How do I integrate expressions with varying mass and air resistance in a rocket?

[BitterX]

Because this is not a 'formal' question I won't use the template.

I was wondering, what if I have a rocket (varying mass) with air resistance acting upon it?
Let's say the $F=-kv$
u is the speed of the rocket relative to the gas, and the rate of mass/second extracted is
b
without g it would look like this:
$\frac{dv}{v} =-u \frac{dm}{mv}-\frac{k}{m}dt$


My problem is that I don't know how I can integrate the expressions in the right side.
I'm sure I can't use the variable v as a constant in dm/mv but maybe I'm wrong.

The only book I have about mechanics is Berkeley's book, if this type of problems are in another, please point me to it. Thanks!

 

[tiny-tim]

Hi BitterX!

u is the speed of the rocket relative to the gas, and the rate of mass/second extracted is b

dm/dt = -b …

does that help? ​

 

[BitterX]

Ok, so:
$F= ub - kv$

$m\frac{dv}{dt}= ub - kv$

$m\frac{dv}{dt}=ub-k\frac{dx}{dt}$

now I'm still stuck

$m\frac{dv}{v}=(\frac{ub}{v}-k)dt$

$vdt=dx \Rightarrow \frac{dt}{v}=\frac{dx}{v^2} = \frac{dv}{v^2 dt}$

how can I isolate v to be only with dv?
should I use $m=M_0 - bt$?

 

[tiny-tim]

Hi BitterX!

Ok, so:
$F= ub - kv$

$m\frac{dv}{dt}= ub - kv$

Why did you introduce x in the next line??

Just separate the variables, and solve! ​

 

[BitterX]

I'm really sorry, but that's exactly my problem.
I can't see how can I separate v and m,
If I divide by v and m I'm still stuck with $\frac{ ub}{mv}dt$
and $\frac{ k}{m}dt$

how can I integrate $\frac{dt}{m}$ or $\frac{dt}{mv}$?

and more generally, is there a text about how to do these things?

Thanks for the help :)

 

[tiny-tim]

$m\frac{dv}{dt}= ub - kv$

mdv/(ub - kv) = dt

have a look at http://www.intmath.com/differential-equations/2-separation-variables.php ​

 

[BitterX]

Thanks! I guess I'm blind :)