Escape Velocity Calculator

Gravitational Physics Guide

What's the key thing to understand about Escape Velocity?

Escape velocity is the minimum speed needed to escape a gravitational field without further propulsion. v_escape = √(2GM/R). From Earth's surface: v = √(2 × 6.67×10⁻¹¹ × 5.97×10²⁴ / 6.37×10⁶) = √(1.25×10⁸) = 11.19 km/s ≈ 40,300 km/h. Note: this assumes no atmosphere. Real spacecraft need more speed to overcome air drag. The escape velocity applies to ballistic trajectories — a rocket with constant thrust does not need to reach 11 km/s instantaneously.

What do I need to know about Gravitational Field Strength?

g = GM/r². At Earth's surface: g = 9.81 m/s². At altitude h: g_h = GM/(R+h)². Gravitational field strength decreases with the square of distance from the centre. At ISS altitude (400km): g = 6.67×10⁻¹¹ × 5.97×10²⁴ / (6.77×10⁶)² = 8.7 m/s². Astronauts experience 'weightlessness' not because there is no gravity (still ~89% of surface gravity) but because they are in free fall — in circular orbit the spacecraft and astronaut fall together, so no normal force acts on them.

What should I know about Orbital Mechanics?

For circular orbit at altitude h: orbital speed v = √(GM/(R+h)). Period T = 2π(R+h)/v = 2π√((R+h)³/(GM)). ISS (400km altitude): v = √(6.67×10⁻¹¹ × 5.97×10²⁴ / 6.77×10⁶) = 7.67 km/s. T = 2π × 6.77×10⁶ / 7670 = 5,541 s = 92.4 minutes. Geostationary orbit: T = 24 hours. r = ∛(GMT²/(4π²)) = 42,164 km from centre = 35,786 km above surface. Kepler's third law: T² ∝ r³ follows directly from the orbital mechanics equations.

What's the key thing to understand about Black Holes and Schwarzschild Radius?

If a mass is compressed within its Schwarzschild radius, it becomes a black hole. r_s = 2GM/c². Earth: r_s = 2 × 6.67×10⁻¹¹ × 5.97×10²⁴ / (3×10⁸)² = 8.87mm. The Earth would need to be compressed to less than 9mm radius to become a black hole. Sun: r_s ≈ 3km. Stellar black holes: typically 10-30 km Schwarzschild radius, 3-20 solar masses. Supermassive black holes: millions to billions of solar masses. M87* (first black hole photographed): 6.5 billion solar masses, r_s ≈ 19 billion km. The escape velocity at the Schwarzschild radius equals the speed of light itself, which is precisely why nothing — not even light — can escape from within this boundary.

What's the difference between Escape Velocity and Orbital Velocity?

Escape velocity is the minimum speed needed to break completely free of a body's gravity and never return (assuming no further propulsion and no other forces), calculated as v_escape = √(2GM/r). Orbital velocity — the speed needed to maintain a stable circular orbit at a given altitude rather than escaping entirely — is a related but distinctly smaller quantity, calculated as v_orbit = √(GM/r). Comparing the two formulas directly shows the relationship: v_escape = v_orbit × √2, meaning escape velocity is always exactly √2 (about 1.414) times larger than the orbital velocity at the same distance from the central body. This is why reaching a stable low-Earth orbit (around 7.8 km/s) requires substantially less speed than fully escaping Earth's gravity (about 11.2 km/s) — a spacecraft only needs enough speed to keep 'falling around' the Earth rather than falling into it, which takes meaningfully less energy than escaping the gravitational field entirely, which is exactly why most space missions (satellites, the International Space Station) aim for orbit rather than escape.

What should I know about Escape Velocities of Different Bodies?

Escape velocity depends only on a body's mass and radius, which is why it varies dramatically across the solar system. Earth's escape velocity is approximately 11.2 km/s (about 40,320 km/h). The Moon, with much less mass and a smaller radius, has a far lower escape velocity of about 2.4 km/s — one of the key reasons it was far more achievable for the Apollo missions to launch back off the lunar surface than it would be to launch from Earth. Mars, smaller and less massive than Earth, has an escape velocity of about 5.0 km/s. At the other extreme, Jupiter's enormous mass gives it an escape velocity of about 59.5 km/s, more than five times Earth's, which is why no spacecraft has ever been designed to launch from Jupiter's surface (and indeed, Jupiter has no solid surface to launch from). These differences in escape velocity are a major factor in mission planning: a return mission from the Moon requires dramatically less fuel than an equivalent return mission would from Mars or a hypothetical landing on a gas giant.

What do I need to know about Deriving Escape Velocity from Energy Conservation?

Escape velocity is found by setting an object's kinetic energy exactly equal to the gravitational potential energy needed to reach infinite distance: ½mv² = GMm/r, which rearranges (after the mass m cancels out) to v_escape = √(2GM/r). That the object's own mass drops out of the equation is why a feather and a cannonball launched from the same point need identical escape velocity, ignoring air resistance — escape velocity depends only on the mass and radius of the body being escaped, never on the mass of what's escaping it.

Worked Example: Escape Velocity from the Surface of Mars

Escape velocity is v_esc = √(2GM/r). For Mars, M = 6.417×10²³ kg and r = 3.3895×10⁶ m (mean radius), with G = 6.674×10⁻¹¹ N·m²/kg². First find 2GM: 2 × 6.674×10⁻¹¹ × 6.417×10²³ = 8.566×10¹³. Dividing by r: 8.566×10¹³ / 3.3895×10⁶ = 2.527×10⁷. Taking the square root: v_esc = √(2.527×10⁷) ≈ 5,027 m/s, or about 5.03 km/s — which matches Mars's well-documented escape velocity, confirming the calculation.

Reference Table: Escape Velocity Decreases with Distance from Earth

Escape velocity isn't fixed for a planet — it depends on how far from the center the escaping object already is. Using Earth's standard gravitational parameter GM = 3.986×10¹⁴ m³/s², escape velocity at several distances from Earth's center is:

This is part of why spacecraft heading beyond Earth orbit benefit from first reaching a high orbit before firing an escape-trajectory burn: escape velocity from geostationary altitude (4.35 km/s) is less than half what's needed from the surface (11.19 km/s), since v_esc falls off as the inverse square root of distance.

Worked Example: Orbital Velocity vs. Escape Velocity at the Same Altitude

A satellite in low Earth orbit at about 400 km altitude (r = 6,771 km from Earth's center) moves at the circular orbital velocity v_orb = √(GM/r), using GM_earth = 3.986×10¹⁴ m³/s²: v_orb = √(3.986×10¹⁴ / 6.771×10⁶) = √(5.885×10⁷) ≈ 7,671 m/s, or 7.67 km/s. The escape velocity at that same altitude was found earlier to be 10.85 km/s. Dividing the two: 10.85 / 7.67 ≈ 1.414 ≈ √2, confirming the general relationship v_esc = √2 × v_orb at any given altitude. Since kinetic energy scales with velocity squared, this means escaping a planet from a given orbit requires exactly twice the kinetic energy of maintaining a circular orbit at that same altitude (because (√2)² = 2) — which is exactly why satellites that only need to stay in orbit carry far less fuel than missions, like Apollo or New Horizons, that need to break free of Earth's gravity entirely.

Worked Example: Escape Velocity from the Moon

The Moon has mass M = 7.342×10²² kg and mean radius r = 1.7374×10⁶ m. Using v_esc = √(2GM/r) with G = 6.674×10⁻¹¹ N·m²/kg²: 2GM = 2 × 6.674×10⁻¹¹ × 7.342×10²² ≈ 9.803×10¹². Dividing by r: 9.803×10¹² / 1.7374×10⁶ ≈ 5.640×10⁶. Taking the square root: v_esc ≈ 2,375 m/s, or about 2.38 km/s — matching the Moon's well-known escape velocity and showing why it's so much easier for a rocket to leave the Moon than to leave Earth (11.19 km/s): the Moon's smaller mass and the absence of any atmosphere both work in a departing spacecraft's favor, which is part of why the Apollo lunar ascent stage could be so much smaller and lighter than the vehicle that launched from Earth.

Worked Example: Escape Velocity from the Sun's Surface

For the Sun, M = 1.989×10³⁰ kg and radius r = 6.957×10⁸ m. Using v_esc = √(2GM/r) with G = 6.674×10⁻¹¹ N·m²/kg²: 2GM = 2 × 6.674×10⁻¹¹ × 1.989×10³⁰ ≈ 2.655×10²⁰. Dividing by r: 2.655×10²⁰ / 6.957×10⁸ ≈ 3.816×10¹¹. Taking the square root: v_esc ≈ 617,700 m/s, or about 618 km/s — consistent with the commonly quoted solar escape velocity of roughly 617.5 km/s. This is more than 50 times Earth's escape velocity, driven almost entirely by the Sun's enormous mass despite its radius being only about 109 times Earth's.