Escape velocity
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- For other senses of this term, see escape velocity (disambiguation).
One somewhat counterintuitive feature of escape velocity is that it is independent of direction, so that "velocity" is a misnomer; it is a scalar quantity and would more accurately be called "escape speed". The simplest way of deriving the formula for escape velocity is to use conservation of energy, thus: in order to escape, an object must have at least as much kinetic energy as the increase of potential energy required to move to infinite height.
Defined a bit more formally, "escape velocity" is the initial speed required to go from an initial point in a gravitational potential field to infinity with a residual velocity of zero, relative to the field. Conversely, an object starting at rest and at infinity, dropping towards the attracting mass, would reach its surface moving at the escape velocity. In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth the escape velocity is about 11.2 kilometres per second. However, at 9000 km altitude in "space", it is slightly less than 7.1 km/s.
The escape velocity from 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 to the east 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 earth 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 earth. 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 in Florida and the European Centre Spatial Guyanais, only 5 degrees from the equator in Guyana.
List of escape velocities
Location with respect to Ve Location with respect to Ve on the Sun, the Sun's gravity: 617.5 km/s on Mercury, Mercury's gravity: 4.4 km/s at Mercury, the Sun's gravity: 67.7 km/s on Venus, Venus' gravity: 10.4 km/s at Venus, the Sun's gravity: 49.5 km/s at the Earth, the Earth's gravity: 11.2 km/s at the Earth/Moon, the Sun's gravity: 42.1 km/s on the Moon, the Moon's gravity: 2.4 km/s at the Moon, the Earth's gravity: 1.4 km/s on Mars, Mars' gravity: 5.0 km/s at Mars, the Sun's gravity: 34.1 km/s on Jupiter, Jupiter's gravity: 59.5 km/s at Jupiter, the Sun's gravity: 18.5 km/s on Saturn, Saturn's gravity: 35.5 km/s at Saturn, the Sun's gravity: 13.6 km/s on Uranus, Uranus' gravity: 21.3 km/s at Uranus, the Sun's gravity: 9.6 km/s on Neptune, Neptune's gravity: 23.5 km/s at Neptune, the Sun's gravity: 7.7 km/s on Pluto, Pluto's gravity: 1.3 km/s at Pluto, the Sun's gravity: 6.7 km/s at the solar system, the Milky Way's gravity: ~1000 km/s
Due to the atmosphere it is not useful and hardly possible to give an object near the surface of the Earth a speed of 11.2 km/s, as these speeds are too far in the hypersonic regime for most practical propulsion systems and would cause most objects to burn up due to atmospheric friction. For an actual escape orbit a spacecraft is first placed in low Earth orbit and then accelerated to the escape velocity at that altitude, which is a little less, ca. 10.9 km/s. The required extra velocity, however, is less because the spacecraft has already been accelerated to about 8 km/s.
Calculating an escape velocity
In the simple case of the escape velocity from a single body, it can be calculated by setting the kinetic energy equal to minus the gravitational potential energy. This is because the positive kinetic energy is needed to increase the negative gravitational potential energy to zero, which applies when the object is at an infinite distance.
- [ \begin\frac12\end mv_e^2=\frac]
- [v_e = \sqrt} = \sqrt}]
The escape velocity at a given height is √2 times the speed in a circular orbit at the same height, compare (14) in circular motion. 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.
For a body with a spherically-symmetric distribution of mass, the escape velocity [v_e] from the surface (in m/s) is approximately 2.364×10−5 m1.5kg−0.5s−1 times the radius r (in metres) times the square root of the average density ρ (in kg/m3), or:
- [v_e \approx 2.364 \times 10^ r \sqrt \rho.\,]
Escape velocity given \"g\" and \"r\"
The Earth's escape speed can be derived from "g", the acceleration at the Earth's surface with calculus. It is not necessary to know the gravitational constant G or the mass M of the Earth. Let
- r = the Earth's radius, and
- g = the acceleration of gravity at the Earth's surface.
- [\int_r^\infty gm (r/s)^2 \, ds=gmr^2 \int_r^\infty s^\,ds=gmr^2 \left[-s^right]_^]
- [=gmr^2\left(0-(-r^)\right)=gmr.]
- [ \begin\frac12\end mv^2=gmr.]
- [v=\sqrt.]
- [v\cong\sqrt/\mathrm^2}\right)(6.4\times 10^6\ \mathrm)}= 11\,200\ \mathrm/\mathrm.]
Multiple sources
The escape velocity from a position in a field with multiple sources is derived from the total potential energy per kg at that position, relative to infinity. The potential energies for all sources can simply be added. For the escape velocity this results in the square root of the sum of the squares of the escape velocities of all sources separately.
For example, at the Earth's surface the escape velocity for the combination Earth and Sun is √(11.2² + 42.1²) = 43.6 km/s. As a result, to leave the solar system requires a speed of 13.6 km/s relative to Earth in the direction of the Earth's orbital motion, since the speed is then added to the speed of 30 km/s of that orbital motion
In the hypothetical case of uniform density, the velocity that an object would achieve when dropped in a hypothetical vacuum hole from the surface of the Earth to the center of the Earth is the escape velocity divided by √2, i.e. the speed in a circular orbit at a low height. Correspondingly, the escape velocity from the center of the Earth would be √1.5 times that from the surface.
A refined calculation would take into account the fact that the Earth's mass is not uniformly distributed as the center is approached. This gives higher speeds.
See also Gravitational potential energy.
Orbit
If a freefalling body at any position has the escape velocity for that position, this is the case for the whole orbit. If the gravity source is a spherically symmetric body the orbit is (part of) a parabola with the center of the source as focus (parabolic trajectory), or part of a straight line through the source. When moving away from the source it is called an escape orbit, otherwise a capture orbit. Both are also known as C3 = 0 orbit.
An actual escape requires that the parabolic orbit does not intersect the celestial body. More generally, for an arbitrarily shaped body it requires that the orbit does not intersect the body. For non-convex bodies, not each point on the surface needs to be a possible starting point of such an orbit.
If the body has the escape velocity with respect to the Earth, this is not enough to escape the Solar System, so near the Earth the orbit resembles a parabola, but further away it bends into an elliptical orbit around the Sun.
See also
- Delta-v budget
- Gravitational slingshot
- Gravity well
- Two body problem.
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All text is available under the terms of the GNU Free Documentation License See Wikipedia Copyrights for details.
