feed me

Posted: November 14, 2010 in Uncategorized
v_e = \sqrt{\frac{2GM}{r}}

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, with all speeds and velocities measured with respect to the field. Additionally, the escape velocity at a point in space is equal to the speed that an object would have if it started at rest from an infinite distance and was pulled by gravity to that point. 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 kilometers per second (~6.96 mi/s), which is approximately 34 times the speed of sound (Mach 34) and at least 10 times the speed of a rifle bullet. 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 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 (latitude 28°28′ N) and the French Guiana Space Centre (latitude 5°14′ N).

The barycentric escape velocity is independent of the mass of the escaping object. It does not matter if the mass is 1 kg or 1,000 kg, what differs is the amount of energy required. For an object of mass m the energy required to escape the Earth’s gravitational field is GMm / r, a function of the object’s mass (where r is the radius of the Earth, G is the gravitational constant, and M is the mass of the Earth).

For a mass equal to a Saturn V rocket, the escape velocity relative to the launch pad is 8 nanometers per year faster than the escape velocity relative to the mutual center of mass. When the mass reaches the Andromeda galaxy the earth will have recoiled a third of a mile.


Planetary or lunar escape velocity is sometimes misunderstood to be the speed a powered vehicle (such as a rocket) must reach to leave orbit; however, this is not the case, as the quoted number is typically the escape velocity at the body’s surface, and vehicles need never achieve that speed. This barycentric escape velocity is the speed required for an object to leave the planet if the object is simply projected from the surface of the planet and then left without any more kinetic energy input: in practice the vehicle’s propulsion system will continue to provide energy after it has left the surface.

In fact a vehicle can leave the Earth’s gravity at any speed less than the speed of light. At higher altitudes, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at that position. As is obvious from the equation, at sufficiently high altitudes this speed approaches 0 as r becomes large.


If an object attains escape velocity, but is not directed straight away from the planet, then it will follow a curved path. Although this path does not form a closed shape, it is still considered an orbit. Assuming that gravity is the only significant force in the system, this object’s speed at any point in the orbit 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 orbit will be a parabola whose focus is located at the center of mass of the planet. An actual escape requires of course that the orbit not intersect the planet nor its atmosphere, since this would cause the object to crash. When moving away from the source, this path is called an escape orbit. Escape orbits are called known as C3 = 0 orbits (where C3 = – μ/a, and a is the semi-major axis).

In reality there are many gravitating bodies in space, so that, for instance, a rocket that travels at escape velocity from Earth will not escape to an infinite distance away because it needs an even higher speed to escape the Sun‘s gravity. In other words, near the Earth, the rocket’s orbit will appear parabolic, but eventually its orbit will become an ellipse around the Sun, except when it is perturbed by the Earth whose orbit it must still intersect.

Dynamic situations

The escape velocity calculation assumes that none of the sources of gravitational potential are moving. In the case of the Earth-Moon system for example the Moon is in orbit around the Earth, and it turns out that an object that passes behind the moon can gain speed from the encounter (a gravitational assist).

Therefore there are certain launch angles and launch windows that will give the assist and this will give an object extra speed in-flight and still permit it to escape Earth’s gravity even though it was launched with less than escape velocity.

Gravity well

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 \scriptstyle\sqrt 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 \scriptstyle\sqrt {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.

A black hole is a region of space from which nothing, not even light, can escape. It is the result of the deformation of spacetime caused by a very compact mass. Around a black hole there is an undetectable surface which marks the point of no return, called an event horizon. It is called “black” because it absorbs all the light that hits it, reflecting nothing, just like a perfect black body in thermodynamics.[1] Quantum mechanics predicts that black holes also emit radiation like a black body with a finite temperature. This temperature decreases with the mass of the black hole, making it unlikely to observe this radiation for black holes of stellar mass.

Whoa ha ha here we we, black hole is just a space that has a greater escape velocity then the speed of light, nothin comes back…..but where does it go ?



Simulated view of a black hole in front of the
Large Magellanic Cloud. The ratio between the black hole Schwarzschild radius and the observer distance to it is 1:9. Of note is the gravitational lensing effect known as an Einstein ring, which produces a set of two fairly bright and large but highly distorted images of the Cloud as compared to its actual angular size.

Despite its invisible interior, a black hole can be observed through its interaction with other matter. A black hole can be inferred by tracking the movement of a group of stars that orbit a region in space. Alternatively, when gas falls into a stellar black hole from a companion star, the gas spirals inward, heating to very high temperatures and emitting large amounts of radiation that can be detected from earthbound and Earth-orbiting telescopes.

Astronomers have identified numerous stellar black hole candidates, and have also found evidence of supermassive black holes at the center of galaxies. In 1998, astronomers found compelling evidence that a supermassive black hole of more than 2 million solar masses is located near the Sagittarius A* region in the center of the Milky Way galaxy, and more recent results using additional data find evidence that the supermassive black hole is more than 4 million solar masses.

Gobble Gobble.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s