(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Taking the equation of motion for a rocket launched from rest in a gravitational field g, [itex] m\dot{v} = -\dot{m}v_{ex} - mg [/itex], and knowing that the rocket ejects mass (fuel) at a constant rate [itex] \dot{m} = -k [/itex] (where k is a positive constant), so that [itex] m = m_{o} - kt [/itex]. Solve the equation [itex] m\dot{v} = -\dot{m}v_{ex} - mg [/itex] for v as a function of t, using separation of variables.

2. Relevant equations

[tex] m\dot{v} = -\dot{m}v_{ex} - mg [/tex]

[tex] \dot{m} = -k [/tex]

[tex] m = m_{o} - kt [/tex].

3. The attempt at a solution

So I've been beating my head at this for a while. It is deceptively simple, and there is probably a simple calculus trick I have forgotten that I need to use to solve this. Here's what I've done so far:

1.) Substituted [itex] m = m_{o} - kt [/itex] for both instances of m, giving:

[tex] (m_{o} - kt)\dot{v} = -\dot{m}v_{ex} - (m_{o} - kt)g [/tex]

2.) Substituted [itex] \dot{m} = -k [/itex], giving:

[tex] m_{o}dv-ktdv+kv_{ex} = -m_{o}g + ktg [/tex]

This is as far as I can get. I can isolate [itex] \dot{v} [/itex] to get this:

[tex] \dot{v}=\frac{kv_{ex}}{m_{o}-kt}-g [/tex]

This doesn't help me much though because I still have [itex] v_{ex} [/itex] on the right side as well as having a less than elegant [itex] m_{o} - kt [/itex] term on the bottom.

Can someone please help me with this separation of variables? Maybe if you can show me the way on this problem you might know of some good online references to practice doing this sort of thing or direct me to a book (I have Stuarts single variable Calculus book from a while ago when I took my 4 semester Calc sequence; time for a refresher maybe?)

Thanks for anyone who can help me.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Separation of Variables for a Rocket Launched in a Gravitational Field

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