Escape Velocity Calculator

Compute the minimum speed needed to escape any planetary body at any altitude, plus the circular orbital velocity and orbital period.

🚀 Escape Velocity Calculator
Planetary Body
Body Mass (× 1024 kg)
×1024 kg
Body Radius
km
Altitude Above Surface200
km
0 kmmax km
Current Speed (optional)
km/s
Target Orbital Speed
km/s
Escape Velocity
Orbital Velocity
vesc / vorb Ratio
Additional Δv Needed
Orbital Speed at Altitude
Escape Velocity at Altitude
Altitude for Target Speed
Orbital Period

🚀 What is Escape Velocity?

Escape velocity is the minimum speed a free-flying object must reach to permanently escape a gravitational field without further propulsion. Once an object is launched at or above escape velocity in any direction, gravity will continuously decelerate it but will never bring it to a complete stop, allowing it to travel infinitely far from the source body. The concept was formalized by Isaac Newton in his 1687 Principia Mathematica and is now a foundational quantity in orbital mechanics, planetary science, and spacecraft mission design.

The formula is derived by equating kinetic energy with gravitational potential energy: setting (1/2)mv squared equal to G x M x m / r gives v_esc = sqrt(2 x G x M / r) = sqrt(2 x mu / r). Three real-world facts follow immediately: escape velocity depends only on the mass of the body and the distance from its center, not on the mass of the escaping object; escape velocity decreases with altitude (doubling the distance from center reduces v_esc by a factor of sqrt(2)); and escape velocity is always exactly sqrt(2) = 1.4142 times the circular orbital velocity at the same altitude.

This calculator covers both major uses. The Escape Velocity mode computes the speed needed to escape from any altitude above eight solar system bodies (including a custom body option for exoplanets, asteroids, or comets) and shows how much additional delta-V a spacecraft would need if it is already moving at a given speed. The Orbital Velocity mode computes the circular orbital speed, the altitude for a desired orbital speed, and the resulting orbital period, all derived from the same gravitational parameter.

Practical applications range from launch vehicle design (what speed must the final stage reach to leave Earth?) to planetary protection (can an asteroid escaping its parent body reach Earth?) to habitability studies (does a planet retain its atmosphere against solar radiation stripping?). For atmosphere retention, the thermal escape criterion states that a planet can hold onto a gas species if the mean thermal speed of that gas is less than about one sixth of the planet's escape velocity.

📐 Formula

vesc  =  √(2μ / r)
vesc = escape velocity at radius r   (km/s)
μ = gravitational parameter = G × M   (m³/s²)
G = 6.674 × 10−11 m³ kg−1 s−2 (universal gravitational constant)
M = mass of the central body   (kg)
r = body radius + altitude above surface   (m)
vorb = √(μ / r)   (circular orbital velocity at the same r)
vesc / vorb = √2 = 1.4142 always (at any altitude, for any body)
Example: Earth at 200 km altitude: r = 6,578,100 m, μ = 3.986e14 m³/s² → vesc = 11.01 km/s, vorb = 7.78 km/s

📖 How to Use This Calculator

Steps

1
Select the body and altitude - Choose the planetary body from the preset list, or select Custom to enter any mass and radius. Enter the altitude above the surface in km where you want to know the escape velocity.
2
Choose the calculation mode - Use Escape Velocity mode to find the speed needed to escape from a given altitude. Use Orbital Velocity mode to find the circular orbital speed, the altitude for a target orbital speed, and the orbital period.
3
Optionally enter current speed - In Escape Velocity mode, enter your current spacecraft speed to see the additional delta-V needed to reach escape velocity from that altitude.
4
Read the results - Results show escape velocity, orbital velocity, and for orbital mode the orbital period and altitude corresponding to the target speed. Pair with the Tsiolkovsky Rocket Equation Calculator to find the propellant needed for the escape burn.

💡 Example Calculations

Example 1 - Earth Escape Velocity at 200 km LEO

Find escape velocity from a 200 km circular parking orbit around Earth

1
Radius at altitude: r = (6,378.1 + 200) km = 6,578.1 km = 6,578,100 m
2
Escape velocity: v_esc = sqrt(2 × 3.986e14 / 6,578,100) = sqrt(1.212e8) = 11.009 km/s
3
Circular orbital velocity: v_orb = sqrt(3.986e14 / 6,578,100) = sqrt(6.061e7) = 7.784 km/s
4
Additional Δv needed from LEO orbit: 11.009 − 7.784 = 3.225 km/s
Escape velocity = 11.009 km/s, requiring 3.225 km/s delta-V from LEO circular orbit
Try this example →

Example 2 - Moon Surface Escape Velocity

Apollo ascent stage: what speed is needed to escape the Moon from the surface?

1
Radius at surface: r = 1,737,400 m
2
Escape velocity: v_esc = sqrt(2 × 4.905e12 / 1,737,400) = sqrt(5.647e6) = 2.376 km/s
3
Circular orbit speed at surface: v_orb = sqrt(4.905e12 / 1,737,400) = 1.680 km/s
4
Apollo ascent stage reached ~1.68 km/s to reach low lunar orbit (not surface escape)
Moon surface escape velocity = 2.376 km/s, just 21% of Earth's surface escape velocity
Try this example →

Example 3 - Jupiter Escape at 50,000 km Altitude

Escape velocity at 50,000 km above Jupiter cloud tops (Galileo probe entry altitude reference)

1
Radius at altitude: r = (71,492 + 50,000) km = 121,492 km = 1.215e8 m
2
Escape velocity: v_esc = sqrt(2 × 1.267e17 / 1.215e8) = sqrt(2.085e9) = 45.66 km/s
3
Orbital velocity at that altitude: v_orb = sqrt(1.267e17 / 1.215e8) = sqrt(1.043e9) = 32.29 km/s
4
Orbital period at 50,000 km altitude: T = 2π × sqrt((1.215e8)³ / 1.267e17) = 23,639 s = 6.57 hours
Jupiter escape velocity at 50,000 km altitude = 45.67 km/s, more than 4 times Earth's LEO escape speed
Try this example →

❓ Frequently Asked Questions

What is escape velocity and how is it calculated?+
Escape velocity is the minimum speed needed to escape a gravitational field without further propulsion. It is derived by equating kinetic energy with gravitational potential energy: (1/2) x m x v squared = G x M x m / r, giving v_esc = sqrt(2 x G x M / r) = sqrt(2 x mu / r), where mu is the gravitational parameter and r is the distance from the body center. For Earth at surface: v_esc = sqrt(2 x 3.986e14 / 6,378,100) = 11.186 km/s. An object launched at exactly this speed will asymptotically approach zero velocity at infinite distance but will never actually stop.
What is the escape velocity of Earth from the surface?+
Earth's escape velocity at sea level is 11.186 km/s (40,270 km/h). This assumes no atmosphere. In practice, rockets do not fly at escape velocity from the surface: they follow a gravity turn to minimize aerodynamic losses, reach low Earth orbit (about 7.78 km/s), then perform a trans-escape burn of about 3.2 km/s. Including gravity and drag losses, the total delta-V from the surface to escape Earth is approximately 11.5 to 14 km/s depending on trajectory.
What is the difference between escape velocity and orbital velocity?+
Orbital velocity (first cosmic velocity) is the speed for a circular orbit just above the surface. Escape velocity (second cosmic velocity) is the speed to leave the gravitational field entirely. Their ratio is always sqrt(2) = 1.4142, regardless of the body or altitude: v_esc = sqrt(2) x v_orb. For Earth at the surface: v_orb = 7.91 km/s, v_esc = 11.19 km/s. A spacecraft already in circular orbit needs to increase its speed by only 41.4% to escape - this is the basis for the departure burn from LEO in most interplanetary missions.
Does escape velocity depend on the launch direction?+
No. Escape velocity is a scalar magnitude: the same speed is needed regardless of launch direction, whether radially outward, tangential, or at any angle. Direction affects the trajectory shape but not the energy requirement. Launching eastward near the equator provides a free boost from Earth's rotation (0.465 km/s at the equator), which is why Kourou and Cape Canaveral are preferred over polar launch sites. For launches into equatorial orbits, this equatorial bonus reduces the propellant needed for the escape burn.
How does altitude affect escape velocity?+
Escape velocity decreases with altitude as v_esc = sqrt(2 x mu / r). Since r appears in the denominator under a square root, doubling r (roughly the distance from the center) reduces v_esc by 1/sqrt(2) = 29.3%. For Earth: surface 11.19 km/s, at 200 km altitude 11.01 km/s, at 400 km 10.84 km/s, at GEO (35,786 km) 4.35 km/s, at lunar distance (384,400 km) 1.44 km/s. A spacecraft at GEO needs only about 4.35 km/s to escape Earth, making GEO an attractive staging point for deep space missions.
What is the escape velocity of Mars compared to Earth?+
Mars surface escape velocity is 5.027 km/s, about 45% of Earth's 11.186 km/s. This lower escape velocity is a key advantage for Mars sample return missions and eventual crewed Mars ascent vehicles. The Mars Ascent Vehicle proposed for Mars Sample Return must achieve approximately 4.1 km/s to reach low Mars orbit at 400 km altitude. In contrast, the Saturn V first stage alone needed to accelerate the entire stack to 2 km/s before staging, illustrating the much lower difficulty of the Mars ascent problem.
What is the third cosmic velocity?+
The three cosmic velocities are defined for Earth. First (7.91 km/s): circular orbital speed at Earth's surface. Second (11.19 km/s): escape velocity from Earth's surface. Third (about 16.6 km/s from Earth's surface, or 12.3 km/s beyond Earth's orbital speed starting from LEO): the speed needed to escape the entire solar system. Voyager 1 achieved the third cosmic velocity through Jupiter and Saturn gravity assists, without the propellant cost of a direct burn. It is now the farthest human-made object from Earth at over 160 AU.
How does escape velocity relate to atmospheric retention?+
A planet can retain a gas in its atmosphere over geological timescales if the mean thermal escape speed of the gas molecules is less than about one sixth of the planet's escape velocity. Mean thermal speed = sqrt(8 x k x T / (pi x m)), where k is Boltzmann's constant, T is temperature, and m is the molecular mass. For Earth (v_esc = 11.2 km/s), the threshold is about 1.9 km/s. Light hydrogen molecules move faster than this at Earth's exosphere temperature, explaining why Earth continuously loses hydrogen to space. The Moon's escape velocity of 2.38 km/s is too low to retain any significant atmosphere.
What is the escape velocity from the surface of Jupiter?+
Jupiter's surface escape velocity (at the cloud tops, taken as the radius of 71,492 km) is 59.54 km/s. This is the highest surface escape velocity of any planet in the solar system and explains why Jupiter retains all gas species including hydrogen and helium. In comparison, the Sun's escape velocity at its surface is 617.7 km/s. A spacecraft arriving at Jupiter from Earth has a hyperbolic excess velocity of about 5 to 9 km/s, meaning it arrives at Jupiter periapsis moving at about 45 to 60 km/s, below Jupiter's escape velocity and therefore captured into a bound orbit (which then requires propulsive capture).
Can objects escape gravity below escape velocity?+
Yes, with continuous thrust. The escape velocity formula assumes a ballistic trajectory with a single instantaneous burn and no further propulsion. Ion drives and solar sails can escape a gravitational field by continuously pushing even at speeds well below the ballistic escape velocity, spiraling slowly outward. The total delta-V for a spiral escape from a circular orbit is slightly greater than the single-burn value (about 3 to 5% more), but the much higher Isp of electric propulsion makes this trade-off worthwhile for missions like Dawn (asteroid belt) and BepiColombo (Mercury), both of which used ion drives for orbital insertions.
What is the escape velocity of the Sun at Earth's orbital distance?+
The Sun's escape velocity at 1 AU (149.6 million km) is 42.1 km/s. Earth's orbital speed is 29.78 km/s. To escape the solar system from Earth's orbit requires reaching sqrt(42.1 squared) in a direction tangential to Earth's orbit, meaning a delta-V of 42.1 minus 29.78 = 12.3 km/s beyond Earth's current orbital speed. This is enormously expensive with chemical rockets, which is why all successful solar system escape missions (Pioneer 10 and 11, Voyager 1 and 2, New Horizons) used planetary gravity assists to achieve the required heliocentric speed.
How do I use the escape velocity for rocket propellant calculations?+
Find the escape velocity at your departure altitude using this calculator, then subtract your current orbital speed to get the required delta-V for the escape burn. Feed that delta-V into the Tsiolkovsky Rocket Equation Calculator with your engine's specific impulse (Isp) to find the required propellant mass fraction: m_prop / m_initial = 1 - exp(-DV / (Isp x g0)). For example, escaping Earth from 200 km LEO requires DV = 3.23 km/s. With LOX/RP-1 (Isp = 311 s), the propellant fraction is 1 - exp(-3230 / 3050) = 65.3% of the wet mass.

What is escape velocity and how is it calculated?

Escape velocity is the minimum speed needed to escape a gravitational field without further propulsion. It is derived by setting kinetic energy equal to gravitational potential energy: (1/2)mv^2 = G*M*m/r, giving v_esc = sqrt(2*G*M/r) = sqrt(2*mu/r), where mu = G*M is the gravitational parameter and r is the distance from the body's center. For Earth at the surface: v_esc = sqrt(2 x 3.986e14 / 6378100) = 11.19 km/s. No thrust is needed after the initial burn since the object continues to slow down but never quite stops.

What is the escape velocity of Earth from the surface?

Earth's escape velocity at the surface (sea level) is 11.186 km/s (40,270 km/h or 25,020 mph). This assumes no atmospheric drag. In practice, rockets do not fly straight up at escape velocity; instead, they follow a gravity turn trajectory to reach orbit, then burn to escape. The actual delta-V needed from Earth's surface to escape the solar system is about 8.8 km/s when launched from LEO, plus 3.2 km/s to reach LEO, totaling around 12 km/s from the ground including gravity and drag losses.

What is the difference between escape velocity and orbital velocity?

Orbital velocity (first cosmic velocity) is the speed needed to maintain a circular orbit just above a planet's surface. Escape velocity (second cosmic velocity) is the speed needed to escape the gravitational field entirely. The ratio is always sqrt(2): v_esc = sqrt(2) x v_orb = 1.4142 x v_orb. For Earth at the surface: v_orb = 7.91 km/s and v_esc = 11.19 km/s. A spacecraft already in low circular orbit needs to increase its speed by only 41.4% to reach escape velocity.

Does escape velocity depend on the direction of launch?

No. Escape velocity is a scalar magnitude, not a vector. A spacecraft launched radially outward, tangentially, or at any angle needs the same total speed to escape. The direction only affects the trajectory shape. However, launching tangentially in the direction of the planet's rotation gives a free boost from the planet's rotational speed, which is why equatorial launch sites like Kourou and Cape Canaveral are preferred. Earth's equatorial rotation adds about 0.465 km/s to an eastward launch.

What is the escape velocity of the Moon?

The Moon's escape velocity at the surface is 2.376 km/s (8,553 km/h). This is about one fifth of Earth's escape velocity, which is why returning from the Moon requires far less propellant than launching from Earth. The Apollo lunar module ascent stage used a single Ascent Propulsion System engine to reach lunar orbit, achieving a velocity of about 1.68 km/s, well below lunar escape velocity, to rendezvous with the command module in low lunar orbit.

What is the escape velocity of Mars?

Mars escape velocity at the surface is 5.027 km/s, about 45% of Earth's. Orbital velocity at Mars surface is 3.555 km/s, giving an orbital period of about 1.76 hours for a surface-skimming orbit. The escape velocity at 300 km altitude (a practical low Mars orbit) is 4.81 km/s. Mars Return Vehicle designs for human exploration must achieve this to leave Mars, which is a significant propellant mass constraint and a key design driver for in-situ propellant production on the Martian surface.

How does altitude affect escape velocity?

Escape velocity decreases with altitude as v_esc = sqrt(2*mu/r), where r = body radius + altitude. Since r appears in the denominator under a square root, v_esc is proportional to 1/sqrt(r). Doubling the distance from the center (roughly doubling altitude for low altitudes) reduces escape velocity by 1/sqrt(2) = 29.3%. For Earth: surface v_esc = 11.19 km/s, at 200 km = 11.01 km/s, at 400 km = 10.84 km/s, at GEO (35,786 km) = 4.35 km/s, at lunar distance (384,400 km) = 1.44 km/s.

What is the escape velocity of Jupiter?

Jupiter's surface escape velocity is 59.54 km/s, the highest of any planet in the solar system. At 50,000 km altitude above the cloud tops: v_esc = 45.65 km/s. This enormous escape velocity is what makes the Oberth effect so powerful for Jupiter flybys: firing even a small burn at Jupiter periapsis is amplified enormously by the high local speed. The Galileo probe that descended into Jupiter's atmosphere was moving at about 48 km/s when it entered the atmosphere.

Can escape velocity be exceeded gradually?

Yes. The escape velocity formula assumes a single instantaneous burn with no further thrust (ballistic trajectory). Ion engines and other low-thrust propulsion systems can escape a gravitational field by thrusting continuously at speeds below escape velocity, gradually spiraling outward. The total delta-V needed for a spiral escape from a circular orbit is slightly more than the single-burn value (about 3-5% more for typical cases), but the advantage is that ion engines have far higher Isp than chemical rockets, making the overall propellant efficiency much better.

What is third cosmic velocity?

The three cosmic velocities are defined for Earth. First cosmic velocity (7.91 km/s) is the circular orbital speed just above Earth's surface. Second cosmic velocity (11.19 km/s) is Earth's surface escape velocity. Third cosmic velocity (16.6 km/s from Earth's surface, or about 12.3 km/s added to Earth's orbital speed from a starting point at Earth's distance from the Sun) is the speed needed to escape the entire solar system. This was achieved by Voyager 1 and 2 using gravity assists from Jupiter and Saturn.

What is the escape velocity of a neutron star or black hole?

For a neutron star of mass 1.4 solar masses and radius 10 km: v_esc = sqrt(2 x 1.327e20 x 1.4 / 10000) = sqrt(3.72e16) = 193,000 km/s, which is 64% of the speed of light. At this level, relativistic effects are significant and the Newtonian formula underestimates the true escape speed. For a black hole, the escape velocity at the event horizon (Schwarzschild radius) equals c (the speed of light), which is why nothing, not even light, can escape. The Schwarzschild radius for an Earth-mass black hole would be about 8.9 mm.

How do I use this calculator for a rocket mission?

Select the destination body, enter the planned periapsis or departure altitude in km, and read the escape velocity. For the Escape Velocity mode, optionally enter your current speed to see the additional delta-V needed. For the Orbital Velocity mode, enter your planned orbital speed to find what altitude that corresponds to and the orbital period. These values are the starting points for Tsiolkovsky rocket equation calculations to determine the required propellant mass fraction for the escape maneuver.

What is the escape velocity of the Sun at Earth's distance?

The Sun's escape velocity at Earth's mean orbital distance of 1 AU (149.6 million km) is 42.1 km/s. Earth's orbital speed is 29.78 km/s, so a spacecraft at Earth's orbit moving purely radially would need 42.1 km/s to escape the Sun. However, since Earth is already orbiting at 29.78 km/s, a tangential burn needs to add only sqrt(42.1 squared minus 29.78 squared) minus 29.78 = 12.3 km/s beyond Earth's orbital speed to escape the solar system. This is why New Horizons needed a large Jupiter gravity assist to reach Pluto within a decade.