Besides the "more accurate at poles" issue, you are also double-counting because you don't need to cover the circumference twice.
An alternative is to find the Earth's surface area (5.1e8 km^2), and divide by 10mX10m. This gives the number of cells. Then take log base 30 (26 alpha + 10 digit - {"O", "I", "S", "U", "V", "Z"}).
This turns out to be 8.60 characters. To detect all single-character errors, you need a full extra character (9.60).
If you allow all alphanumerics (36 possibilities), you end up with 8.16 characters, plus one for errors is 9.16.
Conclusion: if you were willing to give up some accuracy over oceans, you could get away with 9 characters, otherwise, 10 is the best you can do.
An alternative is to find the Earth's surface area (5.1e8 km^2), and divide by 10mX10m. This gives the number of cells. Then take log base 30 (26 alpha + 10 digit - {"O", "I", "S", "U", "V", "Z"}).
This turns out to be 8.60 characters. To detect all single-character errors, you need a full extra character (9.60).
If you allow all alphanumerics (36 possibilities), you end up with 8.16 characters, plus one for errors is 9.16.
Conclusion: if you were willing to give up some accuracy over oceans, you could get away with 9 characters, otherwise, 10 is the best you can do.