Rocket equations of motion w/ drag and gravity

[Adoniram]

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
• Gravity, quadratic 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!

 

[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.

 

[Adoniram]

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.

 

[Adoniram]

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)

 

[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.

 

[Adoniram]

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?

 

[haruspex]

I couldn't get it into the form you proposed:

My post #5 lays it out in detail.

 

[Adoniram]

My post #5 lays it out in detail.

Wow that is really clever. At first I didn't get what you were saying but yeah that works! I would have never thought of that on my own...