escape Velocity?

In practice, rockets never launch straight up at 11.2 km/s and turn off their engines. Instead they use the Hohmann transfer approach: first reaching Low Earth Orbit (LEO, ~7.9 km/s at 400 km altitude), then burning again at the orbit's highest point to add the remaining ΔV needed for escape. This is far more fuel-efficient due to the Oberth effect — rockets gain more energy per kilogram of fuel burned at high speed.

The Tsiolkovsky rocket equation determines the fuel mass needed for any velocity change: Δv = v_exhaust × ln(m_initial/m_final). To achieve Earth escape from LEO (Δv ≈ 3.2 km/s additional), a rocket with exhaust velocity 4.4 km/s needs a mass ratio of e^(3.2/4.4) ≈ 2.1 — meaning 52% of the LEO mass must be fuel. This is why escaping the solar system requires extremely capable rockets like Saturn V and Falcon Heavy.

The four spacecraft that have reached escape velocity from the solar system — Pioneer 10, Pioneer 11, Voyager 1, and Voyager 2 — achieved this using gravitational slingshot manoeuvres around Jupiter, which added enormous velocity without any fuel expenditure, a brilliant application of orbital mechanics.

Black Holes — When Escape Velocity Exceeds c

Setting v_esc = c (speed of light) in the classical formula gives the Schwarzschild radius: r_s = 2GM/c². For an object compressed below this radius, even light cannot escape — it becomes a black hole. For the Sun: r_s = 2 × 6.674×10⁻¹¹ × 1.99×10³⁰ / (3×10⁸)² = 2,954 m ≈ 3 km. If the Sun were somehow compressed to 3 km radius, it would become a black hole.

This classical derivation is historically interesting — John Michell proposed "dark stars" (black holes) using exactly this argument in 1783, over 130 years before Einstein's general relativity provided the rigorous theoretical framework. The Schwarzschild radius formula from general relativity is identical to the classical result, despite the very different physics underlying it.

Frequently Asked Questions

What is escape velocity?
Escape velocity is the minimum speed needed for an object to escape a gravitational field without further propulsion: v_esc = √(2GM/R). For Earth, this is 11.19 km/s (≈40,270 km/h). It is independent of the escaping object's mass — a spacecraft and a cannonball require the same launch speed.
Does escape velocity depend on launch direction?
No — escape velocity is the same in any direction. An object launched horizontally, vertically, or at any angle with escape velocity will have enough energy to escape, provided no atmosphere removes energy through drag. In practice, upward launches minimise atmospheric losses.
Why does the Moon have no atmosphere?
The Moon's escape velocity (2.38 km/s) is comparable to the thermal speeds of gas molecules at typical temperatures. Gas molecules near the lunar surface frequently exceed escape velocity and are lost to space over geological time. Earth retains its atmosphere because 11.2 km/s greatly exceeds typical molecular thermal speeds, so atmospheric loss is negligible.
Is escape velocity the same as orbital velocity?
No. Orbital velocity for a circular orbit at radius r is v_orbit = √(GM/r). Escape velocity from the same point is v_esc = √(2GM/r) = v_orbit × √2 ≈ 1.414 × v_orbit. To escape from LEO, a spacecraft needs to add about 41% more speed beyond its orbital velocity.
What is the event horizon of a black hole?
The event horizon is the boundary at the Schwarzschild radius r_s = 2GM/c² within which escape velocity exceeds the speed of light. Nothing — including light — can escape from within the event horizon. An outside observer never sees an object cross the horizon; time dilation makes the crossing take infinite coordinate time as seen from outside.

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