Model time scale, prototype day and moon

Tides are gravity-dominated, so the Froude law applies: $\dfrac{V}{\sqrt{gL}}$ is the same in model and prototype. The time scale then follows from the length scale and the gravity ratio.

Time scale

Velocity scale $V_r = \sqrt{g_rL_r}$, and time $T = L/V$, so:

$$T_r = \frac{L_r}{V_r} = \sqrt{\frac{L_r}{g_r}}$$

Case 1: model tested on earth ($g_r = 1$)

$$T_r = \sqrt{\frac{1}{400}} = \frac{1}{20}$$

Model time for one prototype day $= 24 \times \dfrac1{20} = 1.2$ h $= 72$ min.

Case 2: model tested on the moon

The prototype is on earth, the model on the moon, so $g_r = g_m/g_p = 1/6$.

$$T_r = \sqrt{\frac{1/400}{1/6}} = \sqrt{\frac{6}{400}} = 0.1225$$

Model time for one prototype day $= 24 \times 0.1225 = 2.94$ h (about 2 h 56 min).

Answer: on earth, 1 prototype day corresponds to 1.2 h of model time ($T_r = 1/20$). On the moon, $T_r = 0.1225$ and 1 prototype day corresponds to about 2.94 h of model time; the model time is $0.1225\times$ the prototype time.