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I am interested in modelling the trajectory of a rocket from the Earth to the Moon by solving a differential equation numerically. Below are some key facts and assumptions I am using. I want to make sure that I have not made any serious mistakes, nor disregarded any necessary facts.

We will consider the following equation, $$ \vec{T} + \vec{c}(\vec{r})\dot{\vec{x}} + \vec{G}(\vec{r})= m(t) \ddot{\vec{x}}, $$

where $T$ is the constant rocket thrust, $c$ denotes air resistance and is a function of radial distance from the earth, and the rocket has mass that drops at a rate that is constant with respect to time (we are assuming that a constant amount of fuel is always used for constant rocket thrust -- is this a valid assumption?).

Now a question:

  • The trajectory of the rocket is not straight; how do we incorporate parabolic motion into the numerics?
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Ah, then T must be a function of atmospheric pressure (i.e. a dependent on distance from the Earth then). But rate of fuel consumption will still be constant, right? This is in response to hwlau's saying that T should not be constant -- don't know where that comment went. – flamearchon Dec 30 '12 at 1:30
Becaues there is also friction, so the net force will change. Realistically, fuel consumption is not necessary constant after it gain energy to orbit around earth in ellipse. There are many consideration and depends which target you want to achieve. – hwlau Dec 30 '12 at 1:33
You should almst certainly treat the ascent phase and the transfer orbit with a different set of assumptions. – dmckee Dec 30 '12 at 1:35
up vote 1 down vote accepted

You seem to have the right idea with your equation, but some practical difficulties. First, I would recommend combining your independent variables into simply $\vec{r}$ rather than the $\vec{x}$ and $\vec{r}$, as you have it right now. Rocket motion is very complicated, however, and consider that nearly all of your forces will be functions of $\vec{r}$ and $\dot{\vec{r}}$ (e.g. air resistance is a function of velocity as well as position - you currently have it as only a function of position). After solving the final equation (presumably numerically, since a symbolic result for such a complex relationship will not be closed-form), you will get a function which models the motion of your rocket at a given time.

To account for "parabolic motion," by which I think you mean "turning," I would make thrust a function of time. For example, you may want to fire the thrusters 60 minutes into the flight to adjust the orbit, so you would define $T$ to have a certain magnitude and direction when $t = 60 \text{ min}$.

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Your reason for asking about parabolic motion is unclear (rockets generally don't travel on parabolic trajectories, especially not when under thrust or in an atmosphere), but you seem to be coming at things the wrong way, trying to directly implement the results of gravitational forces instead of just integrating them.

Since you're already doing a numerical solution, you can't use an analytical one for the gravitational part of the have other forces influencing your trajectory. However, since you're already numerically computing the motion of the rocket due to the forces on it, it should be straightforward to model drag and gravity as just two different forces on the rocket.

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Thanks for bringing this up; I forgot to mention that we will be incorporating gravitational force into the differential equation. I will edit my original post to address this change. Regarding the parabolic reference, what I meant is: what if the rocket needs to turn slightly? How do we model the rocket's turning? Or is it ideal for it to travel in a straight line? – flamearchon Dec 30 '12 at 16:57
@flamearchon: It's not really feasible to make it go in a straight line, and certainly not ideal. I'm not sure what the difficulty is. You model the motions as a result of the forces on the rocket, and the correct trajectories arise as a result of those forces just as they do in don't have to do anything extra to force the rocket to move appropriately. To model turning of the rocket, you simply apply the thrust in a different direction. – Christopher James Huff Dec 30 '12 at 17:44

The trajectory is in 3D. Your equation contains some vectors, $\vec{r}$ and $\vec{x}$. It is not clear what these vectors are, or why the air drag coefficient is a vector, which depends on the vector $\vec{r}$. If $\vec{G}$ is the gravitational force, then it doesn't depend on a vector. And does that include both the attraction due to the Earth and to the Moon?

The rocket won't have a parabolic trajectory, but it will have a curved trajectory. You need to make sure you define your coordinate system correctly (is it attached to the Earth? is it an inertial frame?), and write the separate components of the equation of motion.

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Quick point - gravitational force does depend on $\vec{r}$, which I assume is position relative to some large mass, since $\vec{G}$ has a magnitude related to the magnitude of $\vec{r}$ (i.e. the distance between the rocket and the mass) and a direction of $-\hat{r}$. – Shivam Sarodia Dec 31 '12 at 3:23

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