Search
×

OR

Create a Shvoong account from scratch

×

OR

×

OR

Shvoong Home>Science>Escape Velocity Summary

# Escape Velocity

Book Summary   by:P03     Original Author: P03

Escape velocity

Space Shuttle Atlantis launches on mission STS-71. It has rocket boosters and needs only to get into low-earth orbit, so it does not need to reach escape velocity.
In physics, for a given gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion needs to have to move away indefinitely from the source of the field, as opposed to falling back or staying in an orbit within a bounded distance from the source. The object is assumed to be influenced by no forces except the gravitational field; in particular there is no propulsion (as by a rocket), there is no friction (as between the object and the Earth''s atmosphere; these conditions correspond to freefall), and there is no gravitational radiation. This definition may need modification for the practical problem of two or more sources in some cases. In any case, the object is assumed to be a point with a mass that is negligible compared with that of the source of the field, usually an excellent approximation. It is commonly described as the speed needed to "break free" from a gravitational field.
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 little 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, throughout this trajectory (until it reaches the surface), move at a speed equal to the escape velocity corresponding to its position. 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 (~6.96 mi/s). However, at 9,000 km altitude in "space," it is slightly less than 7.1 km/s.
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 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 Guiana Space Center, only 5 degrees from the equator in French Guiana.
To simply state this, all objects on Earth have the same escape velocity. It does not matter if the mass is 1 kg or 1000 kg, escape velocity is always the same. What differs is the amount of energy needed to accelerate the mass to escape velocity: the energy needed for an object of mass m to escape the Earth''s gravitational field is GMm / r0, a function of the object''s mass (where r0 is the radius of the Earth). More massive objects require more energy to reach escape velocity.

//

List of escape velocities

To leave planet Earth an escape velocity of 11.2 km/s is required, however a speed of 42.1 km/s is required to escape the Sun''s gravity (and exit the solar system) from the same position

Location
with respect to
Ve

Locationrespect 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

on 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

at the solar system,
the Milky Way''s gravity:
~1000 km/s<1>
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 acceleration, however, is generally even less because from that sort of an orbit the spacecraft already has a speed of 8 km/s.

Calculating an escape velocity
In the simple case of escape from a single body, the escape velocity is that corresponding to the kinetic energy equivalent 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.

where ve is the escape velocity, G is the gravitational constant, M is the mass of the body being escaped from, m is the mass of the escaping body (factors out), and r is the distance between the center of the body and the point at which escape velocity is being calculated, and μ is the standard gravitational parameter.<2>
The escape velocity at a given height is 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 ve from the surface (in m/s) is approximately 2.364×10−5 m1.5kg−0.5s−1 times the radius r (in meters) times the square root of the average density ρ (in kg/m3), or:

Derivation using only g and r
the Earth''s escape speed can be derived from "g," the acceleration due to gravity at the Earth''s surface. 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.
Above the Earth''s surface, the acceleration of gravity is governed by Newton''s inverse-square law of universal gravitation. Accordingly, the acceleration of gravity at height s above the center of the Earth (where s > r ) is g(r / s)2. The weight of an object of mass m at the surface is g m, and its weight at height s above the center of the Earth is gm (r / s)2. Consequently the energy needed to lift an object of mass m from height s above the Earth''s center to height s +
Published: April 08, 2007
 Please Rate this Summary : 1 2 3 4 5
Tags:
 Use our Content Translate Send Link Print Share

More

X

.

•