2007-02-15 23:09:19 · 17 answers · asked by muthu s 1 in Science & Mathematics ➔ Astronomy & Space
Escape velocity of the earth is 11.2 km/sec
2007-02-15 23:18:02 · answer #1 · answered by Ray 2 · 0⤊ 2⤋
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.
Escape velocity of Earth is 11.2 km/sec. 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 French Guiana.
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 , a function of the object's mass GMm/r0 (where r0 is the radius of the Earth). More massive objects require more energy to reach escape velocity.
2007-02-17 19:01:50 · answer #2 · answered by ♥ ΛDIƬΥΛ ♥ ııllllııllıı 6 · 0⤊ 1⤋
For Earth, vescape = 11.2 Ã 10 to the power of 3 m/s
OR 11 km/s
2007-02-16 16:23:58 · answer #3 · answered by neumor 2 · 0⤊ 0⤋
Escape velocity of the earth is 11.2km/s
2007-02-16 03:19:59 · answer #4 · answered by ♦Opty misstix♦ 7 · 0⤊ 0⤋
On the surface of the Earth, the escape velocity is about 11.2 kilometres per second (~6.96 mi/s).
2007-02-15 23:18:21 · answer #5 · answered by Rowdy Andy 4 · 0⤊ 0⤋
Well d escape vel of a body on d earth's surface is v=square root of ( 2GM/R) where M nd R are mass nd radius of earth resp. nd G is universal gravitational const.!!!nd it comes out 2 b 11.2km/sec! but f ur askin d escape vel 4 d earth to get out of d orbit of d sun den u got a gud ques!!!
2007-02-16 00:09:54 · answer #6 · answered by Prince 1 · 0⤊ 0⤋
About 7 miles/second, 5 miles/second to get into orbit.
2007-02-15 23:16:58 · answer #7 · answered by Iridflare 7 · 0⤊ 0⤋
my reference books say its 8 km/s not taking friction into account
2007-02-16 23:11:11 · answer #8 · answered by sushobhan 6 · 0⤊ 0⤋
escape velocity, v = (2GM/R)^1/2
v = 11.2 km/s
2007-02-15 23:15:57 · answer #9 · answered by li mei 3 · 2⤊ 0⤋
11.2 Km/sec
2007-02-15 23:17:02 · answer #10 · answered by Anonymous · 0⤊ 0⤋