Man falls to earth at speed of sound?, page 2
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reply posted on 6-2-2006 @ 09:28 AM by Beachcoma
From Wikipedia:

Terminal Velocity

The terminal velocity of an object falling towards the ground, in non-vacuum, is the speed at which the gravitational force pulling it downwards is equal and opposite to the atmospheric drag (also called air resistance) pushing it upwards. At this speed, the object ceases to accelerate downwards and falls at constant speed.

For example, the terminal velocity of a skydiver in a normal free-fall position with a closed parachute is about 195 km/h (120 Mph). This speed increases to about 320 km/h (200 Mph) if the skydiver pulls in his limbs—see also freeflying. This is also the terminal velocity of the Peregrine Falcon diving down on its prey, and of a typical bullet according to a 1920 U.S. Army Ordnance study.



reply posted on 6-2-2006 @ 12:11 PM by Beachcoma
Very very far indeed. The thing is you will still feel Earth's gravity even at a distance. The moon for instance stays where it is because of Earth's gravity. If you want to escape the Earth's gravity you need to reach escape velocity.

From Wikipedia:

Escape Velocity

gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion, at that position, 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.


The article continues to say:

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.

For a body rotating about its axis the escape velocity with respect to the surface does depend on direction e.g., for the Earth the rotational velocity is 465 m/s to the east at the equator, and the escape velocity to the east, with respect to the Earth's surface, is ca. 10.7 km/s.


The rest of the article is somewhat technical for my understanding, but I believe that the further away you are from the source, the less speed it takes to break-away.
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