You described the ring as "orbital" and "at geosynchronous orbit"; therefore, it's in freefall. That's what "orbit" means- in freefall, but moving sideways so fast you miss the planet completely.
The fact that your orbital ring is attached to Earth simply means that if it moves relative to Earth, it'll break the tether and cause problems. The best way to avoid these problems is to put the ring in a circular orbit over the equator at geosynchronous height moving in the same direction that Earth spins- in other words, a geostationary orbit, which you've already done.
The formula for Newtonian gravitational acceleration is
$$ a_g = {GM \over r^2} $$
Plugging in Earth's mass and the radius of geostationary orbit, we get
$$ a_g = {\left(6.674 \times 10^{-11} {N m^2 \over kg^2} \right) \times \left( 5.972 \times 10^{24} kg \right) \over \left(42164000 m\right)^2} = 0.224 {m \over s^2}$$
So people on the ring will experience about 0.02 gees of acceleration from Earth's gravity (1 gee = 9.8 m/s^2), but the ring itself (being in freefall) will be accelerating in exactly the same way. As such, people on the ring will feel weightless- they'll float around in it, just like astronauts on the ISS (or any other spacecraft in orbit).
Now that I've said that, I will note that you've specifically characterized this space station as a "ring". This suggests that you'll be spinning it for artificial gravity. Just to get you started, the formula for centripetal acceleration is
$$ a_c = {v^2 \over r} = {v \over t} = {r \omega \over t} $$
where $a_c$ = centripetal acceleration (i.e. the "gravity" that people standing on the inside rim of the ring will feel), $r$ is the radius of the ring, $v$ is the linear speed of the edge of the ring, $\omega$ is the angular speed of the ring (in radians per second), and $t$ is the period of the ring's rotation (i.e. the time it takes to make one full revolution). You want $a_c$ to be equal to one gee (9.8 meters per second squared); if you know how big you want the ring to be, this will tell you how fast it should spin.
Also, if you spin the ring, you'll probably want it to be in the same plane as Earth's equator. A spinning ring is like a top or a gyroscope- its axis of rotation will want to point in the same direction all the time. If you mount the ring perpendicular to the tether, spin it up, and wait six hours, you'll find that the Earth (and the tether) will have rotated under it, while the ring itself has not. To keep it from crashing into the tether, you'd need to transfer all of its angular momentum somewhere else (no easy task, but doable if you have two identical rings stacked on top of each other, spinning in opposite directions- if the structure connecting them is sufficiently rigid, their angular momenta cancel out, and you can spin the whole assembly any way you want), or abandon the wheel-on-a-stick visual and mount the ring in the same plane as the tether and the Equator, like a unicycle.
Finally, the whole assembly will not have any noticeable effect on the Earth's orbit unless it's a significant fraction of the mass of the Moon- and the Moon is much bigger than most asteroids. Getting something that big into orbit will not be at all feasible any time in the foreseeable future, whether it's lifted up from Earth or dragged down from interplanetary space.
Edit: I should clarify that the above paragraphs were written under the assumption that your "orbital ring" was nothing more than a ring-shaped space station with some space elevators attached. The sort of thing that we could plausibly build once we figure out space elevators. However, I've been informed that this may not be the case, and that instead you may be going for a megastructure of mind-boggling scale, encircling the entire planet at geostationary altitude.
In the case of a megastructure at geosynchronous altitude spinning in time with the Earth, the answer is quite simple: The ring is in freefall, and therefore anything inside will float around like in any normal-sized space station without artificial gravity.
However, if you push the ring outward (by lengthening the tethers) without changing its angular speed (i.e. keeping it at 1 revolution per day), you'll start to get centrifugal gravity. Things inside the ring will "fall" toward its outside edge. We can compute how large the ring needs to be in order to get one gee of artificial gravity this way.
Here's the basic equation:
$$ a_c = a_g + g $$
where $a_c$ is the centripetal acceleration at the ring's edge, $a_g$ is the acceleration due to Earth's gravity at the altitude of the ring, and $g$ i our goal: standard Earth surface gravity, 9.8 m/s^2.
The centripetal acceleration is the acceleration required to put an object on a circular path. Part of that, for a person standing on the ring in in our scenario, will be supplied by Earth's gravity, and the rest by the outside rim of the ring. Its that "the rest" bit that we want to be one gee.
Looked at another way (specifically, from a frame of reference spinning at the same rate as Earth), $a_c$ is the centrifugal acceleration pushing the ring's occupants outward, $a_g$ is the gravitational acceleration pulling them inward, and once again, the difference is what they'll actually feel.
So, plugging in the formulae for $a_c$ and $a_g$:
$$ {r \omega \over t} = {GM \over r^2} + g $$
Rearranging things a bit, trying to solve for r:
$$ {r^3 \omega \over t} = GM + gr^2 $$
$$ {r^3 \omega \over t} - gr^2 - GM = 0 $$
Aaaand that's a cubic equation that I don't know how to solve.
Wolfram|Alpha gives a fantastically complicated solution for r, as well as a couple of equally-complicated solutions with imaginary numbers that can safely be ignored. So I'll just let WolframAlpha puzzle through the numbers on its own... and the solution turns out to be r≈1.15989×10^10 meters. That's about 17 times the radius of the Sun, 30 times farther out than the Moon, and 375 times the radius of geostationary orbits.
Good luck building that!
