I must admit I am not clear on what you mean by "mean radius" of the earth; average, arithmetic, geometric, harmonic, etc. I am making my own interpretation to be the radius that divides the spherical earth into two equal volumes, ignoring the equatorial bulge.
Using an earth radius of 3963 miles, yields a volume of 3.8376x10^22 cub.ft.From the half volume of 1.9188x10^22 cub.ft., I derive a half volume radius of 3145 miles or 16,607,898 ft. The radius of the orbit you defined is therefore 3145 x 3.65 = 11,795.25 miles or 60,610,440 ft.
The orbital circular velocity of a satellite at any altitude derives from Vc = sqrt(µ/Ro) where Vc = the orbital velocity in feet per second, µ = the earth's gravitational constant = Re^2g, Ro = the orbital radius and Re = the earth radius. With µ = 3963(5280)32.19 ~= 1.407974x10^16, the satellite's orbital velocity becomes
Vc = sqrt[{20,924,640)^2 x 32.19}/60,610,440] = 15,249fps = 10,397mph.
If this is not exactly the result you desired, please post your specific wants.
If you should find any inconsistencies in my math-magic, please don't hesitate to let me know.
Find the speed of a satellite in a circular orbit around the Earth with a radius 3.65 times the mean radius of the Earth; RE= 6.37·103 km ME= 5.98·1024 kg
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