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Encyclopedia > Escape velocity
Isaac Newton's analysis of escape velocity. Projectiles A and B fall back to earth. Projectile C achieves a circular orbit, D an elliptical one. Projectile E escapes.

In physics, escape velocity is the speed where the kinetic energy of an object is equal to the magnitude of its gravitational potential energy, as calculated by the equation Ug = − Gm1m2 / 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 a misnomer 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. In more technical terms, escape velocity is a scalar (and not a vector). See: escape velocity (physics) Escape Velocity (computer game) Escape Velocity (band) Escape Velocity Nova is an expandable card-driven board game based on earlier computer game [1],[2] This is a disambiguation pageâ€”a list of articles associated with the same title. ... Image File history File links No higher resolution available. ... Image File history File links No higher resolution available. ... A magnet levitating above a high-temperature superconductor demonstrates the Meissner effect. ... The cars of a roller coaster reach their maximum kinetic energy when at the bottom of their path. ... Potential energy (U, or Ep), a kind of scalar potential, is energy by virtue of matter being able to move to a lower-energy state, releasing energy in some form. ... A scalar may be: Look up scalar in Wiktionary, the free dictionary. ... Look up vector in Wiktionary, the free dictionary. ...

For a given gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion needs to attain in order to "escape" from 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. A gravitational field is a model used within physics to explain how gravity exists in the universe. ... This article does not cite any references or sources. ... A remote camera captures a close-up view of a Space Shuttle Main Engine during a test firing at the John C. Stennis Space Center in Hancock County, Mississippi Spacecraft propulsion is any method used to change the velocity of spacecraft and artificial satellites. ...

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, ve. 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 Ug are the only types of energy that we will deal with, so by the conservation of energy, This article is about the law of conservation of energy in physics. ...

$(K + U_g)_i = (K + U_g)_f. ,$

Kf = 0 because final velocity is zero, and Ugf = 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. This article is about the astronomical term. ... A natural satellite or moon is a celestial body that orbits a planet or smaller body, which is called the primary. ... kilometre per second is an SI derived unit of both speed (scalar) and velocity (vector), signified by the symbol km/s or km s-1. ...

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. The gravitational constant G is a key element in Newtons law of universal gravitation. ...

## 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 direct from 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 m/s.

## 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 an escape orbit; when moving closer to the source, a capture orbit. Both are known as C3 = 0 orbits (where C3 = - μ/a, and a is the semi-major axis). A parabola A graph showing the reflective property, the directrix (light blue), and the lines connecting the focus and directrix to the parabola (blue) In mathematics, the parabola (from the Greek: Ï€Î±ÏÎ±Î²Î¿Î»Î®) (IPA pronunciation: ) is a conic section generated by the intersection of a right circular conical surface and a plane... An escape orbit (also known as C3 = 0 orbit) is the high-energy parabolic orbit around the central body. ... A capture orbit is the high-energy parabolic orbit that allows the capture other than crashing directly to the central bodys surface (or atmospheric re_entry). ... The semi-major axis of an ellipse In geometry, the term semi-major axis (also semimajor axis) is used to describe the dimensions of ellipses and hyperbolae. ...

Remember that 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

To leave planet Earth an escape velocity of 11.2 km/s is required, however a speed of 42.1 km/s is required to escape the Sun's gravity (and exit the solar system) from the same position
Location with respect to Ve[1]     Location with respect to Ve[1]
on the Sun, the Sun's gravity: 617.5 km/s
on Mercury, Mercury's gravity: 4.4 km/s at Mercury, the Sun's gravity: 67.7 km/s
on Venus, Venus' gravity: 10.4 km/s at Venus, the Sun's gravity: 49.5 km/s
on Earth, the Earth's gravity: 11.2 km/s at the Earth/Moon, the Sun's gravity: 42.1 km/s
on the Moon, the Moon's gravity: 2.4 km/s at the Moon, the Earth's gravity: 1.4 km/s
on Mars, Mars' gravity: 5.0 km/s at Mars, the Sun's gravity: 34.1 km/s
on Jupiter, Jupiter's gravity: 59.5 km/s at Jupiter, the Sun's gravity: 18.5 km/s
on Saturn, Saturn's gravity: 35.5 km/s at Saturn, the Sun's gravity: 13.6 km/s
on Uranus, Uranus' gravity: 21.3 km/s at Uranus, the Sun's gravity: 9.6 km/s
on Neptune, Neptune's gravity: 23.5 km/s at Neptune, the Sun's gravity: 7.7 km/s
in the solar system,   the Milky Way's gravity:   ~1,000 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 ve is the escape velocity, G is the gravitational constant, M is the mass of the body being escaped from, m is the mass of the escaping body, r is the distance between the center of the body and the point at which escape velocity is being calculated, g is the gravitational acceleration at that distance, and μ is the standard gravitational parameter.[2] The gravitational constant G is a key element in Newtons law of universal gravitation. ... For other uses, see Mass (disambiguation). ... For other uses, see Mass (disambiguation). ... Distance is a numerical description of how far apart objects are at any given moment in time. ... In physics, gravitational acceleration is the acceleration of an object caused by the force of gravity from another object. ... In astrodynamics, the standard gravitational parameter () of a celestial body is the product of the gravitational constant () and the mass : The units of the standard gravitational parameter are km3s-2 Small body orbiting a central body Under standard assumptions in astrodynamics we have: where: is the mass of the orbiting...

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. In physics, circular motion is rotation along a circle: a circular path or a circular orbit. ...

For a body with a spherically-symmetric distribution of mass, the escape velocity ve from the surface (in m/s) is approximately 2.364×10−5 m1.5kg−0.5s−1 times the radius r (in meters) times the square root of the average density ρ (in kg/m³), or:

$v_e approx 2.364 times 10^{-5} r sqrt rho.,$

## Deriving escape velocity using calculus

These derivations use calculus, Newton's laws of motion and Newton's law of universal gravitation. For other uses, see Calculus (disambiguation). ... Newtons First and Second laws, in Latin, from the original 1687 edition of the Principia Mathematica. ... Isaac Newtons theory of universal gravitation (part of classical mechanics) states the following: Every single point mass attracts every other point mass by a force pointing along the line combining the two. ...

### 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 g (also gee, g-force or g-load) is a non-SI unit of acceleration defined as exactly 9. ... The gravitational constant G is a key element in Newtons law of universal gravitation. ...

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 / sds. 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: This diagram shows how the law works. ... Isaac Newtons theory of universal gravitation (part of classical mechanics) states the following: Every single point mass attracts every other point mass by a force pointing along the line combining the two. ... Infinitesimals have been used to express the idea of objects so small that there is no way to see them or to measure them. ...

$int_r^infty gm (r/s)^2 , ds =gmr^2 int_r^infty s^{-2},ds =gmr^2 left[-s^{-1}right]_{s:=r}^{s:=infty}$
$=gmr^2left(0-(-r^{-1})right)=gmr.$

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)mv². Thus we need

$begin{matrix}frac12end{matrix} mv^2=gmr.$

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

$vcongsqrt{2left(9.8 {mathrm{m}/mathrm{s}^2}right)(6.4times 10^6 mathrm{m})}= 11,201 mathrm{m}/mathrm{s}.$

This is just a bit over 11 kilometers per second, or a bit under 7 miles per second, as Isaac Newton calculated. Sir Isaac Newton FRS (4 January 1643 â€“ 31 March 1727) [ OS: 25 December 1642 â€“ 20 March 1727][1] was an English physicist, mathematician, astronomer, natural philosopher, and alchemist. ...

### 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. The gravitational constant G is a key element in Newtons law of universal gravitation. ...

$ma=mfrac{dv}{dt}=-frac{GMm}{r^2},$
$a=frac{dv}{dt}=-frac{GM}{r^2},$

By applying the chain rule, we get: In calculus, the chain rule is a formula for the derivative of the composite of two functions. ...

$frac{dv}{dt}=frac{dv}{dr} cdot frac{dr}{dt}=-frac{GM}{r^2},$

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

$t rightarrow infty r(t) rightarrow infty$ and $v(t) rightarrow 0$
$-frac{v_0^2}{2} = -frac{GM}{r_0},$
$v_0 = sqrtfrac{2GM}{r_0},$

v0 is the escape velocity and r0 is the radius of the planet. Note that the above derivation relies on the equivalence of inertial mass and gravitational mass. Inertial mass is a measure of the resistance of an entity to a change in its velocity relative to an inertial frame. ... Mass is a property of physical objects that, roughly speaking, measures the amount of matter they contain. ...

### 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 gravity well is the scientific/science fictional term for the distortion in space-time caused by a massive body such as a planet. ...

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.

Potential energy can be thought of as energy stored within a physical system. ... Delta-v budget (or velocity change budget) is a term used in astrodynamics and aerospace industry for velocity change (or delta-v) requirements for the various propulsive tasks and orbital maneuvers over phases of the space mission. ... In orbital mechanics and aerospace engineering, a gravitational slingshot or gravity assist is the use of the gravity of a planet or other celestial body to alter the path and speed of a spacecraft. ... A gravity well is the scientific/science fictional term for the distortion in space-time caused by a massive body such as a planet. ... In classical mechanics, the two-body problem is to determine the motion of two point particles that interact only with each other. ... For other uses, see Black hole (disambiguation). ... The Oberth effect is a feature of astronautics where using a rocket engine close to a gravitational body gives a higher final speed than the same burn executed further from the body. ...

## References

• Roger R. Bate, Donald D. Mueller, and Jerry E. White (1971). Fundamentals of astrodynamics. New York: Dover Publications. ISBN 0-486-60061-0.
1. ^ a b Solar System Data. Georgia State University. Retrieved on 2007-01-21.
2. ^ Bate, Mueller and White, p. 35

Year 2007 (MMVII) was a common year starting on Monday of the Gregorian calendar in the 21st century. ... is the 21st day of the year in the Gregorian calendar. ...

Results from FactBites:

 Escape Velocity - MSN Encarta (322 words) In physics, escape velocity is the speed where the kinetic energy of an object is equal to the magnitude of its gravitational potential energy, as calculated by the equation U g... Escape Velocity, minimum initial velocity required for an object to escape the gravitational attraction of an astronomical body, and to continue traveling away from it without the use of propulsive machinery. The escape velocity of an object from a spherical astronomical body is proportional to the square root of the mass of the body divided by the distance between the object and the center of the body.
 Escape Velocity: Planetary Reckoning (1701 words) It would appear that even the aspect of escape velocities seems to be registered in the historically significant numbers and their fractal expressions. This latter figure is totally feasible and within the range of the escape velocity for the surface of Earth. Remember that we said that the escape velocity most commonly cited for Earth in terms of miles per second is that of 6.94444 mps (in reference to 25000 mph); and that of 11.1 kms/s (and, 11.2 kms/s) for the metric system.
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