To discuss its viability the first thing to figure out is the stress on the cable and see if that is practicle.
I can't easily do the math off the top of my head, so I'm not going to try here. It also varies depending on the specific cable geometry (e.g do you have a cylindrical cable, with the same cross section all the way, or do you design it so it tapers to a point and have it much thicker in the middle (the more viable option in terms of strength).
The key thing to remember is that this cable must be able to hold its entire "weight" (i.e. the force acting on the cable due to gravity and the rotation of Earth), all the way from Earth to a geostationary/geosynchronous orbit, which has a height of 35786 km (which means we need a cable that is 35786 km long, this is the second issue, but is more of a challenge to meet and thus isn't the most important as it is just a case of dealing with a massive cable rather than physical limitations which can't be overcome, and gives us an idea of where to start). The part which makes the math hard is that the weight decreases as you go up and the variable cross section.
The force acting on a particular piece of cable is given by:
F=m*(-GM/r^2+omega^2*r). (taking note that a negative force means being pulled to Earth, a positive force means being flung away).
So lets simplify it a bit.
Lets say we have a constant cross section, A.
We will also make a rough estimate by going half way and using that as the force for the entire cable.
That makes our r for the above equation 24271100 m, and thus the force is -0.548276631*mass N/kg.
We also know the length (l) required (35786 km or 35786000 m).
We can let the density be p (can't be bothered typing rho).
The total mass would be A*l*p.
Thus the total force would be -19620627.52*A*p N m/kg (which can also be taken to mean tension).
To find the stress, that would be the force divided by the area, so -19620627.52*p N m/kg.
(for a sanity check, the force should be in units of N, while the stress should be in units of N/m^2 (or Pa), and the units of density can be kg/m^3.
Using the above, ignoring values and just subbing in units, for force we get m^2 (kg/m^3) N m/kg, after cancelling this gives us N.
For stress, we have (kg/m^3) N m/kg. After cancelling that gives us N/m^2.)
So this means the tensile strength, in Pa, needs to be 19620627.52 times as much as the density, in kg/m^3.
Another option is to look at the speficic strength/stress, which is the stress divided by the density.
This means we need a specific stress of 19620627.52 N m/kg. This can also be expressed as 19620.62752 kN m/kg or 19.62062752 MN m/kg
Some values of this for common materials (rather than list full units, I will just put down the prefix, e.g. k or M):
Steel - 61.1k
Nylon - 69l
Titanium - 76k
Titanium alloy - 260k
Alluminium alloys - 115k
Spider silk - 1.069M
Carbon fibre - 2.4M
Kevlar - 2.5M
So no common material comes close (and i'm fairly confident this is a significant underestimate, as if you just look at it at 10000 km up, if you use the force at that point and the cable below that point, that alone gives you a requirement of 14 M. For the same thing at 1000 km up, it gives 7 M).
The only materials known (at least outside of very niche groups) which come close are things like carbon nanotubes, with a specific strength of 46 M.
However, that is typically single carbon nano-tubes. When you combine it into a rope, that drops. However there are also imperfections in the process.
The theoretical tensile strength is 300 GPa, but typically achieved values are closer to 60 GPa, but when made into ropes it drops to single digit GPa.
So in order for it to be possible, we need to get a lot better at making very long carbon nanotubes.