What is the escape velocity of Earth?

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Escape velocity from Earth is approximately 11.186 km/s, which refers to the speed needed for a projectile to break free from Earth's gravitational pull. This speed is independent of the direction of launch, meaning a projectile can be fired in any direction as long as it reaches this speed. If launched from a height of 1000 meters and at this speed, the projectile will escape Earth's gravity without crashing back. The trajectory must not intersect with Earth's surface to avoid failure. Thus, even a slight downward angle would still allow the projectile to escape successfully.
Stephanus
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What is escape velocity?
According to Wikipedia
http://en.wikipedia.org/wiki/Earth
Earth:
Escape velocity: 11.186km/s?
What does that mean?
Does the projectile should be fired perpendicular with respect to the ground angle?

According to this:
http://en.wikipedia.org/wiki/Escape_velocity
The term escape velocity is a misnomer, and it is referred to as escape speed since it is independent of direction.
According to this, can the projectile be fired in any direction (because it is speed not velocity) as long as its speed is 11.186 Km/s to escape earth?If (supposed Earth has no atmosphere) the projectile is fired parallely to the ground (tangent), what is its escape velocity?

Thanks for any explanation.
 
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The wikipedia article has it right. As long as the trajectory doesn't intersect the surface of the Earth and go "boom!", if it is launched at or above escape velocity the projectile will escape no matter what direction it is aimed. Escape velocity tangent to the surface of an airless non-rotating body is the same as escape velocity vertical to the surface.

There are many ways of showing this, but the easiest is to look at the sum of the projectile's potential and kinetic energy, which must remain constant throughout its trajectory.
 
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Nugatory said:
...as long as the trajectory doesn't intersect the surface of the Earth and go "boom...".
Yes, yes, I understand now. So if the projectile is placed, say..., 1000m above ground (earth).
And it's fired horizontally and 10 to the ground, supposed the Earth has no atmosphere and no mountain (and skyscraper for that matter), I don't know if the math is correct for 10; 1000 m above the ground; and Earth radius. After all Earth is bolate not sphere.
Just supposed in that height and direction the projectile is fired 11.186 km/s, and doesn't go boom, it will travel to space, Right?
 
Stephanus said:
Yes, yes, I understand now. So if the projectile is placed, say..., 1000m above ground (earth).
And it's fired horizontally and 10 to the ground, supposed the Earth has no atmosphere and no mountain (and skyscraper for that matter), I don't know if the math is correct for 10; 1000 m above the ground; and Earth radius. After all Earth is bolate not sphere.
Just supposed in that height and direction the projectile is fired 11.186 km/s, and doesn't go boom, it will travel to space, Right?

Yes. In fact, you could aim the thing slightly downwards so that it just barely missed grazing the surface of the earth, and it would still escape.

All that will change is the direction that it's traveling as it heads away from earth.
 
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