While I do get into link budget calculations, so far I have done them for clear sky, using the graphs from the U.S. Naval Research Laboratory, originally done for radar purposes. Fortunately for me (so far), the attenuation from rain is negligible from S-Band around 2GHz up to the low X-Band at 8GHz. It starts to become important above 9Ghz, where the raindrop size is a sufficient fraction of the wavelength.
I do have some information on rain attenuation, but I have to check it out some more before I would commit to posting information or references here. Getting at the rain attenuation for a region is supposedly "widely published", but I do not find it so easy.
Mainly, I just want to say that the formula you have does not directly correspond to what I understand is a "specific" attenuation concept. Yours seems to be raised to a power alpha.
Total attenuation from rain is A(rain) = alpha*L [in dB]
..where alpha [dB/km] is called the specific rain attenuation, and can be found from (supposedly) widely published tables and/or graphs.
.. and L is the effective path length in km.
This is not the same as the physical path length, because it is modified by rain density values.
Your question is a fair one, and I would need to seek the full answers for myself anyway, and I am sorry I cannot respond with more certainty right now. Eventually, I will have the truth of it, and then I will post again. This stuff is important. Rain attenuation to a satellite can be as low as 0.1dB in Arizona, and as much as 12dB from Oregon, or a site in India.
In the meantime, you might search as well. There are 3 generally recognised rain models..
1. NASA Rain Attenuation Model
2. Crane Rain Attenuation Model
3. CCIR Rain Attenuation Model
More on this later - as time allows..