and

where .

Let us transform to a non-inertial frame of reference which *rotates*, about an axis perpendicular to the orbital plane and passing through , at
the angular velocity . In this reference frame, both masses appear to be *stationary*. Consider mass . In the rotating frame, this mass experiences
a gravitational acceleration

(929) |

(930) |

(931) |

Let the center of the mass distribution lie at , the center of the mass distribution at B, and the center of mass at --see Figure 42. We wish to calculate the centrifugal and gravitational accelerations at some point in the vicinity of point . It is convenient to adopt spherical coordinates, centered on point , and aligned such that the -axis coincides with the line .

Let us assume that the mass distribution
is orbiting around , but is *not* rotating about
an axis passing through its center, in order to
exclude rotational flattening from our analysis. If this is the
case then it is easily seen that each constituent point of executes
circular motion of angular velocity and radius --see Figure 43. Hence, each constituent point experiences the *same*
centrifugal acceleration: *i.e.*,

(932) |

(933) |

(934) |

(935) |

The gravitational acceleration at point due to mass is given by

(936) |

(937) |

Now,

(938) |

(939) |

(940) |

(941) |

Adding and , we obtain

In order to calculate the tidal elongation of the mass distribution we
need to add the potential, , due to the external forces, to
the gravitational potential, , generated by the distribution itself. Assuming that
the mass distribution is spheroidal with mass , mean radius ,
and ellipticity , it follows from Equations (901), (911),
and (942) that the total surface potential
is given by

(943) |

where we have treated and as small quantities. As before, the condition for equilibrium is that the total potential be constant over the surface of the spheroid. Hence, we obtain

(944) |

(945) |

Consider the tidal elongation of the Earth due to the Moon. In this
case, we have
,
,
, and
.
Hence, we calculate that
, or

(946) |

Consider the tidal elongation of the Earth due to the Sun. In this case,
we have
,
,
, and
.
Hence, we calculate that
, or

(947) |

In reality, the amplitude of the tides varies significantly from place to place on the Earth's surface, due to the presence of the continents, which impede the flow of the oceanic tidal bulge around the Earth. Moreover, there is a time-lag of approximately 12 minutes between the Moon being directly overhead (or directly below) and high tide, because of the finite inertia of the oceans. Similarly, the time-lag between a spring tide and a full moon, or a new moon, can be up to 2 days.