- Escape velocity
In

physics ,**escape velocity**is the speed where thekinetic energy of an object is equal to the magnitude of itsgravitational potential energy , as calculated by the equation,:$U\_g\; =\; frac\{-Gm\_1m\_2\}\{r\}$.It is commonly described as the speed needed to "break free" from a gravitational field (without any additional impulse). The term "escape velocity" can be considered amisnomer because it is actually a speed rather than a velocity, i.e. it specifies how fast the object must move but the direction of movement is irrelevant, unless "downward." In more technical terms, escape velocity is ascalar (and not avector ).**Overview**The phenomenon of escape velocity is a consequence of

conservation of energy . For an object with a given total energy, which is moving subject toconservative force s (such as a static gravity fields) it is only possible for the object to reach combinations of places and speeds which have that total energy; and places which have a higher potential energy than this cannot be reached at all.For a given

gravitational potential energy at a given position, the**escape velocity**is the minimumspeed an object without propulsion needs to have sufficient energy to be able to "escape" from the gravity, i.e. so that gravity will never manage to pull it back. For the sake of simplicity, unless stated otherwise, we will assume that the scenario we are dealing with is that an object is attempting to escape from a uniform spherical planet by moving straight up (along a radial line away from the center of the planet), and that the "only" significant force acting on the moving object is the planet's gravity.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. The simplest way of deriving the formula for escape velocity is to use conservation of energy. 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, $v\_e$. At its final state, 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 "U"

_{g}are the only types of energy that we will deal with, so by the conservation of energy,:$(K\; +\; U\_g)\_i\; =\; (K\; +\; U\_g)\_f.\; ,$

"K"

_{"f"}= 0 because final velocity is zero, and "U"_{gf}= 0 because its final distance is infinity, so:$frac\{1\}\{2\}mv\_e^2\; +\; frac\{-GMm\}\{r\}\; =\; 0\; +\; 0$

:$v\_e\; =\; sqrt\{frac\{2GM\}\{r$

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 with a residual velocity of zero, with 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 kilometers per second (~6.96 mi/s), which is approximately 34 times the speed of sound (mach 34) and at least 10 times the speed of a rifle bullet. However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.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 FrenchGuiana Space Centre (latitude 5°14' N).Escape velocity is independent of the mass of the escaping object. It does not matter if the mass is 1 kg or 1000 kg, escape velocity from the same point in the same gravitational field is always the same. What differs is the amount of energy needed to accelerate the mass to achieve escape velocity: the energy needed for an object of mass $m$ 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). More massive objects require more energy to reach escape velocity. All of this, of course, assumes we are neglecting air resistance.**Misconceptions**Planetary or lunar escape velocity is sometimes misunderstood to be the speed a powered vehicle (such as a rocket) "must" reach to leave orbit; however, this is not the case, as the quoted number is typically the "surface" escape velocity, and vehicles never achieve that speed directly from the surface. This surface escape velocity is the speed required for an object to leave the planet if the object is simply projected from the surface of the planet and then left without any more kinetic energy input: in practice the vehicle's propulsion system will continue to provide energy after it has left the surface.

In fact a vehicle can leave the Earth's gravity at any speed. At higher altitude, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at "that" position. At sufficiently high altitude this speed can approach 0.

**Orbit**If an object attains escape velocity, but is not directed straight away from the planet, then it will follow a curved path. Even though this path will not form a closed shape, it is still considered an orbit. Assuming that gravity is the only significant force in the system, this object's speed at any point in the orbit 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 orbit will be a

parabola whose focus is located at the center of mass of the planet. An actual escape requires of course that the orbit not intersect the planet, since this would cause the object to crash. When moving away from the source, this path is called anescape orbit ; when moving closer to the source, acapture orbit . Both are known as "C"3 = 0 orbits (where "C"3 = - "μ/a", and "a" is thesemi-major axis ).In reality there are many gravitating bodies in space, so that, for instance, a rocket that travels at escape velocity from Earth will not escape to an infinite distance away because it needs an even higher speed to escape the Sun's gravity. In other words, near the Earth, the rocket's orbit will appear parabolic, but eventually its orbit will become an ellipse around the Sun.

**List of escape velocities**Because of the atmosphere it is not useful and hardly possible to give an object near the surface of the Earth a speed of 11.2 km/s, as these speeds are too far in the

hypersonic regime for most practical propulsion systems and would cause most objects to burn up due to atmospheric friction. For an actual escape orbit a spacecraft is first placed inlow Earth orbit and then accelerated to the escape velocity at that altitude, which is a little less — about 10.9 km/s. The required acceleration, however, is generally even less because from that sort of an orbit the spacecraft already has a speed of 8 km/s.**Calculating an escape velocity**To expand upon the derivation given in the Overview,

:$v\_e\; =\; sqrt\{frac\{2GM\}\{r\; =\; sqrt\{frac\{2mu\}\{r\; =\; sqrt\{2gr,\}.$

where $v\_e$ is the escape velocity, "G" is the

gravitational constant , "M" is themass of the body being escaped from, "r" is thedistance between the center of the body and the point at which escape velocity is being calculated, "g" is thegravitational acceleration at that distance, and μ is thestandard gravitational parameter . [*Bate, Mueller and White, p. 35*]The escape velocity at a given height is $sqrt\; 2$ times the speed in a circular orbit at the same height (compare this with equation (14) in

circular motion ). 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.For a body with a spherically-symmetric distribution of mass, the escape velocity $v\_e$ from the surface (in m/s) is approximately 2.364×10

^{−5}m^{1.5}kg^{−0.5}s^{−1}times the radius "r" (in meters) times the square root of the average density ρ (in kg/m³), or::$v\_e\; approx\; 2.364\; imes\; 10^\{-5\}\; r\; sqrt\; ho.,$

**Deriving escape velocity using calculus**These derivations use

calculus ,Newton's laws of motion andNewton's law of universal gravitation .**Derivation using only "g" and "r"**the Earth's escape speed can be derived from "g"," the acceleration due to gravity at the Earth's surface. It is not necessary to know the

gravitational constant "G" or the mass "M" of the Earth. Let:"r" = the Earth's radius, and

:"g" = the acceleration of gravity at the Earth's surface.

Above the Earth's surface, the acceleration of gravity is governed by Newton's inverse-square law of universal gravitation. Accordingly, the acceleration of gravity at height "s" above the center of the Earth (where"s" > "r" ) is $g\; (r\; /\; s)^2$.The weight of an object of mass "m" at the surface is "g m", and its weight at height "s" above the center of the Earth is "gm" ("r" / "s")².Consequently the energy needed to lift an object of mass "m" from height "s" above the Earth's center to height "s" + "ds" (where "ds" is an

infinitesimal increment of "s") is "gm" ("r" / "s")² "ds".Since this decreases sufficiently fast as "s" increases, the total energy needed to lift the object to infinite height does not diverge to infinity, but converges to a finite amount. That amount is the integral of the expression above::$int\_r^infty\; gm\; (r/s)^2\; ,\; ds=gmr^2\; int\_r^infty\; s^\{-2\},ds=gmr^2\; left\; [-s^\{-1\}\; ight]\; \_\{s:=r\}^\{s:=infty\}$

:$=gmr^2left(0-(-r^\{-1\})\; ight)=gmr.$

That is how much "kinetic" energy the object of mass "m" needs in order to escape. The kinetic energy of an object of mass "m" moving at speed "v" is (1/2)"mv"². Thus we need

:$egin\{matrix\}frac12end\{matrix\}\; mv^2=gmr.$

The factor "m" cancels out, and solving for "v" we get

:$v=sqrt\{2gr,\}.$

If we take the radius of the Earth to be "r" = 6400 kilometers and the acceleration of gravity at the surface to be "g" = 9.8 m/s², we get

:$vcongsqrt\{2left(9.8\; \{mathrm\{m\}/mathrm\{s\}^2\}\; ight)(6.4\; imes\; 10^6\; mathrm\{m\})\}=\; 11,201\; mathrm\{m\}/mathrm\{s\}.$

This is just a bit over 11 kilometers per second, or a bit under 7 miles per second, as

Isaac Newton calculated.**Derivation using "G" and "M"**Let "G" be the

gravitational constant and let "M" be the mass of the earth or other body to be escaped.:$ma=mfrac\{dv\}\{dt\}=-frac\{GMm\}\{r^2\},$

:$a=frac\{dv\}\{dt\}=-frac\{GM\}\{r^2\},$By applying the

chain rule , we get::$frac\{dv\}\{dt\}=frac\{dv\}\{dr\}\; cdot\; frac\{dr\}\{dt\}=-frac\{GM\}\{r^2\},$

Because $v=frac\{dr\}\{dt\}$:$frac\{dv\}\{dr\}cdot\; v\; =\; -frac\{GM\}\{r^2\},$

:$v\; cdot\; dv\; =\; -frac\{GM\}\{r^2\},dr,$

:$int\_\{v\_0\}^\{v(t)\}\; v,dv\; =\; -int\_\{r\_0\}^\{r(t)\}frac\{GM\}\{r^2\},dr,$

:$frac\{v(t)^2\}\{2\}-frac\{v\_0^2\}\{2\}\; =\; frac\{GM\}\{r(t)\}-frac\{GM\}\{r\_0\},$Since we want escape velocity :$t\; ightarrow\; infty\; r(t)\; ightarrow\; infty$ and $v(t)\; ightarrow\; 0$

:$-frac\{v\_0^2\}\{2\}\; =\; -frac\{GM\}\{r\_0\},$

:$v\_0\; =\; sqrtfrac\{2GM\}\{r\_0\},$"v"

_{0}is the escape velocity and "r"_{0}is the radius of the planet. Note that the above derivation relies on the equivalence ofinertial mass andgravitational mass .**The derivations are consistent**The gravitational acceleration can be obtained from the gravitational constant "G" and the mass of Earth "M":

:$g\; =\; frac\{GM\}\{r^2\},$

where "r" is the radius of Earth. Thus

:$v=sqrt\{2gr,\}=sqrt\{frac\{2GMr\}\{r^2\},\}=sqrt\{frac\{2GM\}\{r\},\},$

so the two derivations given above are consistent.

**Multiple sources**The escape velocity from a position in a field with multiple sources is derived from the total potential energy per kg at that position, relative to infinity. The potential energies for all sources can simply be added. For the escape velocity this results in the square root of the sum of the squares of the escape velocities of all sources separately.

For example, at the Earth's surface the escape velocity for the combination Earth and Sun is $scriptstylesqrt\{11.2^2\; +\; 42.1^2\}\; =\; 43.56\; mathrm\{km\}/mathrm\{s\}$. As a result, to leave the solar system requires a speed of 13.6 km/s relative to Earth in the direction of the Earth's orbital motion, since the speed is then added to the speed of 30 km/s of that orbital motion

**Gravity well**In the hypothetical case of uniform density, the velocity that an object would achieve when dropped in a hypothetical vacuum hole from the surface of the Earth to the center of the Earth is the escape velocity divided by $scriptstylesqrt\; 2$, i.e. the speed in a circular orbit at a low height. Correspondingly, the escape velocity from the center of the Earth would be $scriptstylesqrt\; \{1.5\}$ times that from the surface.

A refined calculation would take into account the fact that the Earth's mass is not uniformly distributed as the center is approached. This gives higher speeds.

**ee also*** Gravitational potential energy

*Delta-v budget - speed needed to perform manoeuvres.

*Gravitational slingshot - 3 body technique for gaining energy

*Gravity well

*Two-body problem

*Black hole - have an escape velocity greater than the speed of light

*Oberth effect - burning fuel deep in a gravity field gives higher velocity**References***cite book |author=Roger R. Bate, Donald D. Mueller, and Jerry E. White |title=Fundamentals of astrodynamics |publisher=Dover Publications |location=New York |year=1971 |isbn=0-486-60061-0 |oclc= |doi=

**External links*** [

*http://www.calctool.org/CALC/phys/astronomy/escape_velocity Web-based numerical escape velocity calculator*]

*Wikimedia Foundation.
2010.*

### Look at other dictionaries:

**Escape Velocity**— Entwickler Ambrosia Software und ATMOS Publisher … Deutsch Wikipedia**escape velocity**— es*cape vel*o ci*ty, n. (Physics) The minimum velocity at which an object must be moving in order for it to overcome the gravitational attraction of a massive celestial body, such as the earth or the sun, and escape beyond its gravitational field … The Collaborative International Dictionary of English**escape velocity**— n. the minimum speed required for a particle, space vehicle, or other body to escape permanently from the gravitational field of a planet, star, etc.: it is c. 11.3 km ( c. 7 mi) per second for escape from the earth … English World dictionary**Escape Velocity**— Cette page d’homonymie répertorie les différents sujets et articles partageant un même nom. Escape Velocity est un terme anglais pour désigner la vitesse de libération. Divers Escape Velocity est un film canadien réalisé par Lloyd A. Simandl en… … Wikipédia en Français**escape velocity**— Physics, Rocketry. the minimum speed that an object at a given distance from a gravitating body must have so that it will continue to move away from the body instead of orbiting about it. [1950 55] * * * Speed sufficient for a body to escape from … Universalium**escape velocity**— escape′ veloc ity n. phs the minimum speed that an object at a given distance from a celestial body must have so that it will escape from orbit around the body • Etymology: 1950–55 … From formal English to slang**escape velocity**— noun the minimum velocity needed to escape a gravitational field • Hypernyms: ↑speed, ↑velocity … Useful english dictionary**escape velocity**— noun Date: 1934 the minimum velocity that a moving body (as a rocket) must have to escape from the gravitational field of a celestial body (as the earth) and move outward into space … New Collegiate Dictionary**escape velocity**— noun The minimum velocity needed to escape the gravitational field of a planet or other body … Wiktionary**escape velocity**— noun the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object … English new terms dictionary