Escape velocity

Luna 1, launched in 1959, was the first man-made object to attain escape velocity from Earth (see below table).[5]

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).

An elegant way to derive the formula for escape velocity is to use the principle of conservation of energy (for another way, based on work, see below). 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 mass m is initially at a distance r from the center of mass of the planet, whose mass is M, and its initial speed is equal to its escape velocity, ${\displaystyle v_{e}}$. At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small. Kinetic energy K and gravitational potential energy U_g 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,

${\displaystyle (K+U_{g})_{initial}=(K+U_{g})_{final}\,}$

We can set K_ƒinal = 0 because final velocity is arbitrarily small, and U_gƒinal = 0 because 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.[6][7]

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.[8] All speeds and velocities are 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. Note that this escape velocity is relative to a non-rotating frame of reference, not relative to the moving surface of the planet or moon (see below).

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 than or equal to zero.

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 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 an escape orbit. Escape orbits are known as C3 = 0 orbits. C3 is the characteristic energy, = −GM/2a, 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. 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:

${\displaystyle v_{\infty }={\sqrt {V^{2}-{v_{e}}^{2}}}={\sqrt {(11.6{\text{ km/s}})^{2}-(11.2{\text{ km/s}})^{2}}}\approx 3.02{\text{ km/s}}.}$

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.

If the speed at periapsis is v, then the eccentricity of the trajectory is given by:

${\displaystyle e=2(v/v_{e})^{2}-1}$

This is valid for elliptical, parabolic, and hyperbolic trajectories. If the trajectory is hyperbolic or parabolic, it will asymptotically approach an angle ${\displaystyle \theta }$ from the direction at periapsis, with

${\displaystyle \sin \theta =1/e.}$

The speed will asymptotically approach

${\displaystyle {\sqrt {v^{2}-v_{e}^{2}}}.}$

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 an orbital speed similar 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, if the rocket just escapes the first body (that is, if it has the escape velocity at periapsis) it will still have the same orbital speed around the second body that the first body has (${\displaystyle v_{o}}$). It needs an excess velocity as it escapes the first body equal to 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:[12]

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

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

Near the first body the trajectory will be hyperbolic with an eccentricity given by:

${\displaystyle e=2(v_{te}/v_{e1})^{2}-1=1+2(kv_{e2}/v_{e1})^{2}}$

This will gradually morph into a parabolic trajectory escaping the second body.

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, V_e refers to the speed relative to the central body (for example the sun), whereas V_te 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).

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[16]

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

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

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

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

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

• Roger R. Bate; Donald D. Mueller; Jerry E. White (1971). Fundamentals of astrodynamics. New York: Dover Publications. ISBN 978-0-486-60061-1.

• Escape velocity calculator
• Web-based numerical escape velocity calculator