# Escape velocity

In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.nb 1 It is the speed needed to "break free" from the gravitational attraction of a massive body, without further propulsion.

For a spherically symmetric body, the escape velocity at a given distance is calculated by the formula1

$v_e = \sqrt{\frac{2GM}{r}},$

where G is the universal gravitational constant (G = 6.67×10−11 m3 kg−1 s−2), M the mass of the planet, star or other body, and r the distance from the center of gravity.nb 2

In this equation 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 do so, but could achieve the same result at any speed with a suitable mode of propulsion and sufficient fuel. Escape velocity only applies to ballistic trajectories.

The term escape velocity is actually a misnomer, and it is often more accurately referred to as escape speed since the necessary speed is a scalar quantity which is independent of direction (assuming a non-rotating planet and ignoring atmospheric friction).

## Overview

Luna 1, launched in 1959, was the first man-made object to attain escape velocity from Earth (see below table).2

A barycentric velocity is a velocity of one body relative to the center of mass of a system of bodies. A relative velocity is the velocity of one body with respect to another. Relative escape velocity is defined only in systems with two bodies. For systems of two bodies the term "escape velocity" is 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.

The existence of escape velocity is a consequence of conservation of energy. 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 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 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). For the sake of simplicity, unless stated otherwise, we will assume that an object is attempting to escape from a uniform spherical planet by moving directly away from it (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 (provided its path does not intersect the planet). 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 Ug are the only types of energy that we will deal with, so by the conservation of energy,

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

Kƒ = 0 because final velocity is zero, and U = 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 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 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).

The barycentric 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 $m$ 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).

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. When the mass reaches the Andromeda galaxy the earth will have recoiled 500 m.

## Orbit

If an object attains escape velocity, but is not directed straight away from the planet, then it will follow a curved path. Although this path does 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 a course with an orbit 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 orbits.

When there are many gravitating bodies, such as in the solar system, a rocket that travels at escape velocity from one body, say Earth, will not travel to an infinite distance because it needs an even higher speed to escape the Sun's gravity. Near the Earth, the rocket's orbit will appear parabolic, but will become an ellipse around the Sun, except when it is perturbed by the Earth, whose orbit it must still intersect, and other bodies.

## List of escape velocities

To leave planet Earth, an escape velocity of 11.2 km/s (approx. 40,320 km/h, or 25,000 mph) 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 (km/s)3 Location with respect to Ve (km/s)3 on the Sun, the Sun's gravity: 617.5 on Mercury, Mercury's gravity: 4.34:230 at Mercury, the Sun's gravity: 67.7 on Venus, Venus's gravity: 10.3 at Venus, the Sun's gravity: 49.5 on Earth, Earth's gravity: 11.24:200 at the Earth/Moon, the Sun's gravity: 42.1 on the Moon, the Moon's gravity: 2.4 at the Moon, the Earth's gravity: 1.4 on Mars, Mars' gravity: 5.04:234 at Mars, the Sun's gravity: 34.1 on Jupiter, Jupiter's gravity: 59.64:236 at Jupiter, the Sun's gravity: 18.5 on Ganymede, Ganymede's gravity: 2.7 on Saturn, Saturn's gravity: 35.64:238 at Saturn, the Sun's gravity: 13.6 on Uranus, Uranus' gravity: 21.34:240 at Uranus, the Sun's gravity: 9.6 on Neptune, Neptune's gravity: 23.84:240 at Neptune, the Sun's gravity: 7.7 on Pluto, Pluto's gravity: 1.2 at Solar System galactic radius, the Milky Way's gravity: 492–594 5 on the event horizon, a black hole's gravity: ≥ 299,792 (Speed of light)

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 (40,320 km/h), 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 aerodynamic heating or be torn apart by atmospheric drag. For an actual escape orbit a spacecraft is first placed in low Earth orbit (160–2,000 km) and then accelerated to the escape velocity at that altitude, which is a little less — about 10.9 km/s. The required change in speed, however, is far less because from a low Earth orbit the spacecraft already has a speed of approximately 8 km/s (28,800 km/h).

## Calculating an escape velocity

To expand upon the derivation given in the Overview,

$v_e = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2\mu}{r}} = \sqrt{2gr\,}$

where $v_e$ is the barycentric escape velocity, G is the gravitational constant, M is the mass of the body being escaped from, 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.6

The escape velocity at a given height is $\sqrt 2$ 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"7

For a body with a spherically-symmetric distribution of mass, the barycentric escape velocity $v_e$ 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.

### Derivation using only g and r

The Earth's escape speed can be derived from "g", the barycentric 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 barycentric 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:

\begin{align} \int_r^\infty gm \left(\frac{r}{s}\right)^2 \, ds & = gmr^2 \int_r^\infty s^{-2}\,ds = gmr^2 \left[-s^{-1}\right]_{s=r}^{s=\infty} \\ & = gmr^2\left(0-(-r^{-1})\right)=gmr. \end{align}

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

$\begin{matrix}\frac12\end{matrix} mv^2=gmr.$

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

$v\cong\sqrt{2\left(9.8\ {\mathrm{m}/\mathrm{s}^2}\right)(6.4\times 10^6\ \mathrm{m})}= 11\,200\ \mathrm{m}/\mathrm{s}.$

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 and 'm' be the mass of the escaping body or projectile.

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

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

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

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.

### Consistency of derivations

The barycentric 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 at rest with respect to each other 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 it can be shown that this gives an escape velocity which is equal to the square root of the sum of the squares of the individual escape velocities due to each source.

For example, at the Earth's surface the escape velocity for the combination Earth and Sun would be $\scriptstyle\sqrt{11.2^2\ +\ 42.1^2}\ \mathrm{km}/\mathrm{s}=\ 43.56\ \mathrm{km}/\mathrm{s}$.

## Notes

1. ^ The gravitational potential energy is negative since gravity is an attractive force and the potential energy has been defined for this purpose to be zero at infinite distance from the centre of gravity.
2. ^ The value GM is called the standard gravitational parameter, or μ, and is often known more accurately than either G or M separately.

## References

1. ^ Khatri, Poudel, Gautam, M.K. , P.R. , A.K. (2010). Principles of Physics. Kathmandu: Ayam Publication. pp. 170, 171. ISBN 9789937903844.
2. ^ NASA – NSSDC – Spacecraft – Details
3. ^ a b "Solar System Data". Georgia State University. Retrieved 2007-01-21.
4. Wimmer, Mark R. Chartrand ; illustrated by Helmut K. (2001). Night sky : a field guide to the heavens. New York: St. Martin's Press. ISBN 9781582381268.
5. ^ Smith, Martin C.; Ruchti, G. R.; Helmi, A.; Wyse, R. F. G. (2007). "The RAVE Survey: Constraining the Local Galactic Escape Speed". Proceedings of the International Astronomical Union 2 (S235): 137. doi:10.1017/S1743921306005692.
6. ^ Bate, Mueller and White, p. 35
7. ^ Teodorescu, P. P. (2007). Mechanical systems, classical models. Springer, Japan. p. 580. ISBN 1-4020-5441-6., Section 2.2.2, p. 580

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