|Buffa, Anthony J and Wilson, Jerry D. College Physics, Fourth Edition. Upper Saddle River: Prentice Hall, 2000.||"Thus if a projectile or spacecraft could be given an initial upward speed of 11 km/s, it would leave the earth and not return."||11.0 km/s|
|Mitton, Jacqueline. Cambridge Dictionary of Astronomy. New York: Cambridge University Press, 2001: 143.||"The velocity of escape from the Earth's surface is about 11.2 km/s."||11.2 km/s|
|Angelo, Joseph A. Jr. The Facts on File Space and Astronomy Handbook. New York: Facts on File, 2002: 269.||"Earth 11.2 km/s"||11.2 km/s|
|Atkins, P.W. The Second Law. New York: Scientific American, 1984: 138.||"The escape velocity is the speed an object must be given to escape from the Earth; it is 11.3 km/sec or 25,300 mph."||11.3 km/s|
|Jones, A. Innovations in Astronomy. Santa Barbara: Helicon Publishing, 1999: 192.||"In the case of the earth, the escape velocity is 11.2kps/6.9mps; the moon, 2.4kps/1.5mps; Mars, 5kps/3.1mps; and Jupiter, 59.6kps/37mps."||11.2 km/s|
Space travel is something we may now take for granted, but there are many factors we first have to look into. One thing we have to do is overcome the force of gravity holding us down.
We must be going at a very high speed, but how high? The escape velocity, as the minimum velocity that will allow a small body to escape from another body, can be calculated using the formula v = sqrt(2Gm/r), where G is the gravitational constant, r is the distance from the center of the body with a mass of m. This formula is derived using the idea of conservation of energy, where the sum of the kinetic energy and the gravitational energy of an object equal the energy at the maximum altitude.
(+K) + (−U) [Earth] = K + U = 0 [at Infinity]
K = −U
½mv2 = GMm/r
v = sqrt(2GM/R)
For the earth, this velocity is 11.2 kilometers per second or 25,950 miles per hour.
Leo Tam -- 2005
|Physics Textbook, Chapter 13 - Gravity and Orbits. Digital Physics Textbook version 0.7.2. Kinetic Books. 20 May 2005.||"What is the minimum speed required to escape the Earth's gravity? v = sqrt2GM/R v = 11,200 m/s"||11.2 km/s|
|Brown, Walt. References and Notes. In the Beginning: Compelling Evidence for Creation and the Flood (7th Edition). Center for Scientific Creation. May 18, 2005.||"For Earth, from its surface, that speed is 11.2 km/sec (7.0 mi/sec). For something at the surface of the Sun to escape the solar system, it is 617.2 km/sec (385.7 mi/sec). For something 1 AU from the Sun to escape the solar system requires 42.3 km/sec (26.4 mi/sec)."||11.2 km/s|
|Cathart, Richard B. Verneshots and Interplanetary Lithopanspermia: A Lava Lite - like Earth life Lift-off Mechanism. Glendale, California. May 16, 2005.||"The first artificial object to escape Earth, according to Robert R. Brownlee’s calculations, was emplaced on 27 August 1957; it was a circular steel mineshaft cover blown into space by the Pascal-B underground nuclear device test explosion in Area U3d of the USA’s Nevada Test Site! By the early-1960s�. [Earth escape velocity = 11.12 km/sec.]"||11.12 km/s|
|What is Escape Velocity? PhysLink.com. 21 May 2005.||"From the surface of the Earth, escape velocity (ignoring air friction) is about 7 miles per second, or 25,000 miles per hour."||11.2 km/s|
Haven't we all wanted to escape the earth at one time or another in our lives? Well, to do that, you would have to go pretty quickly. The speed needed to escape the earth or "escape velocity" is about 11,200 meters per second or 7 miles per second. Think about when you throw something up into the air. It has to come back because of the earth's gravitational pull, right? Well, yes, unless you throw it fast enough, which is highly unlikely no matter how fast you think your "fastball" is. Most commonly, the things that escape the earth are rockets fired by space agencies to explore space. With enough speed, a rocket can escape even the sun's gravity and go outside the solar system.
The first artificial object to escape the earths gravity was in August of 1957. The object was a mineshaft cover sent into space by a nuclear device test explosion. It was sent out in Nevada. The earths rotation has a lot to do with the escape speed. This is why Nevada was a likely test cite rather than the North or South poles. When the object is closer to the equator, about 436 meters per second is added to the velocity. It doesn't seem like much when you look at the total speed, but trust me, it helps.
The formula to find escape velocity is v = sqrt(2GM/r). M is the mass of the earth, G is the gravitational constant, r is the earths radius, and v is escape velocity. Not as much to it as you thought, now is there?
Mayya Kats -- 2005