From the surface of a body An alternative expression for the escape velocity
v particularly useful at the surface on the body is: : 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 from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average density
ρ. : v_\text{e} = Kr\sqrt\rho where K = \sqrt{\frac{8}{3} \pi G} \approx \mathrm{2.364\times 10^{-5} ~m^{1.5}{\cdot}kg^{-0.5}{\cdot}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 an 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 \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 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, for planet) can be expressed in terms of the velocity of the smaller mass (
v, for rocket). We get v_p=-\frac{m}{M}v_r. The 'barycentric' escape velocity now becomes v_r=\sqrt{\frac{2GM^2}{d(M+m)}} \approx \sqrt{\frac{2GM}{d}}, while the 'relative to the other' escape velocity becomes 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 and radius
R will attain a maximum height
h satisfying the equation : v = v_\text{e} \sqrt{\frac{h}{R+h}} \ , which, solving for
h results in : h = \frac{x^2}{1-x^2} \ R \ , where is the ratio of the original speed
v to the escape velocity
v. Unlike escape velocity, the direction (vertically up) is important to achieve maximum height. == Trajectory ==