Editing Escape velocity (section)

== Scenarios ==

=== From the surface of a body ===
An alternative expression for the escape velocity ''v''{{sub|e}} particularly useful at the surface on the body is:
: <math>v_\text{e} = \sqrt{2gr}</math>
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]]).<ref>Bate, Mueller and White, p. 35</ref>

For a body with a spherically symmetric distribution of mass, the escape velocity ''v''{{sub|e}} from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average density ''ρ''.
: <math>v_\text{e} = Kr\sqrt\rho</math>
where <math display="inline">K = \sqrt{\frac{8}{3} \pi G} \approx \mathrm{2.364\times 10^{-5} ~m^{1.5}{\cdot}kg^{-0.5}{\cdot}s^{-1}} </math>.

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&nbsp;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&nbsp;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&nbsp;km/s ''relative to that moving surface''. The surface velocity decreases with the [[trigonometric function|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 Air Force Station|Cape Canaveral]] (latitude 28°28&prime;&nbsp;N) and the French [[Guiana Space Centre]] (latitude 5°14&prime;&nbsp;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&nbsp;km/s, or 40,320&nbsp;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&nbsp;km) and then accelerated to the escape velocity at that altitude, which will be slightly lower (about 11.0&nbsp;km/s at a low Earth orbit of 200&nbsp;km). The required additional [[delta-v|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&nbsp;km/s, or 28,080&nbsp;km/h).

=== From an orbiting body ===
The escape velocity at a given height is <math>\sqrt 2</math> 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'''.<ref>{{Cite book |last=Teodorescu |first=P. P. |url=https://books.google.com/books?id=k4H2AjWh9qQC&pg=PA580 |title=Mechanical systems, classical models |publisher=Springer, Japan |year=2007 |isbn=978-1-4020-5441-9 |page=580}}, [https://books.google.com/books?id=k4H2AjWh9qQC&pg=PA580 Section 2.2.2, p. 580]</ref><ref>{{cite book |title=Physics of Planetary Ionospheres |author1=S. J. Bauer |edition=illustrated |publisher=Springer Science & Business Media |year=2012 |isbn=978-3-642-65555-5 |page=28 |url=https://books.google.com/books?id=ssPyCAAAQBAJ}} [https://books.google.com/books?id=ssPyCAAAQBAJ&pg=PA28 Extract of page 28]</ref><ref>{{cite book |title=Classical Mechanics in Geophysical Fluid Dynamics |author1=Osamu Morita |edition=2nd, illustrated |publisher=CRC Press |year=2022 |isbn=978-1-000-80250-4 |page=195 |url=https://books.google.com/books?id=cqAIEQAAQBAJ}} [https://books.google.com/books?id=cqAIEQAAQBAJ&pg=PA195 Extract of page 195]</ref>

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|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 <math>v_\text{e} = \sqrt{\frac{2GM}{d}} </math>.

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

Because the system has to obey the [[Momentum#Conservation|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''{{sub|p}}, for planet) can be expressed in terms of the velocity of the smaller mass (''v''{{sub|r}}, for rocket). We get <math>v_p=-\frac{m}{M}v_r</math>.

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

=== 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''{{sub|e}} and radius ''R'' will attain a maximum height ''h'' satisfying the equation<ref>{{Cite book |last=Bajaj |first=N. K. |url=https://books.google.com/books?id=OLhyCgAAQBAJ |title=Complete Physics: JEE Main |publisher=[[McGraw-Hill Education]] |year=2015 |isbn=978-93-392-2032-7 |page=6.12}} [https://books.google.com/books?id=OLhyCgAAQBAJ&pg=SA6-PA12 Example 21, page 6.12]</ref>
: <math>v = v_\text{e} \sqrt{\frac{h}{R+h}} \ ,</math>
which, solving for ''h'' results in
: <math>h = \frac{x^2}{1-x^2} \ R \ ,</math>
where {{nowrap|1=''x'' = ''v''/''v''{{sub|e}}}} is the ratio of the original speed ''v'' to the escape velocity ''v''{{sub|e}}.

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