# I Rocket equations of motion w/ drag and gravity

1. Feb 3, 2017

I have seen many examples of the EOM for a rocket derived for the following cases:
• No gravity, No drag
• Gravity, No drag
• No gravity, linear drag (b*v where b is a constant)

I have never seen
• Gravity, linear drag

I took John Taylor's two examples of linear drag and gravity (without drag), and tried to combine them (since quadratic drag would add even more complexity). A couple basic assumptions at first:
The mass of the rocket is described as:
$$m(t)=m_o-kt$$
Where $m_o$ is the initial mass of the fueled up rocket, and $-k$ is the rate at which mass leaves the rocket. Also, $u$ is the velocity of the exhaust (which is taken as a constant), and $v$ is the velocity of the rocket itself.

For linear drag + gravity, the resultant equation of motion would be:

$$\frac{m dv}{dt}=-\frac{u dm}{dt}-mg-bv$$

At this point, for the case without gravity, you can do separation of variables. However, that doesn't seem possible here:
$$\frac{dv}{dt}=\frac{1}{m}\left(ku-bv\right)-g$$
(I've substituted -k in for dm/dt)
So, putting in m->m(t):
$$\frac{dv}{dt}=\frac{\left(ku-bv\right)}{\left(m_o-kt\right)}-g$$

Any help on solving this analytically would be greatly appreciated!

2. Feb 3, 2017

### haruspex

Just doing this in my head, so it might not work...
Looks like you can get it into the form y'=Ay/x+B. Then try substituting y=ux.

3. Feb 5, 2017

I finally found a method in Boyce & DiPrima's diff eq book:
If you can get the diff eq to look like:
$$v'(t)+p(t)v(t)=g(t)$$
Then you can solve it using an integrating factor, $\mu (t)$:
$$\mu(t)=exp\int p(t)dt$$
Then,
$$v(t)=\frac{1}{\mu(t)}\int_{t_o}^{t}\mu(s)g(s)ds$$
In my case, $t_o=0$ which helped simplify it. Anyway, I solved and got a somewhat ugly but completely reasonable formula!
$$v(t)=\frac{ku}{b}\left[1-m_o^{-b/k}\left(m_o-kt\right)^{b/k}\right]-\frac{g}{b-k}\left[m_o-kt-m_o^{1-b/k}\left(m_o-kt\right)^{b/k}\right]$$

Which gives the velocity of the rocket anytime before burnout. I can integrate and get height, and then use conservation of energy (with quadratic drag this time) to calculate final height. Sweet.

4. Feb 5, 2017

Extra cool:

I used mathematica to take the limit of that equation as $b\rightarrow0$ and got exactly the same result as John Taylor's 3.11 which solves for $v(t)$ for the case of gravity but no drag.

Mind = blown. So happy this worked.

(also, same for $g\rightarrow0$, recovers the same as 3.14)

Last edited: Feb 5, 2017
5. Feb 5, 2017

### haruspex

The method I outlined works too. Did you try it?
Writing T=(m0/k)-t and V=(ku/b)-v we have dv=-dV, dt=-dT:
dV/dT=(b/k)(V/T)-g
Now putting V=WT
W' T + W = (b/k)W - g
Writing α=b/k-1
W' T = αW - g
W'/(αW-g)=1/T
Etc.

6. Feb 6, 2017

I did not try that method, since I couldn't get it into the form you proposed:
$$v'=\frac{Av}{t}+B$$

The best I can do with that initial equation is:
$$v'=v\left(\frac{-b}{m(t)}\right)+\frac{uk}{m(t)}-g$$
or
$$v'=-vp(t)+g(t)$$

Does your method work for that form as well?

7. Feb 6, 2017

### haruspex

My post #5 lays it out in detail.

8. Feb 6, 2017