The scientist can calculate the escape velocity of earth throughout the space.(For instance 30km per second)
But which reference surface used for this calculation?
In the space there is no thing which is motionless.Everything in motion.In the earth,our reference surface is ground for calculation speed of something.
I

This isn't escape velocity. 30km/s is how fast we are rotating around the sun, taking the sun as a stationary body.

Escape velocity is the velocity required to 'escape' a gravity well. Earth escape velocity at the equator this is approximately 11180 m/s. The reference is the earths surface, and the rotational speed of the earth is actually used to 'assist' vehicles. That is why launches are generally made to the east, the rotational speed of the earth is added to the velocity of the vehicle.

Escape velocity for the Earth from the Sun would be

v_e=\sqrt {\frac {2GM} r}

v_{eEarth}=\sqrt {\frac {2GM_{Sun}}{r_{EarthSun}}

v_{eEarth}=\sqrt {\frac {2*6.67*10^{-11}*1.99*10^{30}}{150*10^9}

v_{eEarth}=1.8*10^8kms^{-1}

This is faster than the speed of light.
Don't know if that helps you ;)

Capuchin - check you math! The answer is 4.2 x 10^4 m/s. Right equation, but the units are meters/s, and I think you may have forgotten to take the square root. Interesting to note that the earth's escape velocity from the sun is "only" about 50% greater than the its orbital velocity.

Yeah I missed the sq root :) Thanks for checking!