Re: Standing at the sea level at the equator on June 21 how far would I travel

Interesting question. To figure out the answer, we need to know 5 different velocities (see any astronomy textbook for more info about how we get these numbers):

1. Linear speed at the equator from Earth rotation. The earth rotates at an angular frequency of 2pi radians/day, or 7.272x10^-5 rad/s. At the average Earth radius (6.371x10^6 m) this is 463 m/s.

2. Orbital speed of Earth around the Sun. Using v=(GM(sun)/R(ES))^(1/2) we find this to be 29.9 km/s.

3. Speed of the Sun in its orbit around the center of the galaxy. This is harder to calculate but has been measured to be about 220 km/s.

4. Speed of the galaxy around the center of the Local Group of galaxies; I couldn't find a value for this number but I know that it added to the velocity in (3) gives 300 km/s.

5. Speed of the Local group with respect to the cosmic background radiation - measured to be about 600 km/s (see this answer for more detail.

Adding the last 3 effects together (they're not all in the same direction so we have to add the vectors) we find that the velocity of the Sun with respect to the cosmic background radiation is 370 km/s. The rotation of the Earth and motion of the Earth around the Sun are pretty small in comparison to this, so ``to astronomical accuracy'' we could say that you moved 370 km in one second without taking a step.

Don't forget your seat belt..