# Escape velocity

Last updated on 19 May 2017

In physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body.

The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph) at the surface. More generally, escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero; an object which has achieved escape velocity is neither on the surface, nor in a closed orbit (of any radius). With escape velocity in a direction pointing away from the ground of a massive body, the object will move away from the body, slowing forever and approaching, but never reaching, zero speed. Once escape velocity is achieved, no further impulse need be applied for it to continue in its escape. In other words, if given escape velocity, the object will move away from the other body, continually slowing, and will asymptotically approach zero speed as the object's distance approaches infinity, never to return. Note that the minimum escape velocity assumes that there is no friction (e.g., atmospheric drag), which would increase the required instantaneous velocity to escape the gravititational influence, and that there will be no future sources of additional velocity (e.g., thrust), which would reduce the required instantaneous velocity.

For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula

${\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}},}$

where G is the universal gravitational constant (G ≈ 6.67×10−11 m3·kg−1·s−2), M the mass of the body to be escaped, and r the distance from the center of mass of the body to the object. The relationship is independent of the mass of the object escaping the mass body M. Conversely, a body that falls under the force of gravitational attraction of mass M, from infinity, starting with zero velocity, will strike the mass with a velocity equal to its escape velocity.

When given a speed ${\displaystyle V}$ greater than the escape speed ${\displaystyle v_{e},}$ the object will asymptotically approach the hyperbolic excess speed ${\displaystyle v_{\infty },}$ satisfying the equation:

${\displaystyle {v_{\infty }}^{2}=V^{2}-{v_{e}}^{2}.}$

In these equations atmospheric friction (air drag) is not taken into account. A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape. Escape velocity is only required to send a ballistic object on a trajectory that will allow the object to escape the gravity well of the mass M.

## Overview

Luna 1, launched in 1959, was the first man-made object to attain escape velocity from Earth (see below table).

The existence of escape velocity is a consequence of conservation of energy and an energy field of finite depth. For an object with a given total energy, which is moving subject to conservative forces (such as a static gravity field) it is only possible for the object to reach combinations of locations and speeds which have that total energy; and places which have a higher potential energy than this cannot be reached at all. By adding speed (kinetic energy) to the object it expands the possible locations that can be reached, until, with enough energy, they become infinite.

For a given gravitational potential energy at a given position, the escape velocity is the minimum speed an object without propulsion needs to be able to "escape" from the gravity (i.e. so that gravity will never manage to pull it back). Escape velocity is actually a speed (not a velocity) because it does not specify a direction: no matter what the direction of travel is, the object can escape the gravitational field (provided its path does not intersect the planet).

The simplest way of deriving the formula for escape velocity is to use conservation of energy. For the sake of simplicity, unless stated otherwise, we assume that an object is attempting to escape from a uniform spherical planet by moving away from it and that the only significant force acting on the moving object is the planet's gravity. In its initial state, i, imagine that a spaceship of mass m is at a distance r from the center of mass of the planet, whose mass is M. Its initial speed is equal to its escape velocity, ${\displaystyle v_{e}}$. At its final state, f, it will be an infinite distance away from the planet, and its speed will be negligibly small and assumed to be 0. Kinetic energy K and gravitational potential energy Ug are the only types of energy that we will deal with, so by the conservation of energy,

${\displaystyle (K+U_{g})_{i}=(K+U_{g})_{f}\,}$

Kƒ = 0 because final velocity is zero, and U = 0 because its final distance is infinity, so

{\displaystyle {\begin{aligned}\Rightarrow {}&{\frac {1}{2}}mv_{e}^{2}+{\frac {-GMm}{r}}=0+0\\[3pt]\Rightarrow {}&v_{e}={\sqrt {\frac {2GM}{r}}}={\sqrt {\frac {2\mu }{r}}}\end{aligned}}}

where μ is the standard gravitational parameter.

The same result is obtained by a relativistic calculation, in which case the variable r represents the radial coordinate or reduced circumference of the Schwarzschild metric.

Defined a little more formally, "escape velocity" is the initial speed required to go from an initial point in a gravitational potential field to infinity and end at infinity with a residual speed of zero, without any additional acceleration. All speeds and velocities measured with respect to the field. Additionally, the escape velocity at a point in space is equal to the speed that an object would have if it started at rest from an infinite distance and was pulled by gravity to that point.

In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s). However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.

The escape velocity is independent of the mass of the escaping object. It does not matter if the mass is 1 kg or 1,000 kg; what differs is the amount of energy required. For an object of mass ${\displaystyle m}$ the energy required to escape the Earth's gravitational field is GMm / r, a function of the object's mass (where r is the radius of the Earth, G is the gravitational constant, and M is the mass of the Earth, M = 5.9736 × 1024 kg). A related quantity is the specific orbital energy which is essentially the sum of the kinetic and potential energy divided by the mass. An object has reached escape velocity when the specific orbital energy is greater or equal to zero.

## Escape velocity in various situations

### From the surface of a body

An alternative expression for the escape velocity ${\displaystyle v_{e}}$ particularly useful at the surface on the body is:

${\displaystyle v_{e}={\sqrt {2gr\,}}}$

where r is the distance between the center of the body and the point at which escape velocity is being calculated and g is the gravitational acceleration at that distance (i.e., the surface gravity).

For a body with a spherically-symmetric distribution of mass, the escape velocity ${\displaystyle v_{e}}$ from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average density ρ.

${\displaystyle v_{e}=Kr{\sqrt {\rho }}}$

where ${\displaystyle K={\sqrt {{\frac {8}{3}}\pi G}}}$ ≈ 2.364 × 10−5 m1.5·kg−0.5·s−1

### From a rotating body

The escape velocity relative to the surface of a rotating body depends on direction in which the escaping body travels. For example, as the Earth's rotational velocity is 465 m/s at the equator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth. The surface velocity decreases with the cosine of the geographic latitude, so space launch facilities are often located as close to the equator as feasible, e.g. the American Cape Canaveral (latitude 28°28' N) and the French Guiana Space Centre (latitude 5°14' N).

### Practical considerations

In most situations it is impractical to achieve escape velocity almost instantly, because of the acceleration implied, and also because if there is an atmosphere the hypersonic speeds involved (on Earth a speed of 11.2 km/s, or 40,320 km/h) would cause most objects to burn up due to aerodynamic heating or be torn apart by atmospheric drag. For an actual escape orbit, a spacecraft will accelerate steadily out of the atmosphere until it reaches the escape velocity appropriate for its altitude (which will be less than on the surface). In many cases, the spacecraft may be first placed in a parking orbit (e.g. a low Earth orbit, LEO, at 160–2,000 km) and then accelerated to the escape velocity at that altitude, which will be slightly lower (about 11.0 km/s at a LEO of 200 km). The required additional change in speed, however, is far less because the spacecraft already has significant orbital velocity (in low Earth orbit speed is approximately 7.8 km/s, or 28,080 km/h).

### From an orbiting body

The escape velocity at a given height is ${\displaystyle {\sqrt {2}}}$ times the speed in a circular orbit at the same height, (compare this with the velocity equation in circular orbit). This corresponds to the fact that the potential energy with respect to infinity of an object in such an orbit is minus two times its kinetic energy, while to escape the sum of potential and kinetic energy needs to be at least zero. The velocity corresponding to the circular orbit is sometimes called the first cosmic velocity, whereas in this context the escape velocity is referred to as the second cosmic velocity.

For a body in an elliptical orbit wishing to accelerate to an escape orbit the required speed will vary, and will be greatest at periapsis when the body is closest to the central body. However, the orbital speed of the body will also be at its highest at this point, and the change in velocity required will be at its lowest, as explained by the Oberth effect.

### Barycentric escape velocity

Technically escape velocity can either be measured as a relative to the other, central body or relative to center of mass or barycenter of the system of bodies. Thus for systems of two bodies, the term escape velocity can be ambiguous, but it is usually intended to mean the barycentric escape velocity of the less massive body. In gravitational fields, escape velocity refers to the escape velocity of zero mass test particles relative to the barycenter of the masses generating the field. In most situations involving spacecraft the difference is negligible. For a mass equal to a Saturn V rocket, the escape velocity relative to the launch pad is 253.5 am/s (8 nanometers per year) faster than the escape velocity relative to the mutual center of mass.

### Height of lower velocity trajectories

Ignoring all factors other than the gravitational force between the body and the object, an object projected vertically at speed ${\displaystyle v}$ from the surface of a spherical body with escape velocity ${\displaystyle v_{e}}$ and radius ${\displaystyle R}$ will attain a maximum height ${\displaystyle h}$ satisfying the equation

${\displaystyle v=v_{e}{\sqrt {\frac {h}{R+h}}}\ ,}$

which, solving for h results in

${\displaystyle h={\frac {x^{2}}{1-x^{2}}}\ R\ ,}$

where ${\displaystyle x={\frac {v}{v_{e}}}}$ is the ratio of the original speed ${\displaystyle v}$ to the escape velocity ${\displaystyle v_{e}.}$

Unlike escape velocity, the direction (vertically up) is important to achieve maximum height.

## Trajectory

If an object attains exactly escape velocity, but is not directed straight away from the planet, then it will follow a curved path or trajectory. (Although this trajectory does not form a closed shape, it can be referred to as an orbit.) Assuming that gravity is the only significant force in the system, this object's speed at any point in the trajectory will be equal to the escape velocity at that point (due to the conservation of energy, its total energy must always be 0, which implies that it always has escape velocity; see the derivation above). The shape of the trajectory will be a parabola whose focus is located at the center of mass of the planet. An actual escape requires a course with an trajectory that does not intersect with the planet, or its atmosphere, since this would cause the object to crash. When moving away from the source, this path is called an escape orbit. Escape orbits are known as C3 = 0 orbits. C3 is the characteristic energy, = −GM/a, where a is the semi-major axis, which is infinite for parabolic trajectories.

If the body has a velocity greater than escape velocity then its path will form a hyperbolic trajectory and it will have an excess hyperbolic velocity, equivalent to the extra energy the body has. A relatively small extra delta-v above that needed to accelerate to the escape speed can result in a relatively large speed at infinity. For example, at a place where escape speed is 11.2 km/s, the addition of 0.4 km/s yields a hyperbolic excess speed of 3.02 km/s:

${\displaystyle v_{\infty }={\sqrt {V^{2}-{v_{e}}^{2}}}={\sqrt {11.6^{2}-11.2^{2}}}\approx 3.02.}$

If a body in circular orbit (or at the periapsis of an elliptical orbit) accelerates along its direction of travel to escape velocity, the point of acceleration will form the periapsis of the escape trajectory. The eventual direction of travel will be at 90 degrees to the direction at the point of acceleration. If the body accelerates to beyond escape velocity the eventual direction of travel will be at a smaller angle, and indicated by one of the asymptotes of the hyperbolic trajectory it is now taking. This means the timing of the acceleration is critical if the intention is to escape in a particular direction.

## Multiple bodies

When escaping a compound system, such as a moon orbiting a planet or a planet orbiting a sun, a rocket that leaves at escape velocity (${\displaystyle v_{e1}}$) for the first (orbiting) body, (e.g. Earth) will not travel to an infinite distance because it needs an even higher speed to escape gravity of the second body (e.g. the Sun). Near the Earth, the rocket's trajectory will appear parabolic, but it will still be gravitationally bound to the second body and will enter an elliptical orbit around that body, with a similar orbital speed to the first body.

To escape the gravity of the second body once it has escaped the first body the rocket will need to be travelling at the escape velocity for the second body (${\displaystyle v_{e2}}$) (at the orbital distance of the first body). However, when the rocket escapes the first body it will still have the same orbital speed around the second body that the first body has (${\displaystyle v_{o}}$). So its excess velocity as it escapes the first body will need to be the difference between the orbital velocity and the escape velocity. With a circular orbit, escape velocity is 2 times the orbital speed. Thus the total escape velocity ${\displaystyle v_{te}}$ when leaving one body orbiting a second and seeking to escape them both is, under simplified assumptions:

${\displaystyle v_{te}={\sqrt {(v_{e2}-v_{o})^{2}+v_{e1}^{2}}}={\sqrt {\left(kv_{e2}\right)^{2}+v_{e1}^{2}}}}$

where ${\displaystyle k=1-{\frac {1}{\sqrt {2}}}\approx 0.2929}$ for circular orbits.

## List of escape velocities

Location Relative to Ve (km/s) Location Relative to Ve (km/s) System escape Vte (km/s) On the Sun The Sun's gravity 617.5 On Mercury Mercury's gravity 4.25 At Mercury The Sun's gravity ~ 67.7 ~ 20.3 On Venus Venus's gravity 10.36 At Venus The Sun's gravity 49.5 17.8 On Earth Earth's gravity 11.186 At Earth/the Moon The Sun's gravity 42.1 16.6 On the Moon The Moon's gravity 2.38 At the Moon The Earth's gravity 1.4 2.42 On Mars Mars' gravity 5.03 At Mars The Sun's gravity 34.1 11.2 On Ceres Ceres's gravity 0.51 At Ceres The Sun's gravity 25.3 7.4 On Jupiter Jupiter's gravity 60.20 At Jupiter The Sun's gravity 18.5 60.4 On Io Io's gravity 2.558 At Io Jupiter's gravity 24.5 7.6 On Europa Europa's gravity 2.025 At Europa Jupiter's gravity 19.4 6.0 On Ganymede Ganymede's gravity 2.741 At Ganymede Jupiter's gravity 15.4 5.3 On Callisto Callisto's gravity 2.440 At Callisto Jupiter's gravity 11.6 4.2 On Saturn Saturn's gravity 36.09 At Saturn The Sun's gravity 13.6 36.3 On Titan Titan's gravity 2.639 At Titan Saturn's gravity 7.8 3.5 On Uranus Uranus' gravity 21.38 At Uranus The Sun's gravity 9.6 21.5 On Neptune Neptune's gravity 23.56 At Neptune The Sun's gravity 7.7 23.7 On Triton Triton's gravity 1.455 At Triton Neptune's gravity 6.2 2.33 On Pluto Pluto's gravity 1.23 At Pluto The Sun's gravity ~ 6.6 ~ 2.3 At Solar System galactic radius The Milky Way's gravity 492–594 On the event horizon A black hole's gravity 299,792.458 (speed of light)

The last two columns will depend precisely where in orbit escape velocity is reached, as the orbits are not exactly circular (particularly Mercury and Pluto).

## Deriving escape velocity using calculus

Let G be the gravitational constant and let M be the mass of the earth (or other gravitating body) and m be the mass of the escaping body or projectile. At a distance r from the centre of gravitation the body feels an attractive force

${\displaystyle F=G{\frac {Mm}{r^{2}}}.}$

The work needed to move the body over a small distance dr against this force is therefore given by

${\displaystyle dW=Fdr=-G{\frac {Mm}{r^{2}}}\,dr,}$

where the minus sign indicates the force acts in the opposite sense of ${\displaystyle dr}$.

The total work needed to move the body from the surface r0 of the gravitating body to infinity is then

${\displaystyle W=\int _{r_{0}}^{\infty }-G{\frac {Mm}{r^{2}}}\,dr=-G{\frac {Mm}{r_{0}}}=-mgr_{0}.}$

This is the minimal required kinetic energy to be able to reach infinity, so the escape velocity v0 satisfies

${\displaystyle W+K=0\Rightarrow {\frac {1}{2}}mv_{0}^{2}=G{\frac {Mm}{r_{0}}},}$

which results in

${\displaystyle v_{0}={\sqrt {\frac {2GM}{r_{0}}}}={\sqrt {2gr_{0}}}.}$