正二十面体惑星の重力ポテンシャルの\(\rm{Taylor}\)展開

　正二十面体惑星の重力ポテンシャルについて、 \[u_{1}(x, y, z, a)=u_{L}(x, y, z, a)+u_{A}(x, y, z, a)\] \begin{align*} u_{L}(x, y, z, a) =& \frac{a+(3-\sqrt{5})x-z}{3(5-\sqrt{5})} [(a-\dfrac{3+\sqrt{5}}{4}x-\dfrac{3\sqrt{10+2\sqrt{5}}}{4}y-z)\tilde{L}_{12}\\ &+(a+\dfrac{3+\sqrt{5}}{2}x+2z)\tilde{L}_{23} +(a-\dfrac{3+\sqrt{5}}{4}x+\dfrac{3\sqrt{10+2\sqrt{5}}}{4}y-z)\tilde{L}_{31}]\\ \end{align*} \begin{align*} u_{A}(x, y, z, a) =&-\dfrac{5-\sqrt{5}}{60}(4x-(3+\sqrt{5})(z-a))^{2}[S_{1}+S_{2}+S_{3} -\pi\mathrm{sgn}(4x-(3+\sqrt{5})(z-a))] \end{align*} とする。

外部ポテンシャルについての\(\rm{Taylor}\)展開

\(r = \sqrt{x^{2} + y^{2} + z^{2}} \gg a\)のとき、\(\xi =x/r,\ \)\(\eta =y/r,\ \)\(\zeta =z/r,\ \)\(\alpha = a/r\)として、\(\alpha\)について\(\rm{Taylor}\)展開をすると、まず、\(u_{L}(x, y, z, a)\)については、 \[r_{1}=r\sqrt{\xi^{2} + \eta^{2} + (\zeta- \alpha)^{2}}=r\sqrt{1-2\alpha\zeta+\alpha^{2}} =r(1-\zeta\alpha+\frac{1}{2}(1-\zeta^{2})\alpha^{2}+\cdots ), \] \begin{align*} r_{2}=&r\sqrt{(\xi - \tfrac{2}{\sqrt{5}}\alpha\cos (4\pi /5))^{2} + (\eta - \tfrac{2}{\sqrt{5}}\alpha\sin (4\pi /5))^{2} + (\zeta - \tfrac{1}{\sqrt{5}}\alpha)^{2}} =r\sqrt{1-2\alpha(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)+\alpha^{2}}\\ =&r(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)\alpha +\frac{1}{2}(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})\alpha^{2}+\cdots ), \end{align*} \begin{align*} r_{3}=&r\sqrt{(\xi - \tfrac{2}{\sqrt{5}}\alpha\cos (4\pi /5))^{2} + (\eta + \tfrac{2}{\sqrt{5}}\alpha\sin (4\pi /5))^{2} + (\zeta - \tfrac{2}{\sqrt{5}}\alpha)^{2}} =r\sqrt{1-2\alpha(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)+\alpha^{2}}\\ =&r(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)\alpha +\frac{1}{2}(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})\alpha^{2}+\cdots ), \end{align*} \begin{align*} \tilde{L}_{12}=&\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}}\\ =&\ln\dfrac {2+(\tfrac{4}{10+2\sqrt{5}}-\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)-(1+\tfrac{1}{\sqrt{5}})\zeta)\alpha +(1-((\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}+\zeta^{2})/2)\alpha^{2}+\cdots} {2-(\tfrac{4}{10+2\sqrt{5}}+\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+(1+\tfrac{1}{\sqrt{5}})\zeta)\alpha +(1-((\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}+\zeta^{2})/2)\alpha^{2}+\cdots } \\ =&\frac{1}{10}((5-\sqrt{5})\sqrt{10+2\sqrt{5}})\alpha +(-\frac{\sqrt{10+2\sqrt{5}}}{10}\xi+\frac{5-\sqrt{5}}{10}\eta+\frac{\sqrt{10+2\sqrt{5}}}{5}\zeta)\alpha^{2} +(\frac{(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\xi^{2}-\frac{2}{5}\xi\eta \\ &-\frac{(5+2\sqrt{5})\sqrt{10+2\sqrt{5}}}{50}\xi\zeta+\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta^{2} +\frac{3+\sqrt{5}}{10}\eta\zeta+\frac{(25-\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\zeta^{2} +\frac{(\sqrt{5}-9)\sqrt{10+2\sqrt{5}}}{60})\alpha^{3}+\cdots , \end{align*} \begin{align*} \tilde{L}_{23}&=\ln{\frac{r_{2}+r_{3}+r_{23}}{r_{2}+r_{3}-r_{23}}}\\ &=\ln\dfrac {2+(\tfrac{4}{10+2\sqrt{5}}-\tfrac{4}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\zeta)\alpha +(1-((\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}+\tfrac{4}{5}\eta^{2}\sin^{2} (4\pi /5)^{2}))\alpha^{2}+\cdots} {2-(\tfrac{4}{10+2\sqrt{5}}+\tfrac{4}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\zeta)\alpha +(1-((\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}+\tfrac{4}{5}\eta^{2}\sin^{2} (4\pi /5)^{2}))\alpha^{2}+\cdots } \\ &=\frac{1}{10}((5-\sqrt{5})\sqrt{10+2\sqrt{5}})\alpha +(-\frac{\sqrt{10+2\sqrt{5}}}{5}\xi+\frac{(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}}{5}\zeta)\alpha^{2} +(\frac{(15+3\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\xi^{2}-\frac{3\sqrt{5}\sqrt{10+2\sqrt{5}}}{25}\xi\zeta\\ &+\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta^{2}+\frac{(15-3\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\zeta^{2} +\frac{(\sqrt{5}-9)\sqrt{10+2\sqrt{5}}}{60})\alpha^{3}+\cdots , \end{align*} \begin{align*} \tilde{L}_{31}=&\ln{\frac{r_{3}+r_{1}+r_{31}}{r_{3}+r_{1}-r_{31}}}\\ =&\ln\dfrac {2+(\tfrac{4}{10+2\sqrt{5}}-\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)-(1+\tfrac{1}{\sqrt{5}})\zeta)\alpha +(1-((\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}+\zeta^{2})/2)\alpha^{2}+\cdots} {2-(\tfrac{4}{10+2\sqrt{5}}+\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+(1+\tfrac{1}{\sqrt{5}})\zeta)\alpha +(1-((\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}+\zeta^{2})/2)\alpha^{2}+\cdots } \\ =&\frac{1}{10}((5-\sqrt{5})\sqrt{10+2\sqrt{5}})\alpha +(-\frac{\sqrt{10+2\sqrt{5}}}{10}\xi-\frac{5-\sqrt{5}}{10}\eta+\frac{\sqrt{10+2\sqrt{5}}}{5}\zeta)\alpha^{2} +(\frac{(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\xi^{2}+\frac{2}{5}\xi\eta \\ &-\frac{(5+2\sqrt{5})\sqrt{10+2\sqrt{5}}}{50}\xi\zeta+\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta^{2} -\frac{3+\sqrt{5}}{10}\eta\zeta+\frac{(25-\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\zeta^{2} +\frac{(\sqrt{5}-9)\sqrt{10+2\sqrt{5}}}{60})\alpha^{3}+\cdots , \end{align*} より、 \begin{align*} u_{L}(x, y, z, a) =&\frac{(5+\sqrt{5})r^{2}}{60}(\alpha+(3-\sqrt{5})\xi-\zeta)\times \\ &(\alpha-\dfrac{3+\sqrt{5}}{4}\xi-\dfrac{3\sqrt{10+2\sqrt{5}}}{4}\eta-\zeta)\cdot (\frac{1}{10}((5-\sqrt{5})\sqrt{10+2\sqrt{5}})\alpha \\ &+(-\frac{\sqrt{10+2\sqrt{5}}}{10}\xi+\frac{5-\sqrt{5}}{10}\eta+\frac{\sqrt{10+2\sqrt{5}}}{5}\zeta)\alpha^{2} +(\frac{(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\xi^{2}-\frac{2}{5}\xi\eta-\frac{(5+2\sqrt{5})\sqrt{10+2\sqrt{5}}}{50}\xi\zeta \\ &+\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta^{2}+\frac{3+\sqrt{5}}{10}\eta\zeta+\frac{(25-\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\zeta^{2} +\frac{(\sqrt{5}-9)\sqrt{10+2\sqrt{5}}}{60})\alpha^{3}+\cdots)\\ &+(\alpha+\dfrac{3+\sqrt{5}}{2}\xi+2\zeta)\cdot (\frac{1}{10}((5-\sqrt{5})\sqrt{10+2\sqrt{5}})\alpha \\ &+(-\frac{\sqrt{10+2\sqrt{5}}}{5}\xi+\frac{(\sqrt{5}-1)\sqrt{10+2\sqrt{5}}}{5}\zeta)\alpha^{2} +(\frac{(15+3\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\xi^{2}-\frac{3\sqrt{5}\sqrt{10+2\sqrt{5}}}{25}\xi\zeta\\ &+\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta^{2}+\frac{(15-3\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\zeta^{2} +\frac{(\sqrt{5}-9)\sqrt{10+2\sqrt{5}}}{60})\alpha^{3}+\cdots)\\ &+(\alpha-\dfrac{3+\sqrt{5}}{4}\xi+\dfrac{3\sqrt{10+2\sqrt{5}}}{4}\eta-\zeta)\cdot (\frac{1}{10}((5-\sqrt{5})\sqrt{10+2\sqrt{5}})\alpha \\ &+(-\frac{\sqrt{10+2\sqrt{5}}}{10}\xi-\frac{5-\sqrt{5}}{10}\eta+\frac{\sqrt{10+2\sqrt{5}}}{5}\zeta)\alpha^{2} +(\frac{(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\xi^{2}+\frac{2}{5}\xi\eta-\frac{(5+2\sqrt{5})\sqrt{10+2\sqrt{5}}}{50}\xi\zeta \\ &+\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta^{2}-\frac{3+\sqrt{5}}{10}\eta\zeta+\frac{(25-\sqrt{5})\sqrt{10+2\sqrt{5}}}{100}\zeta^{2} +\frac{(\sqrt{5}-9)\sqrt{10+2\sqrt{5}}}{60})\alpha^{3}+\cdots)]\\ =&-\sqrt{10+2\sqrt{5}}(\frac{5+\sqrt{5}}{300}\xi^{3}+\frac{5-2\sqrt{5}}{100}\xi^{2}\zeta +\frac{3-\sqrt{5}}{20}\xi\eta^{2}+\frac{5-3\sqrt{5}}{50}\xi\zeta^{2}-\frac{3-\sqrt{5}}{10}\xi -\frac{1}{20}\eta^{2}\zeta-\frac{5-\sqrt{5}}{150}\zeta^{3}+\frac{1}{10}\zeta)\alpha^{2} \\ &-\sqrt{10+2\sqrt{5}} (-\frac{3+\sqrt{5}}{150}\xi^{4}+\frac{3\sqrt{5}-1}{150}\xi^{3}\zeta-\frac{5-\sqrt{5}}{50}\xi^{2}\eta^{2} +\frac{4\sqrt{5}-3}{100}\xi^{2}\zeta^{2}+\frac{25-2\sqrt{5}}{300}\xi^{2} +\frac{5+\sqrt{5}}{50}\xi\eta^{2}\zeta-\frac{3\sqrt{5}-1}{150}\xi\zeta^{3} \\ &+\frac{5+\sqrt{5}}{150}\xi\zeta -\frac{5+2\sqrt{5}}{100}\eta^{2}\zeta^{2}+\frac{1}{20}\eta^{2}-\frac{3+\sqrt{5}}{150}\zeta^{4} +\frac{10+sqrt{5}}{150}\zeta^{2}-\frac{1}{10})\alpha^{3}+\cdots , \end{align*} 　である。 \[\begin{pmatrix} x_{i} \\ y_{i} \\ z_{i} \end{pmatrix}=Z^{-i}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ \begin{pmatrix} x_{i+5} \\ y_{i+5} \\ z_{i+5} \end{pmatrix}=(Z^{i}Y)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\] \[\begin{pmatrix} x_{i+10} \\ y_{i+10} \\ z_{i+10} \end{pmatrix}=(Z^{i}X)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ \begin{pmatrix} x_{i+15} \\ y_{i+15} \\ z_{i+15} \end{pmatrix}=(Z^{i}YX)^{-1}\begin{pmatrix} x \\ y \\ z \end{pmatrix},\ (i=0,1,2,3,4)\] \[Z=\begin{pmatrix} \cos (2\pi /5) & \sin (2\pi /5) & 0 \\ -\sin (2\pi /5) & \cos (2\pi /5) & 0 \\ 0 & 0 & 1 \end{pmatrix},\ Y=\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix},\ X=\begin{pmatrix} 1/\sqrt{5} & 0 & -2/\sqrt{5} \\ 0 & -1 & 0 \\ -2/\sqrt{5} & 0 & -1/\sqrt{5} \end{pmatrix}\] として、 \[\sum_{i=0}^{19}u_{L}(x_{i}, y_{i}, z_{i}, a) =\sqrt{10+2\sqrt{5}}r^{2}(\frac{2}{5}(\xi^{2}+\eta^{2}+\zeta^{2})^{2}-\frac{4}{3}(\xi^{2}+\eta^{2}+\zeta^{2})+2)\alpha^{3}+\cdots =\frac{16}{15}\sqrt{10+2\sqrt{5}}\cdot\frac{a^{3}}{r}+\cdots \] 　である。次に、\(u_{A}(x, y, z, a)\)については、三角形である各面の頂点の位置ベクトルを\(\mathbf{v}_{1},\ \)\(\mathbf{v}_{2},\ \)\(\mathbf{v}_{3},\ \)として、\(\mathbf{r}=(x,y,z),\ \)\(\mathbf{r}_{1}=\mathbf{r}-\mathbf{v}_{1},\ \)\(\mathbf{r}_{2}=\mathbf{r}-\mathbf{v}_{2},\ \)\(\mathbf{r}_{3}=\mathbf{r}-\mathbf{v}_{3}\ \)、\(\mathbf{n}(|\mathbf{n}|=1)\)を面の法線ベクトルとすると、 \[ u_{A}(x, y, z, a)=-(\mathbf{r}_{1}\cdot\mathbf{n})^{2}\arctan\frac{ [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}} {|\mathbf{r}_{1}||\mathbf{r}_{2}||\mathbf{r}_{3}|+(\mathbf{r}_{1}\cdot\mathbf{r}_{2})|\mathbf{r}_{3}|+(\mathbf{r}_{2}\cdot\mathbf{r}_{3})|\mathbf{r}_{1}|+(\mathbf{r}_{3}\cdot\mathbf{r}_{1})|\mathbf{r}_{2}|} \] と表すこともできて、\(\mathbf{v}_{1}=(0,0,a),\ \)\(\mathbf{v}_{2}=(2a\cos (4\pi /5)/\sqrt{5},2a\sin (4\pi /5)/\sqrt{5},a/\sqrt{5}),\ \)\(\mathbf{v}_{3}=(2a\cos (4\pi /5)/\sqrt{5},-2a\sin (4\pi /5)/\sqrt{5},a/\sqrt{5})\ \)とすると、 \[ \mathbf{n}=(-\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30},0,\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60})\] \begin{align*} \mathbf{r}_{1}\cdot\mathbf{n}&= -\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30}x+\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}(z-a) \\ &=r[-\frac{\sqrt{3}(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{30}\xi+\frac{\sqrt{3}(5+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}(\zeta-\alpha)] \end{align*} \begin{align*} (\mathbf{r}_{1}\cdot\mathbf{r}_{2})&=|\mathbf{r}|^{2}-((\mathbf{v}_{1}+\mathbf{v}_{2})\cdot\mathbf{r})+(\mathbf{v}_{1}\cdot\mathbf{v}_{2})=r^{2}-a(-\frac{5+\sqrt{5}}{10}x+\frac{(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}y+\frac{5+\sqrt{5}}{5}z)+\frac{\sqrt{5}}{5}a^{2} \\ &=r^{2}(1-(-\frac{5+\sqrt{5}}{10}\xi+\frac{(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta+\frac{5+\sqrt{5}}{5}\zeta)\alpha+\frac{\sqrt{5}}{5}\alpha^{2}), \end{align*} \[ (\mathbf{r}_{2}\cdot\mathbf{r}_{3})=|\mathbf{r}|^{2}-((\mathbf{v}_{2}+\mathbf{v}_{3})\cdot\mathbf{r})+(\mathbf{v}_{2}\cdot\mathbf{v}_{3})=r^{2}-a(-\frac{5+\sqrt{5}}{5}x+\frac{2\sqrt{5}}{5}z) +\frac{\sqrt{5}}{5}a^{2}=r^{2}(1-(-\frac{5+sqrt{5}}{5}\xi+\frac{2\sqrt{5}}{5}\zeta) +\frac{\sqrt{5}}{5}\alpha^{2}), \] \begin{align*} (\mathbf{r}_{3}\cdot\mathbf{r}_{1})&=|\mathbf{r}|^{2}-((\mathbf{v}_{3}+\mathbf{v}_{1})\cdot\mathbf{r})+(\mathbf{v}_{3}\cdot\mathbf{v}_{1})=r^{2}-a(-\frac{5+\sqrt{5}}{10}x-\frac{(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}y+\frac{5+\sqrt{5}}{5}z)+\frac{\sqrt{5}}{5}a^{2} \\ &=r^{2}(1-(-\frac{5+\sqrt{5}}{10}\xi-\frac{(5-\sqrt{5})\sqrt{10+2\sqrt{5}}}{20}\eta+\frac{5+\sqrt{5}}{5}\zeta)\alpha+\frac{\sqrt{5}}{5}\alpha^{2}), \end{align*} \[ |\mathbf{r}_{1}|=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{1}\cdot\mathbf{r}+|\mathbf{v}_{1}|^{2})} =\sqrt{r^{2}-2ax+a^{2}} =r\sqrt{1-2\alpha\zeta+\alpha^{2}} =r(1-\zeta\alpha+\frac{1}{2}(1-\zeta^{2})\alpha^{2}+\cdots ), \] \begin{align*} |\mathbf{r}_{2}|=&\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{2}\cdot\mathbf{r}+|\mathbf{v}_{2}|^{2})} =\sqrt{r^{2}-2a(\tfrac{2}{\sqrt{5}}x\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}y\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}z)+a^{2}}\\ =&r\sqrt{1-2\alpha(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)+\alpha^{2}}\\ =&r(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)\alpha +\frac{1}{2}(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})\alpha^{2}+\cdots ), \end{align*} \begin{align*} |\mathbf{r}_{3}|=&\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{3}\cdot\mathbf{r}+|\mathbf{v}_{3}|^{2})} =\sqrt{r^{2}-2a(\tfrac{2}{\sqrt{5}}x\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}y\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}z)+a^{2}}\\ =&r\sqrt{1-2\alpha(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)+\alpha^{2}}\\ =&r(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)\alpha +\frac{1}{2}(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})\alpha^{2}+\cdots ), \end{align*} \begin{align*} [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}=&[\mathbf{v}_{1}\times\mathbf{v}_{2}]\cdot\mathbf{v}_{3} -(\mathbf{v}_{1}\times\mathbf{v}_{2}+\mathbf{v}_{2}\times\mathbf{v}_{3}+\mathbf{v}_{3}\times\mathbf{v}_{1})\cdot\mathbf{r}\\ =&\dfrac{1}{5}\sqrt{10+2\sqrt{5}}(a^{3}+((\sqrt{5}-3)x+z)a^{2}) =\dfrac{1}{5}\sqrt{10+2\sqrt{5}}r^{3}(\alpha^{3}+((\sqrt{5}-3)\xi+\zeta)\alpha^{2}) \end{align*} より、 \[ \arctan\frac{ [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}} {|\mathbf{r}_{1}||\mathbf{r}_{2}||\mathbf{r}_{3}|+(\mathbf{r}_{1}\cdot\mathbf{r}_{2})|\mathbf{r}_{3}|+(\mathbf{r}_{2}\cdot\mathbf{r}_{3})|\mathbf{r}_{1}|+(\mathbf{r}_{3}\cdot\mathbf{r}_{1})|\mathbf{r}_{2}|} \\ =\sqrt{10+2\sqrt{5}}(\frac{3-\sqrt{5}}{20}\xi-\frac{1}{20}\zeta)\alpha^{2} +\sqrt{10+2\sqrt{5}}(-\frac{5-\sqrt{5}}{50}\xi^{2}+\frac{5+\sqrt{5}}{50}\xi\zeta-\frac{5+2\sqrt{5}}{100}\zeta^{2}+\frac{1}{20})\alpha^{3}+\cdots, \] \begin{align*} u_{A}(x, y, z, a)=&-r^{2}\sqrt{10+2\sqrt{5}}[ (\frac{10-4\sqrt{5}}{75}\xi^{3}-\frac{5-\sqrt{5}}{50}\xi^{2}\zeta+\frac{5+\sqrt{5}}{100}\xi\zeta^{2}-\frac{5+2\sqrt{5}}{300}\zeta^{3})\alpha^{2}\\ &+(-\frac{6-2\sqrt{5}}{75}\xi^{4}+\frac{8}{75}\xi^{3}\zeta-\frac{3+\sqrt{5}}{25}\xi^{2}\zeta^{2} +\frac{5-\sqrt{5}}{50}\xi^{2}+\frac{7+3\sqrt{5}}{75}\xi\zeta^{3}-\frac{5+\sqrt{5}}{50}\xi\zeta +\frac{9+4\sqrt{5}}{300}\zeta^{4}+\frac{5+2\sqrt{5}}{100}\zeta^{2})\alpha^{3}+\cdots] \end{align*} \[\sum_{i=0}^{19}u_{A}(x_{i}, y_{i}, z_{i}, a) =-\sqrt{10+2\sqrt{5}}r^{2}(-\frac{3}{5}(\xi^{2}+\eta^{2}+\zeta^{2})^{2}+(\xi^{2}+\eta^{2}+\zeta^{2}))\alpha^{3}+\cdots =-\frac{2}{5}\sqrt{10+2\sqrt{5}}\cdot\frac{a^{3}}{r}+\cdots \] 　である。\(u_{L}(x, y, z, a)\)と\(u_{A}(x, y, z, a)\)の結果をまとめると、 \[ U(x,y,z)=-G\rho [\frac{2}{3}\sqrt{10+2\sqrt{5}}\cdot\frac{a^{3}}{r}+\cdots] \] となる。さらに高次まで\(\rm{Taylor}\)展開を行うと以下のようになる。 \begin{align*} U(x,y,z)=&-G\rho\{\dfrac{2\sqrt{10+2\sqrt{5}}a^{3}}{3r}\\ &-\dfrac{a^{9}}{r^{13}}\cdot\dfrac{11(8+\sqrt{5})\sqrt{10+2\sqrt{5}}}{50400} [x^{6} - \frac{42}{5}x^{5}z + 3x^{4}y^{2} - 18x^{4}z^{2} + 84x^{3}y^{2}z + 3x^{2}y^{4} - 36x^{2}y^{2}z^{2} + 24x^{2}z^{4} - 42xy^{4}z \\ &+ y^{6} - 18y^{4}z^{2} + 24y^{2}z^{4} - \frac{16}{5}z^{6}]\\ &+\dfrac{a^{13}}{r^{21}}\cdot\dfrac{589\sqrt{10+2\sqrt{5}}}{3300000} [x^{10} - \dfrac{495}{31}x^{9}z - \dfrac{4365}{62}x^{8}y^{2} + \dfrac{1575}{62}x^{8}z^{2} + \dfrac{3960}{31}x^{7}y^{2}z + \dfrac{4620}{31}x^{7}z^{3} + \dfrac{9660}{31}x^{6}y^{4} + \dfrac{3150}{31}x^{6}y^{2}z^{2} \\ &- \dfrac{4200}{31}x^{6}z^{4} + \dfrac{6930}{31}x^{5}y^{4}z - \dfrac{41580}{31}x^{5}y^{2}z^{3} - \dfrac{5544}{31}x^{5}z^{5} - \dfrac{9975}{31}x^{4}y^{6} + \dfrac{4725}{31}x^{4}y^{4}z^{2} - \dfrac{12600}{31}x^{4}y^{2}z^{4} + \dfrac{5040}{31}x^{4}z^{6} - \dfrac{23100}{31}x^{3}y^{4}z^{3} \\ &+ \dfrac{55440}{31}x^{3}y^{2}z^{5} + \dfrac{2025}{31}x^{2}y^{8} + \dfrac{3150}{31}x^{2}y^{6}z^{2} - \dfrac{12600}{31}x^{2}y^{4}z^{4} + \dfrac{10080}{31}x^{2}y^{2}z^{6} - \dfrac{1440}{31}x^{2}z^{8} - \dfrac{2475}{31}xy^{8}z + \dfrac{23100}{31}xy^{6}z^{3} \\ &- \dfrac{27720}{31}xy^{4}z^{5} - \dfrac{125}{62}y^{10} + \dfrac{1575}{62}y^{8}z^{2} - \dfrac{4200}{31}y^{6}z^{4} + \dfrac{5040}{31}y^{4}z^{6} - \dfrac{1440}{31}y^{2}z^{8} + \dfrac{64}{31}z^{10}] \\ &+\dfrac{a^{15}}{r^{25}}\dfrac{4471(15+2\sqrt{5})\sqrt{10+2\sqrt{5}}}{312000000} [x^{12} + \dfrac{2860}{263}x^{11}z - \dfrac{23122}{263}x^{10}y^{2} + \dfrac{5764}{263}x^{10}z^{2} - \dfrac{20020}{263}x^{9}y^{2}z - \dfrac{45760}{263}x^{9}z^{3} + \dfrac{78045}{263}x^{8}y^{4} \\ &+ \dfrac{572220}{263}x^{8}y^{2}z^{2} - \dfrac{138600}{263}x^{8}z^{4} - \dfrac{62920}{263}x^{7}y^{4}z + \dfrac{366080}{263}x^{7}y^{2}z^{3} + \dfrac{128128}{263}x^{7}z^{5} - \dfrac{4620}{263}x^{6}y^{6} - \dfrac{2115960}{263}x^{6}y^{4}z^{2} - \dfrac{554400}{263}x^{6}y^{2}z^{4} \\ &+ \dfrac{295680}{263}x^{6}z^{6} - \dfrac{40040}{263}x^{5}y^{6}z + \dfrac{640640}{263}x^{5}y^{4}z^{3} - \dfrac{1153152}{263}x^{5}y^{2}z^{5} - \dfrac{73216}{263}x^{5}z^{7} - \dfrac{84975}{263}x^{4}y^{8} + \dfrac{2448600}{263}x^{4}y^{6}z^{2} - \dfrac{831600}{263}x^{4}y^{4}z^{4} \\ &+ \dfrac{887040}{263}x^{4}y^{2}z^{6} - \dfrac{190080}{263}x^{4}z^{8} + \dfrac{14300}{263}x^{3}y^{8}z - \dfrac{640640}{263}x^{3}y^{4}z^{5} + \dfrac{732160}{263}x^{3}y^{2}z^{7} + \dfrac{20350}{263}x^{2}y^{10} - \dfrac{405900}{263}x^{2}y^{8}z^{2} \\ &- \dfrac{554400}{263}x^{2}y^{6}z^{4} + \dfrac{887040}{263}x^{2}y^{4}z^{6} - \dfrac{380160}{263}x^{2}y^{2}z^{8} + \dfrac{33792}{263}x^{2}z^{10} + \dfrac{14300}{263}xy^{10}z - \dfrac{228800}{263}xy^{8}z^{3} + \dfrac{640640}{263}xy^{6}z^{5} \\ &- \dfrac{366080}{263}xy^{4}z^{7} - \dfrac{725}{263}y^{12} + \dfrac{27500}{263}y^{10}z^{2} - \dfrac{138600}{263}y^{8}z^{4} + \dfrac{295680}{263}y^{6}z^{6} - \dfrac{190080}{263}y^{4}z^{8} + \dfrac{33792}{263}y^{2}z^{10} - \dfrac{1024}{263}z^{12}]+\cdots\} \end{align*} 　多重極展開と同じ結果となる。

内部ポテンシャルについての\(\rm{Taylor}\)展開

\(r = \sqrt{x^{2} + y^{2} + z^{2}} \ll a\)のとき、\(\xi =x/r,\ \)\(\eta =y/r,\ \)\(\zeta =z/r,\ \)\(t = r/a\)として、\(t\)について\(\rm{Taylor}\)展開をすると、まず、\(u_{L}(x, y, z, a)\)については、 \[r_{1}=a\sqrt{(t\xi)^{2} + (t\eta)^{2} + (t\zeta- 1)^{2}}=a\sqrt{1-2t\zeta+t^{2}} =a(1-\zeta t+\frac{1}{2}(1-\zeta^{2})t^{2}+\cdots ), \] \begin{align*} r_{2}=&a\sqrt{(t\xi - \tfrac{2}{\sqrt{5}}\cos (4\pi /5))^{2} + (t\eta - \tfrac{2}{\sqrt{5}}\sin (4\pi /5))^{2} + (t\zeta - \tfrac{1}{\sqrt{5}})^{2}} =a\sqrt{1-2t(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)+t^{2}}\\ =&a(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)t +\frac{1}{2}(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})t^{2}+\cdots ), \end{align*} \begin{align*} r_{3}=&a\sqrt{(t\xi - \tfrac{2}{\sqrt{5}}\cos (4\pi /5))^{2} + (t\eta + \tfrac{2}{\sqrt{5}}\sin (4\pi /5))^{2} + (t\zeta - \tfrac{2}{\sqrt{5}})^{2}} =a\sqrt{1-2t(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)+t^{2}}\\ =&a(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)t +\frac{1}{2}(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})t^{2}+\cdots ), \end{align*} \begin{align*} \tilde{L}_{12}=&\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}}\\ =&\ln\dfrac {2+\tfrac{4}{\sqrt{10+2\sqrt{5}}}-((\tfrac{2}{\sqrt{5}}\cos (4\pi /5)+1)\xi+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)t +\frac{1}{2}(2-\zeta^{2}-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})t^{2}+\cdots} {2-\tfrac{4}{\sqrt{10+2\sqrt{5}}}-((\tfrac{2}{\sqrt{5}}\cos (4\pi /5)+1)\xi+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)t +\frac{1}{2}(2-\zeta^{2}-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})t^{2}+\cdots } \\ =&\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +(\frac{5-\sqrt{5}}{20}\sqrt{10+2\sqrt{5}}(-\xi+2\zeta)+\frac{3-\sqrt{5}}{2}\eta)t \\ &+(\sqrt{10+2\sqrt{5}}(\frac{7-\sqrt{5}}{40}\xi^{2}-\frac{9-\sqrt{5}}{20}\xi\zeta +\frac{45-19\sqrt{5}}{40}\eta^{2}+\frac{19-5\sqrt{5}}{20}\zeta^{2}-\frac{3-\sqrt{5}}{4}) -\frac{10-3\sqrt{5}}{5}\xi\eta+\frac{25-7\sqrt{5}}{10}\eta\zeta)t^{2}\cdots , \end{align*} \begin{align*} \tilde{L}_{23}=&\ln{\frac{r_{2}+r_{3}+r_{23}}{r_{2}+r_{3}-r_{23}}}\\ =&\ln\dfrac {2+\tfrac{4}{\sqrt{10+2\sqrt{5}}}-(\tfrac{4}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\zeta)t +(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}-\tfrac{4}{5}\sin^{2}(4\pi /5)\eta^{2})t^{2}+\cdots} {2-\tfrac{4}{\sqrt{10+2\sqrt{5}}}-(\tfrac{4}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{2}{\sqrt{5}}\zeta)t +(1-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2}-\tfrac{4}{5}\sin^{2}(4\pi /5)\eta^{2})t^{2}+\cdots } \\ =&\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +\sqrt{10+2\sqrt{5}}(-\frac{5-\sqrt{5}}{10}\xi+\frac{3\sqrt{5}-5}{10}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{6-\sqrt{5}}{10}\xi^{2}+\frac{11-7\sqrt{5}}{10}\xi\zeta +\frac{5-2\sqrt{5}}{10}\eta^{2}+\frac{23-9\sqrt{5}}{20}\zeta^{2}-\frac{3-\sqrt{5}}{4})t^{2}\cdots , \end{align*} \begin{align*} \tilde{L}_{31}=&\ln{\frac{r_{3}+r_{1}+r_{31}}{r_{3}+r_{1}-r_{31}}}\\ =&\ln\dfrac {2+\tfrac{4}{\sqrt{10+2\sqrt{5}}}-((\tfrac{2}{\sqrt{5}}\cos (4\pi /5)+1)\xi-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)t +\frac{1}{2}(2-\zeta^{2}-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})t^{2}+\cdots} {2-\tfrac{4}{\sqrt{10+2\sqrt{5}}}-((\tfrac{2}{\sqrt{5}}\cos (4\pi /5)+1)\xi-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)t +\frac{1}{2}(2-\zeta^{2}-(\tfrac{2}{\sqrt{5}}\xi\cos (4\pi /5)-\tfrac{2}{\sqrt{5}}\eta\sin (4\pi /5)+\tfrac{1}{\sqrt{5}}\zeta)^{2})t^{2}+\cdots } \\ =&\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +(\frac{5-\sqrt{5}}{20}\sqrt{10+2\sqrt{5}}(-\xi+2\zeta)+\frac{3-\sqrt{5}}{2}\eta)t \\ &+(\sqrt{10+2\sqrt{5}}(\frac{7-\sqrt{5}}{40}\xi^{2}-\frac{9-\sqrt{5}}{20}\xi\zeta +\frac{45-19\sqrt{5}}{40}\eta^{2}+\frac{19-5\sqrt{5}}{20}\zeta^{2}-\frac{3-\sqrt{5}}{4}) +\frac{10-3\sqrt{5}}{5}\xi\eta-\frac{25-7\sqrt{5}}{10}\eta\zeta)t^{2}\cdots , \end{align*} より、 \begin{align*} u_{L}(x, y, z, a) =& \frac{a^{2}(1+(3-\sqrt{5})\xi t-\zeta t}{3(5-\sqrt{5})}\times \\ &\{(1-\dfrac{3+\sqrt{5}}{4}\xi t-\dfrac{3\sqrt{10+2\sqrt{5}}}{4}\eta t-\zeta t) [\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +(\frac{5-\sqrt{5}}{20}\sqrt{10+2\sqrt{5}}(-\xi+2\zeta)+\frac{3-\sqrt{5}}{2}\eta)t \\ &+(\sqrt{10+2\sqrt{5}}(\frac{7-\sqrt{5}}{40}\xi^{2}-\frac{9-\sqrt{5}}{20}\xi\zeta +\frac{45-19\sqrt{5}}{40}\eta^{2}+\frac{19-5\sqrt{5}}{20}\zeta^{2}-\frac{3-\sqrt{5}}{4}) -\frac{10-3\sqrt{5}}{5}\xi\eta+\frac{25-7\sqrt{5}}{10}\eta\zeta)t^{2}\cdots]\\ &+(1+\dfrac{3+\sqrt{5}}{2}\xi t+2\zeta t) [\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +\sqrt{10+2\sqrt{5}}(-\frac{5-\sqrt{5}}{10}\xi+\frac{3\sqrt{5}-5}{10}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{6-\sqrt{5}}{10}\xi^{2}+\frac{11-7\sqrt{5}}{10}\xi\zeta +\frac{5-2\sqrt{5}}{10}\eta^{2}+\frac{23-9\sqrt{5}}{20}\zeta^{2}-\frac{3-\sqrt{5}}{4})t^{2}\cdots]\\ &+(1-\dfrac{3+\sqrt{5}}{4}\xi t+\dfrac{3\sqrt{10+2\sqrt{5}}}{4}\eta t-\zeta t) [\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +(\frac{5-\sqrt{5}}{20}\sqrt{10+2\sqrt{5}}(-\xi+2\zeta)+\frac{3-\sqrt{5}}{2}\eta)t \\ &+(\sqrt{10+2\sqrt{5}}(\frac{7-\sqrt{5}}{40}\xi^{2}-\frac{9-\sqrt{5}}{20}\xi\zeta +\frac{45-19\sqrt{5}}{40}\eta^{2}+\frac{19-5\sqrt{5}}{20}\zeta^{2}-\frac{3-\sqrt{5}}{4}) +\frac{10-3\sqrt{5}}{5}\xi\eta-\frac{25-7\sqrt{5}}{10}\eta\zeta)t^{2}\cdots]\}\\ =&a^{2}[\frac{5+\sqrt{5}}{20}\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} \\ &+((\frac{5-\sqrt{5}}{10}\xi-\frac{5+\sqrt{5}}{20}\zeta)\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} -\frac{\sqrt{10+2\sqrt{5}}}{15}\xi+\frac{(3+\sqrt{5})\sqrt{10+2\sqrt{5}}}{60}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{37\sqrt{5}-95}{600}\xi^{2}+\frac{-7\sqrt{5}+5}{150}\xi\zeta+\frac{1-3\sqrt{5}}{120}\eta^{2} +\frac{15-7\sqrt{5}}{600}\zeta^{2}+\frac{\sqrt{5}-5}{40})t^{2}+\cdots ], \end{align*} \begin{align*} \sum_{i=0}^{19}u_{A}(x_{i}, y_{i}, z_{i}, a) &=a^{2}[(5+\sqrt{5})\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +\sqrt{10+2\sqrt{5}}(\frac{\sqrt{5}-5}{6}(\xi^{2}+\eta^{2}+\zeta^{2})+\frac{\sqrt{5}-5}{2})t^{2}+\cdots] \\ &=(5+\sqrt{5})\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2}\cdot a^{2} +\frac{2(\sqrt{5}-5)}{3}\sqrt{10+2\sqrt{5}}\cdot r^{2}+\cdots \end{align*} 　である。次に、\(u_{A}(x, y, z, a)\)については、 \begin{align*} S_{1}=&\arctan{\frac{\sqrt{10+2\sqrt{5}}((3-\sqrt{5})x-z+a)r_{1}}{-[\tfrac{1}{2}(3\sqrt{5}-7)x^{2}+4xz-4ax+\tfrac{1}{2}(3\sqrt{5}-15)y^{2}+(1-\sqrt{5})z^{2}+(2\sqrt{5}-2)az+(1-\sqrt{5})a^{2}]}}\\ =&\arctan{\frac{[1 - (\xi - \eta + \zeta)t]\sqrt{3-2t(\xi+\eta+\zeta) +t^{2}}}{-[(1 - (\xi - \eta + \zeta)t) (1 + \zeta t + \xi t) + 3\zeta\xi t^{2} - 3\eta t]}}\\ =&\arctan[\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}-2\sqrt{10+2\sqrt{5}}\xi t +\sqrt{10+2\sqrt{5}}(\frac{13+17\sqrt{5}}{8}\xi^{2}-2\xi\zeta-\frac{15+3\sqrt{5}}{8}\eta^{2} -\frac{1+\sqrt{5}}{8}\zeta^{2}+\frac{1+\sqrt{5}}{8})t^{2}+\cdots]\\ =&\frac{2\pi}{5}-\frac{3-\sqrt{5}}{4}\sqrt{10+2\sqrt{5}}\xi t +\sqrt{10+2\sqrt{5}}(\frac{17-5\sqrt{5}}{32}\xi^{2}+\frac{\sqrt{5}-3}{4}\xi\zeta+\frac{3\sqrt{5}-15}{32}\zeta^{2} +\frac{1-\sqrt{5}}{32}\zeta^{2}+\frac{\sqrt{5}-1}{32})t^{2}+\cdots, \end{align*} \begin{align*} S_{2}=&\arctan{\frac{\sqrt{10+2\sqrt{5}}((3-\sqrt{5})x-z+a)r_{2}} {-[(\sqrt{5}-5)x^{2}-2ax+(\sqrt{5}-5)z^{2}+4az+(\sqrt{5}-5)a^{2}-\sqrt{10+2\sqrt{5}}(xy+(3-\sqrt{5})yz+\tfrac{1}{2}(3-\sqrt{5})ay)]}}\\ =&\arctan{\frac{[1 - (\xi - \eta + \zeta)t]\sqrt{3+2t(\xi+\eta-\zeta) +t^{2}}}{-[(1 - (\xi - \eta + \zeta)t) (1 - \eta t + \zeta t) - 3\eta\zeta t^{2} + 3\xi t]}}\\ =&\arctan[\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1} +\sqrt{10+2\sqrt{5}}(\frac{2\sqrt{5}-5}{5}\xi-(3+\sqrt{5})\eta+\frac{5+\sqrt{5}}{5}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{4\sqrt{5}-37}{20}\xi^{2}+\frac{13\sqrt{5}-35}{10}\xi\eta-\frac{6\sqrt{5}+3}{5}\xi\zeta +\frac{9\sqrt{5}+10}{20}\eta^{2}-\frac{33\sqrt{5}+75}{5}\eta\zeta+\frac{39}{40}(\sqrt{5}+1)\zeta^{2}+\frac{1+\sqrt{5}}{8})t^{2}+\cdots]\\ =&\frac{2\pi}{5}+\sqrt{10+2\sqrt{5}}(\frac{11\sqrt{5}-25}{40}\xi-\frac{1}{2}\eta+\frac{5-\sqrt{5}}{20}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{-45\sqrt{5}+79}{160}\xi^{2}+\frac{17\sqrt{5}-45}{40}\xi\eta-\frac{\sqrt{5}+9}{40}\xi\zeta +\frac{7\sqrt{5}-25}{160}\eta^{2}-\frac{\sqrt{5}+10}{10}\eta\zeta+\frac{23\sqrt{5}-39}{160}\zeta^{2}+\frac{\sqrt{5}-1}{32})t^{2}+\cdots, \end{align*} \begin{align*} S_{3}=&\arctan{\frac{\sqrt{10+2\sqrt{5}}((3-\sqrt{5})x-z+a)r_{3}} {-[(\sqrt{5}-5)x^{2}-2ax+(\sqrt{5}-5)z^{2}+4az+(\sqrt{5}-5)a^{2}+\sqrt{10+2\sqrt{5}}(xy+(3-\sqrt{5})yz+\tfrac{1}{2}(3-\sqrt{5})ay)]}}\\ =&\arctan{\frac{[1 - (\xi - \eta + \zeta)t]\sqrt{3+2t(-\xi+\eta+\zeta)+t^{2}}}{-[(1 - (\xi - \eta + \zeta)t) (1 + \xi t - \eta t) - 3\xi\eta t^{2}+ 3\zeta t]}}\\ =&\arctan[\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1} +\sqrt{10+2\sqrt{5}}(\frac{2\sqrt{5}-5}{5}\xi+(3+\sqrt{5})\eta+\frac{5+\sqrt{5}}{5}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{4\sqrt{5}-37}{20}\xi^{2}-\frac{13\sqrt{5}-35}{10}\xi\eta-\frac{6\sqrt{5}+3}{5}\xi\zeta +\frac{9\sqrt{5}+10}{20}\eta^{2}+\frac{33\sqrt{5}+75}{5}\eta\zeta+\frac{39}{40}(\sqrt{5}+1)\zeta^{2}+\frac{1+\sqrt{5}}{8})t^{2}+\cdots]\\ =&\frac{2\pi}{5}+\sqrt{10+2\sqrt{5}}(\frac{11\sqrt{5}-25}{40}\xi+\frac{1}{2}\eta+\frac{5-\sqrt{5}}{20})t \\ &+\sqrt{10+2\sqrt{5}}(\frac{-45\sqrt{5}+79}{160}\xi^{2}-\frac{17\sqrt{5}-45}{40}\xi\eta-\frac{\sqrt{5}+9}{40}\xi\zeta +\frac{7\sqrt{5}-25}{160}\eta^{2}+\frac{\sqrt{5}+10}{10}\eta\zeta+\frac{23\sqrt{5}-39}{160}\zeta^{2}+\frac{\sqrt{5}-1}{32})t^{2}+\cdots, \end{align*} \begin{align*} &S_{1}+S_{2}+S_{3}-\pi\mathrm{sgn}(4x-(3+\sqrt{5})(z-a)) \\ =&\frac{\pi}{5}+\sqrt{10+2\sqrt{5}}(\frac{4\sqrt{5}-10}{5}\xi+\frac{5-\sqrt{5}}{10}\zeta)t \\ &+\sqrt{10+2\sqrt{5}}(\frac{-115\sqrt{5}+243}{160}\xi^{2}-\frac{6-\sqrt{5}}{5}\xi\zeta +\frac{29\sqrt{5}-125}{160}\eta^{2}+\frac{41\sqrt{5}-73}{160}\zeta^{2}+\frac{3\sqrt{5}-3}{32})t^{2}+\cdots \end{align*} より、 \begin{align*} u_{A}(x, y, z, a)=&-a^{2}[\frac{(5+2\sqrt{5})\pi}{150}+ (\pi(\frac{5+\sqrt{5}}{75}\xi-\frac{5+2\sqrt{5}}{75}\zeta)+ \sqrt{10+2\sqrt{5}}(-\frac{1}{15}\xi+\frac{3+\sqrt{5}}{60}\zeta))t \\ &(\pi(\frac{5-\sqrt{5}}{75}\xi^{2}-\frac{5+\sqrt{5}}{75}\xi\zeta+\frac{5+2\sqrt{5}}{150}\zeta^{2}) \\ &+\sqrt{10+2\sqrt{5}}(\frac{551\sqrt{5}-1855}{4800}\xi^{2}+\frac{20-7\sqrt{5}}{150}\xi\zeta -\frac{21\sqrt{5}+67}{960}\eta^{2}-\frac{101\sqrt{5}+435}{4800}\zeta^{2}+\frac{3\sqrt{5}+5}{320}))t^{2}+\cdots] \end{align*} \begin{align*} \sum_{i=0}^{19}u_{A}(x_{i}, y_{i}, z_{i}, a) =&-a^{2}[\frac{(10+4\sqrt{5})\pi}{15}+(\frac{2\pi}{3}+\frac{2(\sqrt{5}-5)}{3}\sqrt{10+2\sqrt{5}})t^{2}+\cdots ] \\ =&-\frac{(10+4\sqrt{5})\pi}{15}\cdot a^{2}-(\frac{2\pi}{3}+\frac{2(\sqrt{5}-5)}{3}\sqrt{10+2\sqrt{5}})r^{2}+\cdots , \end{align*} 　である。\(u_{L}(x, y, z, a)\)と\(u_{A}(x, y, z, a)\)の結果をまとめると、 \[ U(x,y,z)=G\rho [(-(5+\sqrt{5})\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +\frac{(10+4\sqrt{5})\pi}{15})a^{2}+\frac{2\pi}{3}r^{2}+\cdots] \] となる。さらに高次まで\(\rm{Taylor}\)展開を行うと以下のようになる。 \begin{align*} U(x,y,z)=&G\rho \{(-(5+\sqrt{5})\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +\frac{(10+4\sqrt{5})\pi}{15})a^{2}+\frac{2\pi}{3}r^{2} \\ &+\dfrac{1}{a^{4}}\cdot\dfrac{(\sqrt{5}-2)\sqrt{10+2\sqrt{5}}}{16} [x^{6} - \frac{42}{5}x^{5}z + 3x^{4}y^{2} - 18x^{4}z^{2} + 84x^{3}y^{2}z + 3x^{2}y^{4} - 36x^{2}y^{2}z^{2} + 24x^{2}z^{4} - 42xy^{4}z \\ &+ y^{6} - 18y^{4}z^{2} + 24y^{2}z^{4} - \frac{16}{5}z^{6}]\\ &+\dfrac{1}{a^{8}}\cdot\dfrac{(22909\sqrt{5}- 49693)\sqrt{10+2\sqrt{5}}}{288000} [x^{10} - \dfrac{495}{31}x^{9}z - \dfrac{4365}{62}x^{8}y^{2} + \dfrac{1575}{62}x^{8}z^{2} + \dfrac{3960}{31}x^{7}y^{2}z + \dfrac{4620}{31}x^{7}z^{3} + \dfrac{9660}{31}x^{6}y^{4} \\ &+ \dfrac{3150}{31}x^{6}y^{2}z^{2} - \dfrac{4200}{31}x^{6}z^{4} + \dfrac{6930}{31}x^{5}y^{4}z - \dfrac{41580}{31}x^{5}y^{2}z^{3} - \dfrac{5544}{31}x^{5}z^{5} - \dfrac{9975}{31}x^{4}y^{6} + \dfrac{4725}{31}x^{4}y^{4}z^{2} - \dfrac{12600}{31}x^{4}y^{2}z^{4} \\ &+ \dfrac{5040}{31}x^{4}z^{6} - \dfrac{23100}{31}x^{3}y^{4}z^{3} + \dfrac{55440}{31}x^{3}y^{2}z^{5} + \dfrac{2025}{31}x^{2}y^{8} + \dfrac{3150}{31}x^{2}y^{6}z^{2} - \dfrac{12600}{31}x^{2}y^{4}z^{4} + \dfrac{10080}{31}x^{2}y^{2}z^{6} - \dfrac{1440}{31}x^{2}z^{8} \\ &- \dfrac{2475}{31}xy^{8}z + \dfrac{23100}{31}xy^{6}z^{3} - \dfrac{27720}{31}xy^{4}z^{5} - \dfrac{125}{62}y^{10} + \dfrac{1575}{62}y^{8}z^{2} - \dfrac{4200}{31}y^{6}z^{4} + \dfrac{5040}{31}y^{4}z^{6} - \dfrac{1440}{31}y^{2}z^{8} + \dfrac{64}{31}z^{10}] \\ &+\dfrac{1}{a^{10}}\cdot\dfrac{(2610801- 1157463\sqrt{5})\sqrt{10+2\sqrt{5}}}{7040000} [x^{12} + \dfrac{2860}{263}x^{11}z - \dfrac{23122}{263}x^{10}y^{2} + \dfrac{5764}{263}x^{10}z^{2} - \dfrac{20020}{263}x^{9}y^{2}z \\ &- \dfrac{45760}{263}x^{9}z^{3} + \dfrac{78045}{263}x^{8}y^{4} + \dfrac{572220}{263}x^{8}y^{2}z^{2} - \dfrac{138600}{263}x^{8}z^{4} - \dfrac{62920}{263}x^{7}y^{4}z + \dfrac{366080}{263}x^{7}y^{2}z^{3} + \dfrac{128128}{263}x^{7}z^{5} - \dfrac{4620}{263}x^{6}y^{6} \\ &- \dfrac{2115960}{263}x^{6}y^{4}z^{2} - \dfrac{554400}{263}x^{6}y^{2}z^{4} + \dfrac{295680}{263}x^{6}z^{6} - \dfrac{40040}{263}x^{5}y^{6}z + \dfrac{640640}{263}x^{5}y^{4}z^{3} - \dfrac{1153152}{263}x^{5}y^{2}z^{5} - \dfrac{73216}{263}x^{5}z^{7} \\ &- \dfrac{84975}{263}x^{4}y^{8} + \dfrac{2448600}{263}x^{4}y^{6}z^{2} - \dfrac{831600}{263}x^{4}y^{4}z^{4} + \dfrac{887040}{263}x^{4}y^{2}z^{6} - \dfrac{190080}{263}x^{4}z^{8} + \dfrac{14300}{263}x^{3}y^{8}z - \dfrac{640640}{263}x^{3}y^{4}z^{5} \\ &+ \dfrac{732160}{263}x^{3}y^{2}z^{7} + \dfrac{20350}{263}x^{2}y^{10} - \dfrac{405900}{263}x^{2}y^{8}z^{2} - \dfrac{554400}{263}x^{2}y^{6}z^{4} + \dfrac{887040}{263}x^{2}y^{4}z^{6} - \dfrac{380160}{263}x^{2}y^{2}z^{8} + \dfrac{33792}{263}x^{2}z^{10} \\ &+ \dfrac{14300}{263}xy^{10}z - \dfrac{228800}{263}xy^{8}z^{3} + \dfrac{640640}{263}xy^{6}z^{5} - \dfrac{366080}{263}xy^{4}z^{7} - \dfrac{725}{263}y^{12} + \dfrac{27500}{263}y^{10}z^{2} - \dfrac{138600}{263}y^{8}z^{4} \\ &+ \dfrac{295680}{263}y^{6}z^{6} - \dfrac{190080}{263}y^{4}z^{8} + \dfrac{33792}{263}y^{2}z^{10} - \dfrac{1024}{263}z^{12}]+\cdots\} \end{align*} 　内部ポテンシャルについても、外部ポテンシャルと同様に正二十面体の対称性をもつ球面調和関数を用いて表すと以下のようになる。 \begin{align*} U(x,y,z)=&G\rho a^{2}[(-(5+\sqrt{5})\ln\frac{\sqrt{10+2\sqrt{5}}+2}{\sqrt{10+2\sqrt{5}}-2} +\frac{(10+4\sqrt{5})\pi}{15}) +\frac{2\pi}{3}\Bigl(\dfrac{r}{a}\Bigr)^{2} \\ &-\Bigl(\dfrac{r}{a}\Bigr)^{6}\cdot 2(\sqrt{5}-2)\sqrt{10+2\sqrt{5}}\sqrt{\dfrac{\pi}{143}}Ic_{6}(\theta,\varphi) +\Bigl(\dfrac{r}{a}\Bigr)^{10}\cdot\dfrac{(739\sqrt{5}-1603)\sqrt{10+2\sqrt{5}}}{450}\sqrt{\dfrac{\pi}{1729}}Ic_{10}(\theta,\varphi) \\ &-\Bigl(\dfrac{r}{a}\Bigr)^{12}\cdot\dfrac{(6618\sqrt{5}-14670)\sqrt{10+2\sqrt{5}}}{1375}\sqrt{\dfrac{\pi}{119}}Ic_{12}(\theta,\varphi)+\cdots ] \end{align*}
「正二十面体惑星の重力ポテンシャルの多重極展開」へ戻る 目次へ戻る 「正十二面体惑星の重力ポテンシャル」へ進む