正八面体惑星の重力ポテンシャルの\(\rm{Taylor}\)展開

　正八面体惑星の重力ポテンシャルについて、 \[u_{1}(x, y, z, a)=u_{L}(x, y, z, a)+u_{A}(x, y, z, a)\] \[u_{L}(x, y, z, a) = \frac{a-(x+y+z)}{6\sqrt{2}}[(-x-y+2z+a)\tilde{L}_{12}+(2x-y-z+a)\tilde{L}_{23}+(-x+2y-z+a)\tilde{L}_{31}]\] \[u_{A}(x, y, z, a) =-\frac{(a-(x+y+z))^2}{6}[S_{1}+S_{2}+S_{3}-\pi\mathrm{sgn}(a-(x+y+z))]\] 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、 \[r_{1}=r\sqrt{(\xi -\alpha)^{2}+\eta^{2}+\zeta ^{2}}=r\sqrt{1-2\alpha\xi +\alpha^{2}} =r(1-\xi\alpha+\frac{1}{2}(1-\xi^{2})\alpha^{2}+\cdots ), \] \[r_{2}=r\sqrt{\xi ^{2}+(\eta -\alpha)^{2}+\zeta^{2}}=r\sqrt{1-2\alpha\eta +\alpha^{2}} =r(1-\eta\alpha+\frac{1}{2}(1-\eta^{2})\alpha^{2}+\cdots ), \] \[r_{3}=r\sqrt{\xi ^{2}+\eta ^{2}+(\zeta -\alpha)^{2}}=r\sqrt{1-2\alpha\zeta +\alpha^{2}} =r(1-\zeta\alpha+\frac{1}{2}(1-\zeta^{2})\alpha^{2}+\cdots ), \] \begin{align*} \tilde{L}_{12}&=\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}} =\ln\dfrac{2+(\sqrt{2}-\xi -\eta)\alpha+(1-(\xi^{2}+\eta^{2})/2)\alpha^{2}+\cdots} {2-(\sqrt{2}+\xi +\eta)\alpha+(1-(\xi^{2}+\eta^{2})/2)\alpha^{2}+\cdots } \\ &=\sqrt{2}\alpha+\frac{1}{\sqrt{2}}(\xi+\eta)\alpha^{2}+\frac{1}{\sqrt{2}}(\xi^{2}+\xi\eta+\eta^{2}-\frac{2}{3})\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+(\sqrt{2}-\eta -\zeta)\alpha+(1-(\eta^{2}+\zeta^{2})/2)\alpha^{2}+\cdots} {2-(\sqrt{2}+\eta +\zeta)\alpha+(1-(\eta^{2}+\zeta^{2})/2)\alpha^{2}+\cdots } \\ &=\sqrt{2}\alpha+\frac{1}{\sqrt{2}}(\eta+\zeta)\alpha^{2}+\frac{1}{\sqrt{2}}(\eta^{2}+\eta\zeta+\zeta^{2}-\frac{2}{3})\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+(\sqrt{2}-\zeta -\xi)\alpha+(1-(\zeta^{2}+\xi^{2})/2)\alpha^{2}+\cdots }{2-(\sqrt{2}+\zeta +\xi)\alpha+(1-(\zeta^{2}+\xi^{2})/2)\alpha^{2}+\cdots } \\ &=\sqrt{2}\alpha+\frac{1}{\sqrt{2}}(\zeta+\xi)\alpha^{2}+\frac{1}{\sqrt{2}}(\zeta^{2}+\zeta\xi+\xi^{2}-\frac{2}{3})\alpha^{3}+\cdots , \end{align*} より、 \begin{align*} u_{L}(x, y, z, a) =&\frac{r^{2}}{6\sqrt{2}}(\alpha-(\xi +\eta +\zeta ))\times \\ &[(-\xi -\eta+2\zeta+\alpha)(\sqrt{2}\alpha+\frac{1}{\sqrt{2}}(\xi+\eta)\alpha^{2}+\frac{1}{\sqrt{2}}(\xi^{2}+\xi\eta+\eta^{2}-\frac{2}{3})\alpha^{3}+\cdots)\\ &+( 2\xi-\eta -\zeta +\alpha)(\sqrt{2}\alpha+\frac{1}{\sqrt{2}}(\eta+\zeta)\alpha^{2}+\frac{1}{\sqrt{2}}(\eta^{2}+\eta\zeta+\zeta^{2}-\frac{2}{3})\alpha^{3}+\cdots)\\ &+(-\xi+2\eta -\zeta +\alpha)(\sqrt{2}\alpha+\frac{1}{\sqrt{2}}(\zeta+\xi)\alpha^{2}+\frac{1}{\sqrt{2}}(\zeta^{2}+\zeta\xi+\xi^{2}-\frac{2}{3})\alpha^{3}+\cdots)]\\ =&r^{2}[(\frac{1}{6}(\xi^{3}+\eta^{3}+\zeta^{3})-\frac{1}{2}(\xi\eta\zeta+\xi+\eta+\zeta))\alpha^{2}\\ &+(\frac{1}{6}(\xi^{3}+\eta^{3}+\zeta^{3})(\xi+\eta+\zeta)-\frac{1}{2}\xi\eta\zeta(\xi+\eta+\zeta)-\frac{1}{6}(\xi\eta+\eta\zeta+\zeta\xi)+\frac{1}{6})\alpha^{3}+\cdots]\\ \end{align*} \begin{align*} \sum_{i,j,k=0,1}u_{L}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a) &=r^{2}[(\frac{4}{3}(\xi^{4}+\eta^{4}+\zeta^{4})+\frac{4}{3})\alpha^{3}+\cdots]\\ &=(\frac{4}{3}(\xi^{4}+\eta^{4}+\zeta^{4})+\frac{4}{3})a^{3}r^{-1}+\cdots] \end{align*} 　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、\(\mathbf{v}_{1}=(a,0,0),\ \)\(\mathbf{v}_{2}=(0,a,0),\ \)\(\mathbf{v}_{3}=(0,0,a)\ \)とすると、 \[ \mathbf{n}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}),\ \mathbf{r}_{1}\cdot\mathbf{n}=\frac{1}{\sqrt{3}}(x+y+z-a)=\frac{1}{\sqrt{3}}r(\xi+\eta+\zeta-\alpha) \] \[ (\mathbf{r}_{1}\cdot\mathbf{r}_{2})=|\mathbf{r}|^{2}-((\mathbf{v}_{1}+\mathbf{v}_{2})\cdot\mathbf{r})+(\mathbf{v}_{1}\cdot\mathbf{v}_{2})=r^{2}-a(x+y)=r^{2}(1-\alpha(\xi+\eta)), \] \[ (\mathbf{r}_{2}\cdot\mathbf{r}_{3})=|\mathbf{r}|^{2}-((\mathbf{v}_{2}+\mathbf{v}_{3})\cdot\mathbf{r})+(\mathbf{v}_{2}\cdot\mathbf{v}_{3})=r^{2}-a(y+z)=r^{2}(1-\alpha(\eta+\zeta)), \] \[ (\mathbf{r}_{3}\cdot\mathbf{r}_{1})=|\mathbf{r}|^{2}-((\mathbf{v}_{3}+\mathbf{v}_{1})\cdot\mathbf{r})+(\mathbf{v}_{3}\cdot\mathbf{v}_{1})=r^{2}-a(z+x)=r^{2}(1-\alpha(\zeta+\xi)), \] \[ |\mathbf{r}_{1}|=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{1}\cdot\mathbf{r}+|\mathbf{v}_{1}|^{2})} =\sqrt{r^{2}-2ax+a^{2}}=r\sqrt{1-2\alpha\xi+\alpha^{2}}=r(1-\xi\alpha+\frac{1}{2}(1-\xi^{2})\alpha^{2}+\cdots), \] \[ |\mathbf{r}_{2}|=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{2}\cdot\mathbf{r}+|\mathbf{v}_{2}|^{2})} =\sqrt{r^{2}-2ay+a^{2}}=r\sqrt{1-2\alpha\eta+\alpha^{2}}=r(1-\eta\alpha+\frac{1}{2}(1-\eta^{2})\alpha^{2}+\cdots), \] \[ |\mathbf{r}_{3}|=\sqrt{|\mathbf{r}|^{2}-2\mathbf{v}_{3}\cdot\mathbf{r}+|\mathbf{v}_{3}|^{2})} =\sqrt{r^{2}-2az+a^{2}}=r\sqrt{1-2\alpha\zeta+\alpha^{2}}=r(1-\zeta\alpha+\frac{1}{2}(1-\zeta^{2})\alpha^{2}+\cdots), \] \begin{align*} [\mathbf{r}_{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}\\ =&a^{3}-(x+y+z)a^{2}=r^{3}(\alpha^{3}-(\xi+\eta+\zeta)\alpha^{2}) \end{align*} より、 \[ \arctan\frac{ [\mathbf{r}_{1}\times\mathbf{r}_{2}]\cdot\mathbf{r}_{3}} {|\mathbf{r}_{1}||\mathbf{r}_{2}||\mathbf{r}_{3}|+(\mathbf{r}_{1}\cdot\mathbf{r}_{2})|\mathbf{r}_{3}|+(\mathbf{r}_{2}\cdot\mathbf{r}_{3})|\mathbf{r}_{1}|+(\mathbf{r}_{3}\cdot\mathbf{r}_{1})|\mathbf{r}_{2}|} \\ =-\frac{1}{4}(\xi+\eta+\zeta)\alpha-\frac{1}{2}(\xi\eta+\eta\zeta+\zeta\xi)\alpha^{2}+\cdots \] \[ u_{A}(x, y, z, a)=r[\frac{1}{12}(\xi+\eta+\zeta)^{3}\alpha+(\frac{1}{12}(\xi+\eta+\zeta)^{4}-\frac{1}{4}(\xi+\eta+\zeta)^{2})\alpha^{2}+\cdots] \] \begin{align*} &\sum_{i,j,k=0,1}u_{A}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a)\\ =&r^{2}[(\frac{2}{3}(\xi^{4}+\eta^{4}+\zeta^{4})+4(\xi^{2}\eta^{2}+\eta^{2}\zeta^{2}+\zeta^{2}\xi^{2}) -2(\xi^{2}+\eta^{2}+\zeta^{2}))\alpha^{3}+\cdots] \\ =&(\frac{2}{3}(\xi^{4}+\eta^{4}+\zeta^{4})+4(\xi^{2}\eta^{2}+\eta^{2}\zeta^{2}+\zeta^{2}\xi^{2}) -2(\xi^{2}+\eta^{2}+\zeta^{2}))a^{3}r^{-1}+\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、 \begin{align*} U(x,y,z)=&-G\rho\sum_{i,j,k=0,1}[u_{L}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a)+u_{A}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a)]\\ =&-G\rho [(2(\xi^{2}+\eta^{2}+\zeta^{2})^{2}-2(\xi^{2}+\eta^{2}+\zeta^{2})+\frac{4}{3})a^{3}r^{-1}+\cdots]\\ =&-G\rho [\frac{4a^{3}}{3}r^{-1}+\cdots] \end{align*} 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{4a^{3}}{3r} -\dfrac{a^{7}}{r^{9}}\cdot\dfrac{1}{30}[(x^{4}+y^{4}+z^{4})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})] \\ &-\dfrac{a^{9}}{r^{13}}\cdot\dfrac{1}{504}[(x^{6}+y^{6}+z^{6})-\frac{15}{2}(x^{4}y^{2}+x^{2}y^{4}+y^{4}z^{2}+y^{2}z^{4}+z^{4}x^{2}+z^{2}x^{4})+90x^{2}y^{2}z^{2}] \\ &-\dfrac{a^{11}}{r^{17}}\cdot\dfrac{1}{160}[(x^{8}+y^{8}+z^{8})-14(x^{6}y^{2}+x^{2}y^{6}+y^{6}z^{2}+y^{2}z^{6}+z^{6}x^{2}+z^{2}x^{6})+35(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})] \\ &-\dfrac{a^{13}}{r^{21}}\cdot\dfrac{1}{1056}[(x^{10}+y^{10}+z^{10})-\frac{45}{2}(x^{8}y^{2}+x^{2}y^{8}+y^{8}z^{2}+y^{2}z^{8}+z^{8}x^{2}+z^{2}x^{8}) \\ &+21(x^{6}y^{4}+x^{4}y^{6}+y^{6}z^{4}+y^{4}z^{6}+z^{6}x^{4}+z^{4}x^{6})+504(x^{6}y^{2}z^{2}+y^{6}z^{2}x^{2}+z^{6}x^{2}y^{2})-630(x^{4}y^{4}z^{2}+y^{4}z^{4}x^{2}+z^{4}x^{4}y^{2})]\\ &-\dfrac{a^{15}}{r^{25}}\cdot\dfrac{731}{698880}[(x^{12}+y^{12}+z^{12})-33(x^{10}y^{2}+x^{2}y^{10}+y^{10}z^{2}+y^{2}z^{10}+z^{10}x^{2}+z^{2}x^{10}) \\ &+\frac{9745}{43}(x^{8}y^{4}+x^{4}y^{8}+y^{8}z^{4}+y^{4}z^{8}+z^{8}x^{4}+z^{4}x^{8})-\frac{17353}{43}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})+\frac{5385}{43}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &-\frac{12565}{43}(x^{6}y^{4}z^{2}+x^{6}y^{2}z^{4}+y^{6}z^{4}x^{2}+y^{6}z^{2}x^{4}+z^{6}x^{4}y^{2}+z^{6}x^{2}y^{4})+\frac{62825}{43}x^{4}y^{4}z^{4}]+\cdots\} \end{align*}

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、 \[r_{1}=a\sqrt{(1-t\xi)^{2}+(t\eta )^{2}+(t\zeta)^{2}}=a\sqrt{1-2t\xi +t^{2}}=a(1-\xi t+\frac{1}{2}(1-\xi ^{2})t^{2}+\cdots ),\] \[r_{2}=a\sqrt{(t\xi)^{2}+(1-t\eta )^{2}+(t\zeta )^{2}}=a\sqrt{1-2t\eta +t^{2}}=a(1-\eta t+\frac{1}{2}(1-\eta ^{2})t^{2}+\cdots ),\] \[r_{3}=a\sqrt{(t\xi)^{2}+(t\eta)^{2}+(1-t\zeta )^{2}}=a\sqrt{1-2t\zeta +t^{2}}=a(1-\zeta t+\frac{1}{2}(1-\zeta ^{2})t^{2}+\cdots ),\] \begin{align*} \tilde{L}_{12}&=\ln{\frac{r_{1}+r_{2}+r_{12}}{r_{1}+r_{2}-r_{12}}} =\ln\dfrac{2+\sqrt{2}-(\xi +\eta) t+(1-(\xi^{2}+\eta^{2})/2)t^{2}+\cdots} {2-\sqrt{2}-(\xi +\eta) t+(1-(\xi^{2}+\eta^{2})/2)t^{2}+\cdots } \\ &=\ln(3+2\sqrt{2})+\sqrt{2}(\xi+\eta)t+\sqrt{2}(\frac{3}{2}(\xi^{2}+\eta^{2})+2\xi\eta-1)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+\sqrt{2}-(\eta +\zeta) t+(1-(\eta^{2}+\zeta^{2})/2)t^{2}+\cdots} {2-\sqrt{2}-(\eta +\zeta) t+(1-(\eta^{2}+\zeta^{2})/2)t^{2}+\cdots } \\ &=\ln(3+2\sqrt{2})+\sqrt{2}(\eta+\zeta)t+\sqrt{2}(\frac{3}{2}(\eta^{2}+\zeta^{2})+2\eta\zeta-1)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+\sqrt{2}-(\zeta +\xi) t+(1-(\zeta^{2}+\xi^{2})/2)t^{2}+\cdots }{2-\sqrt{2}-(\zeta +\xi) t+(1-(\zeta^{2}+\xi^{2})/2)t^{2}+\cdots } \\ &=\ln(3+2\sqrt{2})+\sqrt{2}(\zeta+\xi)t+\sqrt{2}(\frac{3}{2}(\zeta^{2}+\xi^{2})+2\zeta\xi-1)t^{2}+\cdots , \end{align*} より、 \begin{align*} u_{L}(x, y, z, a) =& \frac{a^{2}}{6\sqrt{2}}(1-(\xi +\eta +\zeta )t)\times \\ &\{[((-\xi -\eta+2\zeta)t +1)(\ln(3+2\sqrt{2})+\sqrt{2}( \xi +\eta)t+\sqrt{2}(\frac{3}{2}(\xi^{2}+\eta^{2})+2\xi\eta-1)t^{2}+\cdots)]\\ &+[(( 2\xi-\eta -\zeta)t +1)(\ln(3+2\sqrt{2})+\sqrt{2}(\eta+\zeta)t+\sqrt{2}(\frac{3}{2}(\eta^{2}+\zeta^{2})+2\eta\zeta-1)t^{2}+\cdots)]\\ &+[((-\xi+2\eta -\zeta)t +1)(\ln(3+2\sqrt{2})+\sqrt{2}(\zeta+\xi)t+\sqrt{2}(\frac{3}{2}(\zeta^{2}+\xi^{2})+2\zeta\xi-1)t^{2}+\cdots)]\}\\ =& a^{2}[\frac{\sqrt{2}}{4}\ln(3+2\sqrt{2}) -(\frac{\sqrt{2}}{4}\ln(3+2\sqrt{2})-\frac{1}{3})(\xi+\eta+\zeta)t-\frac{2}{3}t^{2}+\cdots] \\ \end{align*} \begin{align*} \sum_{i,j,k=0,1}u_{L}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a) =&a^{2}[2\sqrt{2}\ln(3+2\sqrt{2})-\frac{16}{3}t^{2}+\cdots] \\ =&2\sqrt{2}\ln(3+2\sqrt{2})a^{2}-\frac{16}{3}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{(x+y+z-a)r_{1}}{-[(y+z)(x+y+z-a)-3yz]}} =\frac{\pi}{2}-\arctan{\frac{-[(y+z)(x+y+z-a)-3yz]}{(x+y+z-a)r_{1}}}\\ &=\frac{\pi}{2}-\arctan{\frac{-[(\eta+\zeta)((\xi+\eta+\zeta)t^{2}-t)-3\eta\zeta t^{2}]}{((\xi+\eta+\zeta)t-1)\sqrt{1-2t\xi +t^{2}}}}\\ &=\frac{\pi}{2}-[(\eta+\zeta)t+(\xi\eta+3\eta\zeta+\zeta\xi)t^{2}+\cdots], \end{align*} \begin{align*} S_{2}&=\arctan{\frac{(x+y+z-a)r_{2}}{-[(z+x)(x+y+z-a)-3zx]}} =\frac{\pi}{2}-\arctan{\frac{-[(z+x)(x+y+z-a)-3zx]}{(x+y+z-a)r_{2}}}\\ &=\frac{\pi}{2}-\arctan{\frac{-[(\zeta+\xi)((\xi+\eta+\zeta)t^{2}-t)-3\zeta\xi t^{2}]}{((\xi+\eta+\zeta)t-1)\sqrt{1-2t\eta +t^{2}}}}\\ &=\frac{\pi}{2}-[(\zeta+\xi)t+(\xi\eta+\eta\zeta+3\zeta\xi)t^{2}+\cdots], \end{align*} \begin{align*} S_{3}&=\arctan{\frac{(x+y+z-a)r_{3}}{-[(x+y)(x+y+z-a)-3xy]}} =\frac{\pi}{2}-\arctan{\frac{-[(x+y)(x+y+z-a)-3xy]}{(x+y+z-a)r_{3}}}\\ &=\frac{\pi}{2}-\arctan{\frac{-[(\xi+\eta)((\xi+\eta+\zeta)t^{2}-t)-3\xi\eta t^{2}]}{((\xi+\eta+\zeta)t-1)\sqrt{1-2t\zeta +t^{2}}}}\\ &=\frac{\pi}{2}-[(\xi+\eta)t+(3\xi\eta+\eta\zeta+\zeta\xi)t^{2}+\cdots], \end{align*} \[S_{1}+S_{2}+S_{3}-\pi\mathrm{sgn}(a-(x+y+z)) =\frac{\pi}{2}-[2(\xi+\eta+\zeta)t+5(\xi\eta+\eta\zeta+\zeta\xi)t^{2}+\cdots]\] より、 \begin{align*} u_{A}(x, y, z, a)=&-\frac{a^{2}}{6}(1-(\xi+\eta+\zeta)t)^{2}\{\frac{\pi}{2}-[2(\xi+\eta+\zeta)t+5(\xi\eta+\eta\zeta+\zeta\xi)t^{2}+\cdots]\}\\ =&a^{2}[-\frac{\pi}{12}(1-(\xi+\eta+\zeta)t)^{2}-\frac{1}{3}(\xi+\eta+\zeta)t+(\frac{2}{3}+\frac{1}{2}(\xi\eta+\eta\zeta+\zeta\xi))t^{2}+\cdots], \end{align*} \[ \sum_{i,j,k=0,1}u_{A}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a) =-\frac{2\pi}{3}(a^{2}+r^{2})+\frac{16}{3}r^{2}+\cdots \] 　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、 \begin{align*} U(x,y,z)=&-G\rho\sum_{i,j,k=0,1}[u_{L}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a)+u_{A}((-1)^{i}x, (-1)^{j}y, (-1)^{k}z, a)]\\ =&G\rho[(-2\sqrt{2}\ln(3+2\sqrt{2})+\frac{2\pi}{3})a^{2}+\frac{2\pi}{3}r^{2}+\cdots] \end{align*} 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\{(-2\sqrt{2}\ln(3+2\sqrt{2})+\frac{2\pi}{3})a^{2}+\frac{2\pi}{3}r^{2} \\ &-\dfrac{1}{a^{2}}\cdot\dfrac{2}{3}[(x^{4}+y^{4}+z^{4})-3(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})] \\ &+\dfrac{1}{a^{4}}\cdot\dfrac{1}{5}[(x^{6}+y^{6}+z^{6})-\frac{15}{2}(x^{4}y^{2}+x^{2}y^{4}+y^{4}z^{2}+y^{2}z^{4}+z^{4}x^{2}+z^{2}x^{4})+90x^{2}y^{2}z^{2}] \\ &-\dfrac{1}{a^{6}}\cdot\dfrac{31}{168}[(x^{8}+y^{8}+z^{8})-14(x^{6}y^{2}+x^{2}y^{6}+y^{6}z^{2}+y^{2}z^{6}+z^{6}x^{2}+z^{2}x^{6})+35(x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})] \\ &+\dfrac{1}{a^{8}}\cdot\dfrac{23}{144}[(x^{10}+y^{10}+z^{10})-\frac{45}{2}(x^{8}y^{2}+x^{2}y^{8}+y^{8}z^{2}+y^{2}z^{8}+z^{8}x^{2}+z^{2}x^{8}) \\ &+21(x^{6}y^{4}+x^{4}y^{6}+y^{6}z^{4}+y^{4}z^{6}+z^{6}x^{4}+z^{4}x^{6})+504(x^{6}y^{2}z^{2}+y^{6}z^{2}x^{2}+z^{6}x^{2}y^{2})-630(x^{4}y^{4}z^{2}+y^{4}z^{4}x^{2}+z^{4}x^{4}y^{2})]\\ &-\dfrac{1}{a^{10}}\cdot\dfrac{59}{320}[(x^{12}+y^{12}+z^{12})-33(x^{10}y^{2}+x^{2}y^{10}+y^{10}z^{2}+y^{2}z^{10}+z^{10}x^{2}+z^{2}x^{10}) \\ &+\frac{18405}{118}(x^{8}y^{4}+x^{4}y^{8}+y^{8}z^{4}+y^{4}z^{8}+z^{8}x^{4}+z^{4}x^{8})-\frac{12138}{59}(x^{6}y^{6}+y^{6}z^{6}+z^{6}x^{6})+\frac{32400}{59}(x^{8}y^{2}z^{2}+y^{8}z^{2}x^{2}+z^{8}x^{2}y^{2}) \\ &-\frac{75600}{59}(x^{6}y^{4}z^{2}+x^{6}y^{2}z^{4}+y^{6}z^{4}x^{2}+y^{6}z^{2}x^{4}+z^{6}x^{4}y^{2}+z^{6}x^{2}y^{4})+\frac{378000}{59}x^{4}y^{4}z^{4}]+\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四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(r,\theta,\varphi)=&G\rho a^{2}[(-2\sqrt{2}\ln(3+2\sqrt{2})+\frac{2\pi}{3}) +\frac{2\pi}{3}\Bigl(\dfrac{r}{a}\Bigr)^{2} -\Bigl(\dfrac{r}{a}\Bigr)^{4}\cdot\dfrac{1}{6}\sqrt{\dfrac{\pi}{21}}Te_{4}(\theta,\varphi)\\ &+\Bigl(\dfrac{r}{a}\Bigr)^{6}\cdot\dfrac{1}{40}\sqrt{\dfrac{\pi}{26}}Te_{6,o}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{8}\cdot\dfrac{31}{2688}\sqrt{\dfrac{\pi}{561}}Te_{8}(\theta,\varphi) -\Bigl(\dfrac{r}{a}\Bigr)^{10}\cdot\dfrac{23}{96}\sqrt{\dfrac{\pi}{910}}Te_{10,o}(\theta,\varphi)\\ &-\Bigl(\dfrac{r}{a}\Bigr)^{12}\cdot( \dfrac{729}{8200}\sqrt{\dfrac{41\pi}{11}}Te_{12,o8}(\theta,\varphi) +\dfrac{6911}{2099200}\sqrt{\dfrac{246\pi}{676039}}Te_{12,o12}(\theta,\varphi))+\cdots] \end{align*}
「正八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 「立方体惑星の重力ポテンシャルの多重極展開」へ進む