Equations of motion in Earth-centered, Earth-fixed coordinates

Question

I'm trying to write down the equation of motion for projectile motion in a coordinate system fixed to the rotating Earth, accounting for as much as possible - say, a ballistic missile. It's proving a little confusing. Here is my attempt:

$$ \underbrace{\dot{m} \dot{\mathbf{r}}_{P/R}}_{\text{thrust}} - \underbrace{\frac{GMm}{r^2} \hat{\mathbf{r}}}_{\text{gravity}} - \underbrace{\frac{1}{2} C_D \rho A v^2 \hat{\dot{\mathbf{r}}}}_{\substack{\text{atmospheric} \\ \text{drag}}} + \underbrace{\frac{1}{2} C_L \rho A v^2 \hat{\mathbf{e}}_L}_{\substack{\text{atmospheric} \\ \text{lift}}} + \underbrace{\mathbf{F}_{ext}}_{\substack{\text{external} \\ \text{forces}}} = m ( \ddot{\mathbf{r}} - \underbrace{\boldsymbol{\Omega} \times \boldsymbol{\Omega} \times \mathbf{r}}_{\substack{\text{centrifugal} \\ \text{acceleration}}} - \underbrace{2 \boldsymbol{\Omega} \times \dot{\mathbf{r}}}_{\substack{\text{Coriolis} \\ \text{acceleration}}} - \underbrace{\dot{\boldsymbol{\Omega}} \times \mathbf{r}}_{\substack{\text{Euler} \\ \text{acceleration}}} ) $$

For example, if the coordinate system were chosen as spherical coords:

• $ (r, \theta, \phi) $ represents radial distance from the centre $r$, longitude $\theta$ and colatitude $\phi$.
• $ \mathbf{r} = r \hat{\mathbf{r}} $ is the position vector of the object.
• $ \dot{\mathbf{r}} = \dot{r} \hat{\mathbf{r}} + r \dot{\theta} \sin \phi \hat{\boldsymbol{\theta}} + r \dot{\phi} \hat{\boldsymbol{\phi}} $ is the velocity vector.
• $ \ddot{\mathbf{r}} = \left ( \ddot{r} - r \dot{\theta}^2 \sin^2 \phi - r \dot{\phi}^2 \right ) \hat{\mathbf{r}} + \left ( (r \ddot{\theta} + 2 \dot{r} \dot{\theta}) \sin \phi + 2r \dot{\theta} \dot{\phi} \cos \phi \right ) \hat{\boldsymbol{\theta}} + \left ( r \ddot{\phi} + 2 \dot{r} \dot{\phi} - r \dot{\theta}^2 \sin \phi \cos \phi \right ) \hat{\boldsymbol{\phi}} $ is the acceleration vector. These three are all standard results.
• $ \boldsymbol{\Omega} $ is the angular velocity vector of the Earth.
• The left hand side is the sum of the forces on the object, in the Earth's rotating frame.
• The right hand side is the rate of change of momentum, $ ma + \dot{m} v $, but with extra terms added to $ a $ and $ v $ to account for the rotating reference frame.

My main concerns are -

1. Was I right to include the 'fictitious forces' in $ a $ and $ v $? (centrifugal, Coriolis, Euler, tangential velocity)? I got their expressions from here. That page has some pretty confusing notation though.

2. If so, do those terms have the correct sign? Or should they be negative?

3. I am not confident at all with the variable mass term. I've never seen a single example doing variable mass problems in rotating reference frames so I just kept it the same.

Thanks!

Answer 1

Based on the corrections in the comments I believe the corrected force equation is:

$$ \underbrace{\dot{m} \mathbf{v}_{P/R}}_{\text{thrust}} - \underbrace{\frac{GMm}{r^2} \hat{\mathbf{r}}}_{\text{gravity}} - \underbrace{\frac{1}{2} C_D \rho A v^2 \hat{\dot{\mathbf{r}}}}_{\substack{\text{atmospheric} \\ \text{drag}}} + \underbrace{\frac{1}{2} C_L \rho A v^2 \hat{\mathbf{e}}_L}_{\substack{\text{atmospheric} \\ \text{lift}}} + \underbrace{\mathbf{F}_{ext}}_{\substack{\text{external} \\ \text{forces}}} = m ( \ddot{\mathbf{r}} + \underbrace{\boldsymbol{\Omega} \times \boldsymbol{\Omega} \times \mathbf{r}}_{\substack{\text{centrifugal} \\ \text{acceleration}}} + \underbrace{2 \boldsymbol{\Omega} \times \dot{\mathbf{r}}}_{\substack{\text{Coriolis} \\ \text{acceleration}}}) $$

where I have gotten rid of the Euler acceleration term (Earth's rotation axis and rate are both practically constant) and changed the signs of the fictitious forces on the RHS.

This equation doesn't constrain the orientation of the object - a separate equation applying angular momentum would be necessary for that.

Answer 2

In physics education it is very common to simplify cases to make them less hard to solve.

My best guess is that the overwhelming majority of textbook authors will only give a treatment of the no-atmospheric-friction case - a much simpler case.

My best guess is that only the actual manufacturers of actual real-life ballistic rockets will have engineers develop the full scope modeling that you have in mind.

And those models will not be made public. No doubt those full scope models are secret.

Of course the way to deploy any model is to implement it in a simulation.

That is: I expect that you will be developing a simulation anyway. If that is the case then my recommendation is to develop the simulation in stages, adding features one at a time.

In the no-atmospheric-friction and no-propulsion case the mass drops out of the equation. It drops out of the equation because gravitational mass and inertial mass are equivalent. The terms representing gravitational effect has gravitational mass $m$ in it, and the centrifugal term and the coriolis terms have the inertial mass $m$ in them. Gravitational mass and inertial mass are equivalent, therefore you can divide all of the terms by the mass $m$, making that factor drop out of the equation

But of course for your purpose you want to accomodate that as the ballistic missile is burning propellent the mass of the missile is decreasing.

The thrust term must accomodate the instantaneous inertial mass of the ballistic missile. As we know: if the thrust remains the same over some time interval then magnitude of the acceleration will not be constant, it will increase, because the inertial mass of the missile keeps getting decreasing.

Now, in order for the equation to be mathematically consistent all of the terms must be dimensionally the same. (See: dimensional analysis)

Here is a discussion of the case of a satellite orbiting the Earth in low Earth orbit, along the Equator, in west-to-east direction.

Relative to the inertial coordinate system the satellite is circumnavigating the Earth in about 90 minutes. At that velocity the Earth's gravity provides the required centripetal force to sustain that circumnavigating motion.

Relative to a coordinate system that is co-moving with the Earth the satellite is moving slower than orbital velocity. But the amount of gravitational force exerted by the Earth is a given. So: relative to a coordinate system that is co-moving with the Earth it appears as if there is a surplus of centripetal force.

Relative to a coordinate system that is co-moving with the Earth the surplus of centripetal force is accounted for by the coriolis term. The satellite has a velocity relative to the coordinate system; the coriolis term is proportional to the velocity of the object with respect to the coordinate system that is used.

In the above described case: the coriolis term specifies an outward directed acceleration.