Missile with Coriolis force, air resistance and variable gravity

[aldom]

Homework Statement

Make a model the motion of a missile in a case when it's high enough that gravity has to be taken as variable, Earth rotation effects are significant, and there's drag

I am modeling the missile as cylindrical for simplicity.
I understand that the equation modeling this should be
$$m\frac{d\mathbf{v}}{dt}=-\frac{GMm}{(R_E+z)^2}\hat{k}-\frac{1}{8}\rho_0e^{-\frac{z}{H_1}}\pi D^2C_d(\mathrm{Re})|\mathbf{v}_R|\mathbf{v}_R+2m\mathbf{v}\times\mathbf{\omega}+m\mathbf{\omega}\times(\mathbf{r}\times \mathbf{\omega}),$$
where:
$$\mathbf{v}$$ is the time derivative of the position vector with origin at the Earth's center, $$\rho_0 = 1.2\ \mathrm{kg/m^3}$$, $$H_1 = 8.4 \ \mathrm{km}$$, $$\mathbf{v}_R=\mathbf{v}-\mathbf{\omega} \times \mathbf{r}$$ and $$|\mathbf{\omega}|=7.3\times10^{-5} \ \mathrm{s^{-1}}$$.
I'm confused about what should the $x$, $y$ and $z$ components of this equation be. I'm also confused about what $C_d(\mathrm{Re})$ should I use in this case.

 

[kuruman]

Why is gravity in the $\mathbf{\hat k}$ direction? Don't forget that the Earth is approximated as a perfect sphere and "up" is the direction away from the Earth's center, i.e. the radial direction $\mathbf{\hat r}.$

I would put the $z-$axis along the axis of rotation and write $\boldsymbol {\omega}=\omega~\mathbf{\hat z}$. I would label as $xz$ the plane defined by vectors $\boldsymbol {\omega}$ and the position vector $\mathbf r = r \sin\theta ~\mathbf {\hat x}+ r \cos\theta ~\mathbf {\hat z}$ where $\theta$ is the angle between the angular velocity and the position vector. Of course, I would also rewrite the force of gravity as a radial vector $\mathbf F_g= F_g(r)~\mathbf{\hat r}.$

If I were solving this problem, I would "turn off" the drag ($C_d=0$) just to see how far I can go with the mathematical handling of the radially-dependent gravitational force and the Coriolis force by themselves. I have not attempted to solve this problem, but I suspect that getting an analytical solution without approximations, even in the absence of drag, is not easy.