Escape velocity

Scenarios

From the surface of a body

An alternative expression for the escape velocity v_e particularly useful at the surface on the body is:

${\displaystyle v_{\text{e}}=\sqrt{2gr}}$

where r is the distance between the center of the body and the point at which escape velocity is being calculated and g is the gravitational acceleration at that distance (i.e., the surface gravity).

For a body with a spherically symmetric distribution of mass, the escape velocity v_e from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average density ρ.

${\displaystyle v_{\text{e}}=Kr\sqrt{\rho }}$

where $K=\sqrt{\frac{8}{3}\pi G}\approx 2.364\times {10}^{-5}{\mathrm{m}}^{1.5}\cdot \mathrm{k}{\mathrm{g}}^{-0.5}\cdot {\mathrm{s}}^{-1}$.

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.

From a rotating body

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 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/s relative to that moving surface. 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).

Practical considerations

In most situations it is impractical to achieve escape velocity almost instantly, because of the acceleration implied, and also because if there is a substantial 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 to aerodynamic heating or be torn apart by atmospheric 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 a parking orbit (e.g. a low 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 additional change in speed, however, is far less because the spacecraft already has a significant orbital speed (in low Earth orbit speed is approximately 7.8 km/s, or 28,080 km/h).

For an orbiting body

The escape velocity at a given height is ${\displaystyle \sqrt{2}}$ 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.

For a body in an elliptical orbit wishing to accelerate to an escape orbit the required speed will vary, and will be greatest at periapsis 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 the Oberth effect.

Barycentric escape velocity

Escape velocity can either be measured as relative to the other, central body or relative to center of mass or barycenter of the system of bodies. Thus for systems of two bodies, the term escape 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 mass test particles. For zero mass test particles we have that the 'relative to the other' and the 'barycentric' escape velocities are the same, namely ${\displaystyle v_{\text{e}}=\sqrt{\frac{2GM}{d}}}$.

But when we can't neglect the smaller mass (say m) we arrive at slightly different formulas.

Because the system has to obey the law 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 (v_p, for planet) can be expressed in terms of the velocity of the smaller mass (v_r, for rocket). We get ${\displaystyle v_p=-\frac{m}{M}{v}_r}$.

The 'barycentric' escape velocity now becomes ${\displaystyle v_r=\sqrt{\frac{2G{M}^2}{d(M+m)}}\approx \sqrt{\frac{2GM}{d}}}$, while the 'relative to the other' escape velocity becomes ${\displaystyle v_r-v_p=\sqrt{\frac{2G(m+M)}{d}}\approx \sqrt{\frac{2GM}{d}}}$.

Height of lower-velocity trajectories

Ignoring all factors other than the gravitational force between the body and the object, an object projected vertically at speed v from the surface of a spherical body with escape velocity v_e and radius R will attain a maximum height h satisfying the equation

${\displaystyle v=v_{\text{e}}\sqrt{\frac{h}{R+h}},}$

which, solving for h results in

${\displaystyle h=\frac{x^2}{1-x^2}R,}$

where x = v/v_e is the ratio of the original speed v to the escape velocity v_e.

Unlike escape velocity, the direction (vertically up) is important to achieve maximum height.

Trajectory

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 length, 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:

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

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:

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

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

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

List of escape velocities

In this table, the left-hand half gives the escape velocity from the visible surface of the planet or moon (which may be gaseous, for example Jupiter) relative to its centre (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). Moons are indented below their planet. The values in the last column are variable, depending on precisely where in a body’s orbit escape velocity is reached, as the orbits are elliptical, not circular (particularly those of Mercury and Pluto).

Deriving escape velocity using calculus

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

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

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

${\displaystyle dW=Fdr=G\frac{Mm}{r^2}dr.}$

The total work needed to move the body from the surface r₀ of the gravitating body to infinity is then

${\displaystyle W={\int }_{r_0}^{\mathrm{\infty }}G\frac{Mm}{r^2}dr=G\frac{Mm}{r_0}=mg{r}_0.}$

In order to do this work to reach infinity, the body's minimal kinetic energy at departure must match this work, so the escape velocity v₀ satisfies

${\displaystyle \frac{1}{2}m{v_0}^2=G\frac{Mm}{r_0},}$

which results in

${\displaystyle v_0=\sqrt{\frac{2GM}{r_0}}=\sqrt{2g{r}_0}.}$