Escape velocity earth confusion

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Escape velocity on Earth is 11 km/s and is defined relative to the Earth itself, meaning it refers to the minimum speed needed for an object to break free from Earth's gravitational pull. While the Earth rotates at about 30 km/s at the equator, this rotation does not affect the escape velocity, but launching a rocket eastward can utilize this rotational speed to reduce fuel requirements. In the upper atmosphere, only a small fraction of molecules can reach or exceed escape velocity due to their random motion. The discussion clarifies that escape velocity is based on a non-rotating frame centered on Earth, and projectiles launched in the direction of Earth's rotation benefit from this initial speed. Understanding these concepts resolves the confusion surrounding escape velocity and atmospheric retention.
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If escape velocity on Earth is 11 km/s and velocity of Earth is 30 km/s how is that the atmosphere doesn't escape the pull of earth.
Is the escape velocity with reference to Earth ? I don't think so because when finding the formula we add kinetic and potential energy and then equate to zero. If it is not then doesn't everything on Earth has velocity greater than escape velocity...
Please help me out.
 
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The escape velocity on Earth is relative to the earth. In the outer fringes of the atmosphere, a very tiny percentage of molecules will randomly end up with >= escape velocity. Escape velocity from the solar system, is relative to the Sun (with some effect from orbiting planets). There are a few satellites that have exceeded escape velocity from the solar system. A list of them from wiki:

http://en.wikipedia.org/wiki/List_of_artificial_objects_escaping_from_the_Solar_System
 
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If the escape velocity is relative to Earth then why is it that( i read in my book) that projectiles are projected in the direction of rotation of Earth ?
 
AakashPandita said:
If the escape velocity is relative to Earth then why is it that( i read in my book) that projectiles are projected in the direction of rotation of Earth ?
It's relative to the Earth centered non-rotating frame.
 
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AakashPandita said:
If escape velocity on Earth is 11 km/s and velocity of Earth is 30 km/s
At the equator, the surface speed of the Earth relative to it's center of mass is about 0.464 km/s. As posted by A.T., the escape velocity from the Earth is not affected by the Earth rotation, but by launching a rocket to the east at the equator the rocket would start with 0.464 km/s speed relative to the center of mass of the earth, reducing the fuel it would need to reach orbital or escape velocity speeds.
 
Thanks. It makes sense to me now,
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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