Here’s how to get MathJax up and running for your blog: part I, part II. The three tests below are text lifted from elsewhere.

Test 1

Consider first what we shall call the direct geometry case, in which we use only the zenith angle $z$ and bypass the geocentric angle $\theta$. The length of side $\overline{CM}$ follows from the right triangle $\widehat{CMP}$:

$$\begin{equation}\begin{array}[b]{ccl}\left(R+H\right)^{2} & = & \left(D\sin z\right)^{2}+\left(R+h+D\cos z\right)^{2}\\ \\ & = & D^{2}+\left(R+h\right)^{2}+2\left(R+h\right)D\cos z\end{array}\label{eq:R+H-test}\end{equation}$$

or

\begin{equation}D^{2}+2\left(R+h\right)D\cos z-\left[\left(R+H\right)^{2}-\left(R+h\right)^{2}\right]=0\label{eq:D eqn-test}\end{equation}

with solution

\begin{equation}\begin{array}[b]{ccl}D & = & -\left(R+h\right)\cos z\pm\sqrt{\left(R+h\right)^{2}\cos^{2}z+\left[\left(R+H\right)^{2}-\left(R+h\right)^{2}\right]}\\ \\& = & \left(R+h\right)\left(\sqrt{\cos^{2}z+\dfrac{\left(R+H\right)^{2}-\left(R+h\right)^{2}}{\left(R+h\right)^{2}}}-\cos z\right)\end{array}\label{eq:D soln quadratic ugly-test}\end{equation}

where the geometry of the problem requires the positive root. For convenience, define

\begin{equation}\epsilon\equiv\dfrac{H}{R}\quad\mathrm{and}\quad\xi\equiv\dfrac{h}{R}\label{eq:eps and xsi defs-test}\end{equation}

Then we can write eq. \eqref{eq:D soln quadratic ugly-test} as

\begin{equation}D=\left(R+h\right)\left(\sqrt{\cos^{2}z+\left(\dfrac{1+\epsilon}{1+\xi}\right)^{2}-1}-\cos z\right)\label{eq:D soln quadratic-test}\end{equation}

Eq. \eqref{eq:D soln quadratic-test} has the disadvantage of subtraction of two nearly equal numbers.

Test 2

We would like to know what is the radius $\bar{r}$ of the center of mass

of a grid cell of inner radius $r_{1}$ and outer radius $r_{2}$. In polar coordinates $\left(r,\theta\right)$ an infinitesimal area element is $dA=r\,dr\,d\theta$, so

\begin{equation}\bar{r}=\frac{1}{\Delta A}\intop_{0}^{\Delta\theta}\intop_{r_{1}}^{r_{2}}r\,dA=\frac{1}{\Delta A}\intop_{0}^{\Delta\theta}\intop_{r_{1}}^{r_{2}}r^{2}dr\,d\theta\label{eq: area-weighted r integral-test}\end{equation}

where $\Delta A=\frac{\Delta\theta}{2\pi}\cdot\pi\left(r_{2}^{2}-r_{1}^{2}\right)$.

Thus,

\begin{equation}\Delta A=\frac{\Delta\theta}{2}\left(r_{2}^{2}-r_{1}^{2}\right)\label{eq: cell area-test}\end{equation}

and

\begin{equation}\bar{r}=\frac{1}{3}\frac{\Delta\theta}{\Delta A}\left(r_{2}^{3}-r_{1}^{3}\right)=\frac{2}{3}\frac{r_{2}^{2}+r_{1}r_{2}+r_{1}^{2}}{r_{1}+r_{2}}\label{eq: area-weighted r-test}\end{equation}

[…]

Thus, we have the bootstrapping scheme

\begin{equation}\begin{array}{rclcrcl}\bar{r}_{0} & = & \dfrac{2}{3\Delta^{2}}\left(r_{2,0}^{3}-r_{1,0}^{3}\right) & & r_{2,0} & = & \sqrt{r_{1,0}^{2}+\Delta^{2}}\\& \vdots & & & & \vdots\\\bar{r}_{k} & = & \dfrac{2}{3\Delta^{2}}\left(r_{2,\,k}^{3}-r_{2,\,k-1}^{3}\right) & & r_{2,\,k} & = & \sqrt{r_{2,\,k-1}^{2}+\Delta^{2}}\\& \vdots & & & & \vdots\\\bar{r}_{N_{r}-1} & = & \dfrac{2}{3\Delta^{2}}\left(r_{2,\,N_{r}-1}^{3}-r_{2,\,N_{r}-2}^{3}\right) & & r_{2,\,N_{r}-1} & = & \sqrt{r_{2,\,N_{r}-2}^{2}+\Delta^{2}}\end{array}\label{eq: bootstrap scheme}\end{equation}

where, again, we start with $r_{1,0}=r_{min}$ .

Test 3

Now, $-\widehat{z}\times{\left(\widehat{z}\times\overrightarrow{r}\right)}=\overrightarrow{r}-{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}$, so

\begin{equation}\overrightarrow{r}^{\prime\prime}+2\widehat{z}\times\overrightarrow{r}^{\prime}=\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}{\left(\overrightarrow{r}+\overrightarrow{\nabla}U\right)}-{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}\label{}\end{equation}

Define a new effective potential

\begin{equation}\mathrm{\Omega}=\frac{1}{2}r^{2}+U=\frac{1}{2}r^{2}+\frac{{1-\mathrm{\mu}}}{r_{1}}+\frac{\mathrm{\mu}}{r_{2}}\label{EQUATION.5d0b51dc-3a17-4d57-95ed-8e8768257778}\end{equation}

where

\begin{equation}r_{1}=\sqrt{{{\left(x+\mathrm{\mu}\right)}^{2}+y^{2}+z^{2}}}\hspace{2em}r_{2}=\sqrt{{{\left(x-1+\mathrm{\mu}\right)}^{2}+y^{2}+z^{2}}}\label{EQUATION.10d1bacb-a0cf-4bdc-8b6d-c72d845b975b}\end{equation}

Then we find the satisfying result

\begin{equation}\overrightarrow{r}^{\prime\prime}+2\widehat{z}\times\overrightarrow{r}^{\prime}+{\left(\widehat{z}\cdot\overrightarrow{r}\right)}\widehat{z}=\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\overrightarrow{\nabla}\mathrm{\Omega}\label{EQUATION.7aeaeb03-1226-46ab-815a-4b28e71a84a5}\end{equation}

The individual components of \eqref{EQUATION.7aeaeb03-1226-46ab-815a-4b28e71a84a5} are

\begin{equation}\begin{aligned}x^{\prime\prime}-2y^{\prime} & =\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\frac{{\partial\mathrm{\Omega}}}{{\partial x}}\\y^{\prime\prime}+2x^{\prime} & =\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\frac{{\partial\mathrm{\Omega}}}{{\partial y}}\\z^{\prime\prime}+z\hspace{0.9em} & =\frac{1}{{1+e_{p}\mathrm{cos}\mathrm{\theta}}}\frac{{\partial\mathrm{\Omega}}}{{\partial z}}\end{aligned}\label{}\end{equation}

where

\begin{equation}\begin{array}{rcl}\overrightarrow{\nabla}\mathrm{\Omega} & = & \left[\begin{matrix}x-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}\left(x+\mathrm{\mu}\right)-\dfrac{\mathrm{\mu}}{r_{2}^{3}}\left(x-1+\mathrm{\mu}\right)\\y\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mathrm{\mu}}{r_{2}^{3}}\right)\\z\left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mathrm{\mu}}{r_{2}^{3}}\right)\end{matrix}\right]\\ \\& = & \left(1-\dfrac{1-\mathrm{\mu}}{r_{1}^{3}}-\dfrac{\mu}{r_{2}^{3}}\right)\overrightarrow{r}-\mathrm{\mu}\left(1-\mathrm{\mu}\right)\left(\dfrac{1}{r_{1}^{3}}-\dfrac{1}{r_{2}^{3}}\right)\widehat{x}\end{array}\label{}\end{equation}