Escape velocity
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- Ballistic trajectory - no other forces are acting on the object, including propulsion and friction
- No other gravity-producing objects exist
Although the term escape velocity is common, it is more accurately described as a
Escape speed varies with distance from the center of the primary body, as does the velocity of an object traveling under the gravitational influence of the primary. If an object is in a circular or elliptical orbit, its speed is always less than the escape speed at its current distance. In contrast if it is on a
Escape velocity calculations are typically used to determine whether an object will remain in the
Precise trajectory calculations require taking into account small forces like
Calculation
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:
- G is the universal gravitational constant (G ≈ 6.67×10−11 m3·kg−1·s−2)
- g = GM/d2 is the local gravitational acceleration (or the surface gravity, when d = r).
The value GM is called the standard gravitational parameter, or μ, and is often known more accurately than either G or M separately.
When given an initial speed greater than the escape speed the object will asymptotically approach the hyperbolic excess speed satisfying the equation:[4]
Earth
For example, at the Earth's surface, the surface gravity is about 9.8 m/s2 (9.8 N/kg, 32 ft/s2), and the escape speed for a small object is about 11.186 km/s (40,270 km/h; 25,020 mph; 36,700 ft/s).[5] This 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). At 9,000 km altitude, escape speed is slightly less than 7.1 km/s. These velocities are relative to a non-rotating frame of reference; launching near the equator rather than the poles can actually provide a boost.
In this context, when taking Earth as the primary body, escape velocity is sometimes called "second cosmic velocity"[6]
Energy required
For an object of mass the energy required to escape the Earth's gravitational field is GMm / r, a function of the object's mass (where r is radius of the Earth, nominally 6,371 kilometres (3,959 mi), 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.
Conservation of energy
The existence of escape velocity can be thought of as 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; places which have a higher potential energy than this cannot be reached at all. Adding speed (kinetic energy) to an object expands the region of locations it can reach, until, with enough energy, everywhere to infinity becomes accessible.
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. 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, . At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small.
We can set Kfinal = 0 because final velocity is arbitrarily small, and Ug final = 0 because final gravitational potential energy is defined to be zero a long distance away from a planet, so
Relativistic
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.[8][9]
Scenarios
From the surface of a body
An alternative expression for the escape velocity 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).[10]
For a body with a spherically symmetric distribution of mass, the escape velocity from the surface is proportional to the radius assuming constant density, and proportional to the square root of the average density ρ.
where
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
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
From an orbiting body
The escape velocity at a given height is 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.[11]
For a body in an elliptical orbit wishing to accelerate to an escape orbit the required speed will vary, and will be greatest at
Barycentric escape velocity
Escape velocity can either be measured as relative to the other, central body or relative to
But when we can't neglect the smaller mass (say ) 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 ( , for planet) can be expressed in terms of the velocity of the smaller mass (, for rocket). We get .
The 'barycentric' escape velocity now becomes : while the 'relative to the other' escape velocity becomes : .
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 from the surface of a spherical body with escape velocity and radius will attain a maximum height satisfying the equation[12]
which, solving for h results in
where is the ratio of the original speed to the escape velocity
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
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:
If a body in circular orbit (or at the
If the speed at periapsis is v, then the eccentricity of the trajectory is given by:
This is valid for elliptical, parabolic, and hyperbolic trajectories. If the trajectory is hyperbolic or parabolic, it will
The speed will asymptotically approach
List of escape velocities
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, Ve refers to the speed relative to the central body (for example the sun), whereas Vte is the speed (at the visible surface of the smaller body) relative to the smaller body (planet or moon).
Location | Relative to | Ve (km/s)[13] | Location | Relative to | Ve (km/s)[13] | System escape, Vte (km/s) | |
---|---|---|---|---|---|---|---|
On the Sun | The Sun's gravity | 617.5 | |||||
On Mercury | Mercury's gravity | 4.25 | At Mercury | The Sun's gravity | ~ 67.7 | ~ 20.3 | |
On Venus | Venus's gravity | 10.36 | At Venus | The Sun's gravity | 49.5 | 17.8 | |
On Earth | Earth's gravity | 11.186 | At Earth | The Sun's gravity | 42.1 | 16.6 | |
On the Moon | The Moon's gravity | 2.38 | At the Moon | Earth's gravity | 1.4 | 2.42 | |
On Mars | Mars' gravity | 5.03 | At Mars | The Sun's gravity | 34.1 | 11.2 | |
On Ceres | Ceres's gravity | 0.51 | At Ceres | The Sun's gravity | 25.3 | 7.4 | |
On Jupiter | Jupiter's gravity | 60.20 | At Jupiter | The Sun's gravity | 18.5 | 60.4 | |
On Io | Io's gravity | 2.558 | At Io | Jupiter's gravity | 24.5 | 7.6 | |
On Europa | Europa's gravity | 2.025 | At Europa | Jupiter's gravity | 19.4 | 6.0 | |
On Ganymede | Ganymede's gravity | 2.741 | At Ganymede | Jupiter's gravity | 15.4 | 5.3 | |
On Callisto | Callisto's gravity | 2.440 | At Callisto | Jupiter's gravity | 11.6 | 4.2 | |
On Saturn | Saturn's gravity | 36.09 | At Saturn | The Sun's gravity | 13.6 | 36.3 | |
On Titan | Titan's gravity | 2.639 | At Titan | Saturn's gravity | 7.8 | 3.5 | |
On Uranus | Uranus' gravity | 21.38 | At Uranus | The Sun's gravity | 9.6 | 21.5 | |
On Neptune | Neptune's gravity | 23.56 | At Neptune | The Sun's gravity | 7.7 | 23.7 | |
On Triton | Triton's gravity | 1.455 | At Triton | Neptune's gravity | 6.2 | 2.33 | |
On Pluto | Pluto's gravity | 1.23 | At Pluto | The Sun's gravity | ~ 6.6 | ~ 2.3 | |
200 AU from the Sun | The Sun's gravity | 2.98[14] | |||||
1774 AU from the Sun | The Sun's gravity | 1[14] | |||||
At Solar System galactic radius | The Milky Way's gravity | 492–594[15][16] | |||||
On the event horizon | A black hole's gravity | 299,792.458 (speed of light) |
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).
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
The work needed to move the body over a small distance dr against this force is therefore given by
The total work needed to move the body from the surface r0 of the gravitating body to infinity is then[17]
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
which results in
See also
- Black hole – an object with an escape velocity greater than the speed of light
- Characteristic energy (C3)
- Delta-v budget – speed needed to perform maneuvers.
- Gravitational slingshot – a technique for changing trajectory
- Gravity well
- List of artificial objects in heliocentric orbit
- List of artificial objects leaving the Solar System
- Newton's cannonball
- Oberth effect – burning propellant deep in a gravity field gives higher change in kinetic energy
- Two-body problem
Notes
- ^ Gravitational potential energy is defined to be zero at an infinite distance.
References
- ISBN 978-0-13-149508-1.
- ISBN 9789937903844.)
{{cite book}}
: CS1 maint: multiple names: authors list (link - ISBN 978-0-486-60061-1.
- ISBN 978-1-4008-3909-4.
- ^ "NASA – NSSDC – Spacecraft – Details". Archived from the original on 2 June 2019. Retrieved 21 August 2019.
- ISBN 978-0-19-966646-1.
- ^ Bate, Mueller and White, p. 35
- ^ a b For planets: "Planets and Pluto : Physical Characteristics". NASA. Retrieved 18 January 2017.
- ^ a b "To the Voyagers and escaping from the Sun". Initiative for Interstellar Studies. 25 February 2015. Retrieved 3 February 2023.
- S2CID 125255461.
- S2CID 119040135.