Tidal potential expressed as the Legendre polynomial
Let us derive the tidal potential by expanding the moon's gravitational potential using the Legendre polynomial.
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Fig. 1. Vectors used in the equations.
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Let \({\boldsymbol{r}}\) be the position of point
P seen from the earth center and
\({\boldsymbol{a}}\) be the position of the moon
seen from the earth center.
Then the gravitational potential per unit mass due to the moon at point
P can be written as
\[
{V_{\text{m}}} = - \frac{{Gm}}{{\left| {{\boldsymbol{r}} - {\boldsymbol{a}}} \right|}}. \tag{1}
\]
By the way, when \(r < a\) ,
it is known that the potential of this form can be expanded
using the Legendre polynomial as follows:
\[
\frac{1}{{\left| {{\boldsymbol{r}} - {\boldsymbol{a}}} \right|}}
= \frac{1}{a}\left[ {1 + \frac{r}{a}{P_1}(\cos \psi )
+ {{\left( {\frac{r}{a}} \right)}^2}{P_2}(\cos \psi ) + \cdots } \right]. \tag{2}
\]
Thus the gravitational potential by the moon can be written as
\[
{V_{\text{m}}} = - \frac{{Gm}}{a}\left[ {1 + \frac{r}{a}{P_1}(\cos \psi )
+ {{\left( {\frac{r}{a}} \right)}^2}{P_2}(\cos \psi ) + \cdots } \right] , \tag{3}
\]
where \(\psi\) is the angle between
\({\boldsymbol{r}}\) and
\({\boldsymbol{a}}\), i.e.
\[
\cos \psi = \frac{{{\boldsymbol{a}} \cdot {\boldsymbol{r}}}}{{ar}}.
\]
Moreover,
\[
\begin{gathered}
\begin{align}
&{P_1}(\cos \psi ) = \cos \psi , \hfill \\
&{P_2}(\cos \psi ) = \frac{1}{2}(3{\cos ^2}\psi - 1) . \hfill \\
\end{align}
\end{gathered}
\]
In the potential of Eq. (3),
the first-order term of \(r \)
represents the gravity that the entire earth receives from the moon,
and the terms of the second and higher orders of
\(r \) correspond to the tidal force due to the moon.
If the direction of the moon as seen from the earth (or the direction of
\({\boldsymbol{a}}\)) is
\(x \),
and Eq. (3) is expressed in terms of
\(x \),
\(y \), and
\(z \),
we obtain Eq. (4) on the main page of this article.
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T. Fujiwara, updated 2024/01