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I'm trying to understand the equations of motion for a rocket launched from earth, through the atmosphere, and into LEO. I found a set of ordinary differential equations which describe the motion of the rocket (see this reference, slide 3). Below is a diagram of which should help to describe the variables in the equation:

enter image description here

The differential equations are:

$$\dot{v}=-\frac{ksin(\gamma)}{r^2}-\frac{D}{m}+\frac{Tcos(\phi_T)}{m}$$ $$\dot{\gamma}=-\frac{kcos(\gamma)}{vr^2}+\frac{L}{mv}+\frac{Tsin(\phi_T)}{mv}+\frac{vcos(\gamma)}{r}$$ $$\dot{\nu}=v\frac{cos(\gamma)}{r}$$ $$\dot{r}=vsin(\gamma)$$ $$\dot{m}=-T/I_{sp}$$

I recognize the first equation as the equation for linear acceleration, where I think $k=\mu=GM$ is the gravitational constant times the mass of earth (please correct me if I'm wrong). I don't know what the second equation represents or how it was obtained. Can anyone give me some insight into how this equation is derived? Is it the statement of the conservation of some forces? If so, what forces in particular?

I'm not even entirely sure that this is the proper form of the dynamic model for the rocket's trajectory. I tried cross-validating this model with other models to see if this is correct. In this process, I found an answer on another stack exchange question which almost describes the same equation, but with an additional $\frac{v}{r}$ term in the $\dot{\gamma}$ equation. If I were to add this term to the $\dot{\gamma}$ equation, what would it mean? What force does it describe?

Also, I'm having a little bit of trouble understanding the notation, since the variables are not explicitly defined anywhere in the document. I think the D and L represent drag and lift forces. If so, I'm not sure why lift forces are not included in the $\dot{v}$ equation or why they appear only in the $\dot{\gamma}$ equation. The term T seems to represent the thrust of the rocket, which acts at an angle $\phi_T$ from the normal direction to the surface of the (perfectly circular) earth. The angle $\gamma$ seems to be the angle from the velocity vector to the local horizon. Have I correctly summarized these variables?

Any help understanding this equation would be greatly appreciated.

Paul
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    The v˙ equation is dealing only with forces that act along the direction of the instantaneous velocity vector, that is why L is not included. The γ˙equation deals only with forces that act perpendicular to the instantaneous velocity vector, that is why D is not included. I think ϕT is not what you describe, but instead is the angle between the thrust vector and the instantaneous velocity vector. Set up your frame of reference in regards to the velocity vector and this will make more sense to you. – Organic Marble Sep 08 '17 at 02:03
  • @OrganicMarble: If $\phi_T=0$, then the thrust contributes only in the direction of the velocity vector, so what you say makes a lot of sense! The lift vector is, by definition, orthogonal to the direction of drag, so that makes sense too. Still, this would suggest that the acceleration in the direction normal to the velocity is somehow characterized by $mv\dot{\gamma}$. What would this term represent? Some sort of angular acceleration? – Paul Sep 08 '17 at 03:33
  • @Paul $\dot{\gamma}$ is the speed with which the velocity's direction changes. It is equal to $a_\bot/v$, where $a_\bot$ is the (linear) acceleration in the direction perpendicular to the velocity, so $mv\dot{\gamma} = ma_\bot$ is the component of the total force perpendicular to the velocity. – Litho Sep 08 '17 at 08:32
  • @Paul By the way, you forgot to divide the first term in the second formula by $v$. This $v$ is there in the reference you linked. You also skipped the term $v\cos(\gamma)/r$, which describes the change in the angle between the velocity and the horizontal direction due to the change of the horizontal direction. – Litho Sep 08 '17 at 08:46
  • @litho: Thanks! I corrected those two terms. I'm curious though... If $vcos(\gamma)/r$ describes the change in angle due to normal forces, shouldn't the first equation have a term of $v^2sin(\gamma)/r$ To describe a similar force in the direction parallel to the velocity vector? – Paul Sep 08 '17 at 13:01
  • @Paul Sorry, what I wrote before was wrong. The value $\dot{\gamma}$ would be equal to $a_\bot/v$ if the angle $\gamma$ was measured relative to a constant (in an inertial frame) direction. But it is measured relative to the horizontal direction, and this direction depends on the rocket's position. So the formula for $\dot{\gamma}$ consists of two parts: one is equal to $a_\bot/v$ and describes the change of the velocity's direction (in an inertial frame) due to normal forces (lift and the normal components of gravity and thrust)... – Litho Sep 08 '17 at 15:11
  • @Paul ...and the other, $v\cos(\gamma)/r$, describes the change of the horizontal direction at the rocket's position due to the change of the rocket's position. This change is not a result of any forces acting on the rocket; it depends on the rocket's horizontal velocity and doesn't depend on the vertical one, that's why you get $\cos \gamma$ there. The rotation of the horizontal and vertical directions doesn't affect the magnitude of rocket's velocity, so there is no similar term in the first formula. – Litho Sep 08 '17 at 15:19
  • @Litho: You should really post these comments as an answer. It has really helped me understand the problem better. On a side note: if the second equation was multiplied by $mv$ on both sides, would the resulting term $m\frac{v^2}{r}cos(\gamma)$ represent a fictitious 'centrifrugal' force acting in the velocity direction? Since the velocity vector may not necessarily align with the local horizon, I think it that term would appear as well. If so, I would think the 1st equation should also have such a term $m\frac{v^2}{r}sin(\gamma)$. – Paul Sep 08 '17 at 18:57
  • Let us continue this discussion in chat. – Paul Sep 08 '17 at 18:57

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