The moon revolves in its orbit around the earth with an angular velocity of approximately 12.2o per day, in the same direction in which the earth is rotating on its axis with an angular velocity of 360o per day.
In each day, therefore, a point on the rotating earth must complete a rotation of 360o plus 12.2o, or 372.2o, in order to “catch up” with the moon.
Since 15o is equal to one hour of time, this extra amount of rotation equal to 12.2o each day would require a period of time equal to 12.2o/15o x 60 min/hr., or 48.8 minutes – if the moon revolved in a circular orbit, and its speed of revolution did not vary.
On the average it requires about 50 minutes longer each day for a sub lunar point on the rotating earth to regain this position directly along the major axis of the moon’s tidal force envelope, where the tide-raising influence is a maximum.
In consequence, the recurrence of a tide of the same phase and similar rise would take place at an interval of 24 hours 50 minutes after the preceding occurrence, if this single astronomical factor known as lunar retardation were considered. This period of 24 hours 50 minutes has been established as the tidal day.