Hello. In my book with examples of astronomy, I found this: Calculate, around which the known celestial bodies in the solar system (with a diameter greater than 1000 km) do a probe one complete cycle fastest (without the use of engines). Other movements of these bodies is negligible. If cycle must be fast, it must be a celestial body with small diameter (but greater than 1000 km). It is Tethys (1060 km) or something smaller? It is true that the circulation period is equal to: 2 * π * (radius of celestial body + height probe above the surface) / speed the probe It is good? Sorry for my bad English.
This would probably fit better in the homework section, even though it's not exactly homework. You might try asking this kind of questions there in the future. I assume you mean to calculate orbital velocity at minimal height - that is, as if the satellite were orbiting just above the surface, with no atmospheric drag, or mountains to worry about. You want minimal height, because that's where the orbit is the fastest. You want to find out how the orbital period changes with radius of the planet the probe is orbiting. The equation for the period you've provided is indeed good, but you can drop the "height above surface" part, since we're trying to imagine it going around the planet/moon at zero height. You still need to find out what is the speed, as it will not be the same for all bodies. To do so, look up the force of gravity, and the centripetal force, and write an equation that compares them: F_{G}=F_{c} solve for velocity. The important bit is to express mass of the central body in terms of volume of a sphere and density. Plug in the velocity to the equation for the period, and tell us what do you think it now tells you about the relationship between the radius and the period.
Smaller objects have a smaller circumference, but they also have a lower mass. There is a nice relation between the time for an orbit and one specific, single parameter of the object. This allows to find the "best" one in our solar system.