Assume that a wheel has an outside diameter of 0.800 km and an inside diameter of 0.720 m.
Use mathematical calculations to show how a simulated gravitational force equal to that on Earth could be established
I'm assuming your wheel has an inside diameter of 0.720 kilometers, not meters, since the outside diameter is in kilometers.....The gravitational force on Earth is 9.8 meters per second squared, or 9.8 m/sec^2. That force can be induced upon your wheel if its spinning at a specific rate. The relationship between the gravitational force (or acceleration), the angular velocity of the wheel, and the radius is given by the formula
a = W^2r
where a is acceleration in m/sec^2, W is the angular velocity in radians per second (a radian is a measure where the radius length equals an arc length of a circle) and r is the radius. In this case,
9.8 m/sec^2 = W^2 (800 m)
.01225 m^2/sec^2 = W^2
.1106 m/sec = W
so your wheel would have to be spinning at about a tenth of a meter per second for a radius of 800 m.
See the link for an actual gravitational calculator: