Earths escape velocity is 11.1km/s
That 11.1 km/s is measured from the surface of Earth.
This is the velocity required such that the acceleration due to gravity is unable to bring it back to the planet, i.e. as it travels "up" and slows down, it will continue moving, with the velocity tending to 0 as the distance tends to infinity.
This means close to Earth, the velocity of the object will decrease significantly as it rises. As it gets further away, due to the signifantly reduced gravitational attraction, its velocity will decrease much less in a given period of time.
This also means that if you start from further away, the escape velocity is less, as it corresponds to the speed the object would be going at if launched from the surface at the 11.1 km/s, and slowed down on its way.
Escape velocity can be determined with the simple formula:
v=sqrt(2GM/r)
e.g. with G=6.67E-11 kg m/s^2, Earth with a mass of 5.97E+24 kg and r=6371000 m (6371 km), we end up with v=11186.14 m.s
However, if you move to r=100 000 km, you end up with v=2823 m/s, i.e. 2.8 km/s
For v=5.4 km/s, it would need to be 27 000 km from the centre of Earth.
So it seems fine.
Editted in Fun Fact: to escape the solar system from the distance of Earth, ignoring the gravitational potential of Earth, you need a velocity of roughly 42 km/s
To do so from the surface of the sun you need an escape velocity of over 600 km/s.