Incelestial mechanics,escape velocity orescape speed is the minimum speed needed for an object to escape from contact with or orbit of aprimary body, assuming:
Although the termescape velocity is common, it is more accurately described as aspeed than as avelocity because it is independent of direction. Because gravitational force between two objects depends on their combined mass, the escape speed also depends on mass. Forartificial satellites and small natural objects, the mass of the object makes a negligible contribution to the combined mass, and so is often ignored.
Escape speed varies with distance from the center of the primary body, as does the velocity of an object traveling under the gravitational influence of the primary. If an object is in a circular or elliptical orbit, its speed is always less than the escape speed at its current distance. In contrast if it is on ahyperbolic trajectory its speed will always be higher than the escape speed at its current distance. (It will slow down as it gets to greater distance, but do soasymptotically approaching a positive speed.) An object on aparabolic trajectory will always be traveling exactly the escape speed at its current distance. It has precisely balanced positivekinetic energy and negativegravitational potential energy;[a] it will always be slowing down, asymptotically approaching zero speed, but never quite stop.[1]
Escape velocity calculations are typically used to determine whether an object will remain in thegravitational sphere of influence of a given body. For example, insolar system exploration it is useful to know whether a probe will continue to orbit the Earth or escape to aheliocentric orbit. It is also useful to know how much a probe will need to slow down in order to begravitationally captured by its destination body. Rockets do not have to reach escape velocity in a single maneuver, and objects can also use agravity assist to siphon kinetic energy away from large bodies.
Precise trajectory calculations require taking into account small forces likeatmospheric drag,radiation pressure, andsolar wind. A rocket under continuous or intermittent thrust (or an object climbing aspace elevator) can attain escape at any non-zero speed, but the minimum amount of energy required to do so is always the same.
Escape speed at a distanced from the center of a spherically symmetric primary body (such as a star or a planet) with massM is given by the formula[2][3]
When given an initial speedV greater than the escape speedve, the object will asymptotically approach thehyperbolic excess speedv∞, satisfying the equation:[5]
For example, with the definitional value forstandard gravity of 9.80665 m/s2 (32.1740 ft/s2),[6] the escape velocity from Earth is 11.186 km/s (40,270 km/h; 25,020 mph; 36,700 ft/s).[7]
For an object of mass the energy required to escape the Earth's gravitational field isGMm / r, a function of the object's mass (wherer isradius of the Earth, nominally 6,371 kilometres (3,959 mi),G is thegravitational constant, andM is the mass of theEarth,M = 5.9736 × 1024 kg). A related quantity is thespecific 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 than or equal to zero.
The existence of escape velocity can be thought of as a consequence ofconservation of energy and an energy field of finite depth. For an object with a given total energy, which is moving subject toconservative 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; places which have a higher potential energy than this cannot be reached at all. Adding speed (kinetic energy) to an object expands the region of locations it can reach, until, with enough energy, everywhere to infinity becomes accessible.
The formula for escape velocity can be derived from the principle of conservation of energy. For the sake of simplicity, unless stated otherwise, we assume that an object will escape the gravitational field of 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. Imagine that a spaceship of massm is initially at a distancer from the center of mass of the planet, whose mass isM, and its initial speed is equal to its escape velocity,ve. At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small.Kinetic energyK andgravitational potential energyUg are the only types of energy that we will deal with (we will ignore the drag of the atmosphere), so by the conservation of energy,
We can setKfinal = 0 because final velocity is arbitrarily small, andUgfinal = 0 because final gravitational potential energy is defined to be zero a long distance away from a planet, so
The same result is obtained by arelativistic calculation, in which case the variabler represents theradial coordinate orreduced circumference of theSchwarzschild metric.[9][10]
For a body with a spherically symmetric distribution of mass, the escape velocityve from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average densityρ.
where.
This escape velocity is relative to a non-rotating frame of reference, not relative to the moving surface of the planet or moon, as explained below.
The escape velocityrelative 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 theequator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/srelative to the moving surface at the point of launch to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/srelative to that moving surface. The surface velocity decreases with thecosine of the geographic latitude, so space launch facilities are often located as close to the equator as feasible, e.g. the AmericanCape Canaveral (latitude 28°28′ N) and the FrenchGuiana Space Centre (latitude 5°14′ N).
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 toaerodynamic heating or be torn apart byatmospheric 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 aparking orbit (e.g. alow Earth orbit 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 low Earth orbit of 200 km). The required additionalchange in speed, however, is far less because the spacecraft already has a significantorbital speed (in low Earth orbit speed is approximately 7.8 km/s, or 28,080 km/h).
The escape velocity at a given height is times the speed in a circular orbit at the same height, (compare this with the velocity equation incircular 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 thefirst cosmic velocity, whereas in this context the escape velocity is referred to as thesecond cosmic velocity.[12][13][14]
For a body in an elliptical orbit wishing to accelerate to an escape orbit the required speed will vary, and will be greatest atperiapsis 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 theOberth effect.
Escape velocity can either be measured as relative to the other, central body or relative tocenter of mass or barycenter of the system of bodies. Thus for systems of two bodies, the termescape velocity can be ambiguous, but it is usually intended to mean the barycentric escape velocity of the less massive body. Escape velocity usually refers to the escape velocity of zero masstest particles. For zero mass test particles we have that the 'relative to the other' and the 'barycentric' escape velocities are the same, namely.
But when we can't neglect the smaller mass (saym) we arrive at slightly different formulas.
Because the system has to obey thelaw of conservation of momentum we see that both the larger and the smaller mass must be accelerated in the gravitational field. Relative to the center of mass the velocity of the larger mass (vp, for planet) can be expressed in terms of the velocity of the smaller mass (vr, for rocket). We get.
The 'barycentric' escape velocity now becomes, while the 'relative to the other' escape velocity becomes.
Ignoring all factors other than the gravitational force between the body and the object, an object projected vertically at speedv from the surface of a spherical body with escape velocityve and radiusR will attain a maximum heighth satisfying the equation[15]
which, solving forh results in
wherex =v/ve is the ratio of the original speedv to the escape velocityve.
Unlike escape velocity, the direction (vertically up) is important to achieve maximum height.
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 velocityat 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 aparabola whose focus is located at the center of mass of the planet. An actual escape requires a course with a 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 anescape orbit. Escape orbits are known asC3 = 0 orbits.C3 is thecharacteristic energy, −GM/2a, wherea is thesemi-major axis length, which is infinite for parabolic trajectories.
If the body has a velocity greater than escape velocity then its path will form ahyperbolic trajectory and it will have an excess hyperbolic velocity, equivalent to the extra energy the body has. A relatively small extradelta-v above that needed to accelerate to the escape speed can result in a relatively large speed at infinity.Some orbital manoeuvres make use of this fact. 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:
If a body in circular orbit (or at theperiapsis 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.
If the speed at periapsis isv, then theeccentricity of the trajectory is given by:
This is valid for elliptical, parabolic, and hyperbolic trajectories. If the trajectory is hyperbolic or parabolic, it willasymptotically approach an angle from the direction at periapsis, with
In this table, the left-hand half gives the escape velocity from the visible surface (which may be gaseous as with Jupiter for example), relative to the centre of the planet or moon (that is, not relative to its moving surface). In the right-hand half,Ve refers to the speed relative to the central body (for example the sun), whereasVte is the speed (at the visible surface of the smaller body) relative to the smaller body (planet or moon).
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).
LetG be thegravitational constant and letM be themass of the earth (or other gravitating body) andm be the mass of the escaping body or projectile. At a distancer from the centre of gravitation the body feels an attractive force
The work needed to move the body over a small distancedr against this force is therefore given by
The total work needed to move the body from the surfacer0 of the gravitating body to infinity is then[20]
In order to do this work to reach infinity, the body's minimal kinetic energy at departure must match this work, so the escape velocityv0 satisfies
^Bureau International des Poids et Mesures (1901). "Déclaration relative à l'unité de masse et à la définition du poids; valeur conventionnelle degn".Comptes Rendus des Séances de la Troisième Conférence· Générale des Poids et Mesures (in French). Paris: Gauthier-Villars. p. 68.Le nombre adopté dans le Service international des Poids et Mesures pour la valeur de l'accélération normale de la pesanteur est 980,665 cm/sec², nombre sanctionné déjà par quelques législations. Déclaration relative à l'unité de masse et à la définition du poids; valeur conventionnelle degn.
