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.

 

[aldom]

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.

I have finally figured out. I was looking at this article; they are dealing with the same problem just without the drag. I then saw that the way to go was to use the same coordinate they are using: taking gravity to go along the radial direction and setting the $z$ axis to be direction of Earth's rotation, so that $\vec{\omega}=\omega\hat{k}$. As for the drag, from this article I was convinced that I could take $C_d$ to be constant in within a certain set of values for the Mach number, so I decided to do so for to make computations easier. I also figured out why I should use the ordinary velocity in the drag term instead of $\mathbf{v}_R$. With all of this, the components of the equations are
$$m\ddot{x}=-\frac{GMm}{|\mathbf{r}|^3}x-\frac{1}{2}\rho_0e^{-\frac{r-R_{\text{Earth}}}{H_1}}C_dA|\dot{\mathbf{r}}|\dot{x}+2m\omega\dot{y}+m\omega^2x$$
$$m\ddot{y}=-\frac{GMm}{|\mathbf{r}|^3}y-\frac{1}{2}\rho_0e^{-\frac{r-R_{\text{Earth}}}{H_1}}C_dA|\dot{\mathbf{r}}|\dot{y}-2m\omega\dot{x}+m\omega^2y$$
$$m\ddot{z}=-\frac{GMm}{|\mathbf{r}|^3}z-\frac{1}{2}\rho_0e^{-\frac{r-R_{\text{Earth}}}{H_1}}C_dA|\dot{\mathbf{r}}|\dot{z}$$

 

[kuruman]

A couple of questions and a comment.

1. Why is the altitude-dependent density $\rho_0e^{-\frac{z}{H_1}}$ instead of $\rho_0e^{-\frac{r}{H_1}}~?$
2. Now what? Are you going to find the trajectory using machine code or are you going to try solving this analytically? If by code, how are you going to test that it gives the right result?

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