That CAT excavator probably has a center of gravity that is eight feet above the ground.

It is the ratio of the height of the COG to the horizontal distance out to the outer edge of the tire (or track) that counts.
You didn't say what model Husqvarna lawn tractor but if you pick a typical one you can find dimensions on like the YTH2348LS you will find it weighs 458 lbs and is 42 inches high and 50 inches wide. That 50 inches is the mower deck so the tires out to out would be less, maybe 46 inches. It has 20x8-8 rear tires so the rear axle is 18 inches off the ground and the COG with you sitting on it is about 24 inches off the ground.
So your ratio would be 23 inches of width divided by 24 of height would be 0.96.
My tractor as set up is 94 inches wide so half of that is 47 inches and it measures 32 inches from the ground to the top of the center casting between the operators feet which is above where the COG is with the fluid in the tires but using that. 47 / 32 =1.47 so mine is more stable then yours. :silly:
Now if I take the spacers off and set the tires in to the narrowest workable setting the half width is just 35 inches and I'd get 35/32=1.09 which is a lot closer to your tractor so the set up is what makes the difference. Notice the weight of the tractors doesn't factor into this ratio except for determining exactly where the COG of the tractor is. If your adding weight you want to add it somewhere low to lower the COG not pile it on top and increase the COG.