Differential equation for a rocket

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SUMMARY

The discussion focuses on solving a separable differential equation for the velocity (v) of a rocket moving through interstellar dust without gravity, where drag is represented by the equation Drag = -bv. The key equation derived is $$\frac{dv}{-u - \frac{b}{k} v} = \frac{dm}{m}$$, which allows for separation of variables. The solution hints at a logarithmic relationship involving the mass ratio (m/m0) raised to a power, indicating the dependence of velocity on mass.

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DontDeimos
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Homework Statement


A rocket through interstellar dust, no gravity. Solve for v with respect to mass.

gif.latex?%5Cdot%7Bm%7D%3D%5Cfrac%7Bdm%7D%7Bdt%7D%3Dk.gif

k is a constant.
Drag = -bv

Homework Equations


[/B]
gif.latex?m%5Cfrac%7Bdv%7D%7Bdt%7D%3D-%5Cdot%7Bm%7Du-bv.gif


The Attempt at a Solution



To start:

%7Bdv%7D%7Bdm%7D%3D-ku-bv%20%5Crightarrow%20m%5Cfrac%7Bdv%7D%7Bdm%7D%3D-u-%5Cfrac%7Bb%7D%7Bk%7Dv.gif


It should become a separable differential equation, but I am having a lot of trouble solving it for v(m). A hint that was given, is in the final equation you should have has m/m0 to a power.
 
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You could divide by ## -u - \frac{b}{k} v## and also by ##m## and then multiply by ##dm## to get

$$\frac{dv}{-u - \frac{b}{k} v} = \frac{dm}{m}$$

This is solvable, see if it gets you the answer you need--I feel like both terms' logarithmic solutions would cancel but because of the coefficients perhaps you will get the correct dependence when you solve for it.
 
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