Escape velocity
(It will slow down as it gets to greater distance, but do so asymptotically approaching a positive speed.)
An object on a parabolic trajectory will always be traveling exactly the escape speed at its current distance.
[1] Escape velocity calculations are typically used to determine whether an object will remain in the gravitational sphere of influence of a given body.
It is also useful to know how much a probe will need to slow down in order to be gravitationally captured by its destination body.
Rockets do not have to reach escape velocity in a single maneuver, and objects can also use a gravity assist to siphon kinetic energy away from large bodies.
Precise trajectory calculations require taking into account small forces like atmospheric drag, radiation pressure, and solar wind.
A rocket under continuous or intermittent thrust (or an object climbing a space elevator) can attain escape at any non-zero speed, but the minimum amount of energy required to do so is always the same.
Escape speed at a distance d from the center of a spherically symmetric primary body (such as a star or a planet) with mass M is given by the formula[2][3] where: The value GM is called the standard gravitational parameter, or μ, and is often known more accurately than either G or M separately.
An object has reached escape velocity when the specific orbital energy is greater than or equal to zero.
The formula for escape velocity can be derived from the principle of conservation of energy.
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.
At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small.
[9][10] An alternative expression for the escape velocity ve particularly useful at the surface on the body is: 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).
[11] For a body with a spherically symmetric distribution of mass, the escape velocity ve from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average density ρ. where
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.
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.
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.
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.
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 ve and radius R will attain a maximum height h satisfying the equation[15] which, solving for h results in where x = v/ve is the ratio of the original speed v to the escape velocity ve.
Unlike escape velocity, the direction (vertically up) is important to achieve maximum height.
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.
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.
C3 is the characteristic energy, −GM/2a, where a is the semi-major axis length, which is infinite for parabolic trajectories.
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.
from the direction at periapsis, with The speed will asymptotically approach 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).
