Non constant acceleration due to one constant force, and one non constant force

AI Thread Summary
The discussion focuses on the dynamics of a rocket experiencing constant gravitational force and a non-constant thrust force. The net force is expressed as the difference between the booster force and gravitational force, leading to a non-constant acceleration that requires integration. A key point raised is that the rocket's mass decreases over time due to fuel consumption, complicating the force calculations. The solution involves formulating a differential equation that accounts for changing mass and velocity. The conversation emphasizes the importance of understanding differential equations to solve the problem effectively, especially after the fuel is exhausted.
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Homework Statement


This is actually related to another https://www.physicsforums.com/showthread.php?t=531596"
basically, I have a Constance force of gravity, and a non constant force of a rocket going the opposite way.

the net force would be something like F(booster) -Force(gravity). Assumed (-) direction, and Force(booster) is smaller. I know when acceleration isn't constant we have to integrate the equation.



Homework Equations


a = dx/dt


The Attempt at a Solution



a=( F(booster)- F(gravity) )/m

a=( F(booster)- mg) )/m

but I'm not sure how i should go about integrating this mess.
 
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Hi there...
This is a tad more complex than you put it.
A rocket, in your case has to have a constant rate of ejection of its gases, let's call it, u;
Now, recall the proper definition of a force; You're right to state that F=ma, when the mass is constant, but in our case, our rocket propels upwards, and in doing that it loses its fuel, therefore, F has to be taken as: \large \vec{F}=\frac{d\vec{p}}{dt}=-mg, p = m(t)\cdot\vec{v}
Loss of mass means: m = m_0-u*t;
Differentiating:
\large \vec{F}= \frac{d}{dt}m(t)\vec{v}(t) = \frac{dm}{dt}v(t)+m(t)\cdot\frac{d\vec{v}}{dt} = -uv+ma=mg
Assuming motion only on the vertical axis, which we'll call y, we get a very pleasant differential equation:
-uy'+(m_0+ut)*y'' = (m0-ut)g;
That you can solve, either by exponential substitution, or, simply through this:
http://www.wolframalpha.com/input/?i=-u*y'[t]+++(m0+-+u*t)*y''[t]+==+(m0-u*t)*g"
But you probably know better(feel free to ask about the technique of resolution to be employed here, if you care to do this manually).
Don't forget to specify your initial conditions,
Daniel
 
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Also, note that this solution does not account for what happens after all the fuel is exhausted. So its only valid till t=m_0/u; afterwards, standard projectile motion with an initial velocity laws apply...
Daniel
 
Thanks this helps a lot more than the lack of info in my text!

And yes, my book doesn't have any of that in it. I'm "Supposed to figure that out"
 
Glad to be of service,
Do you know how to slove such differential equations? It might be worth your while to learn, if you're in the mood?
Daniel
 
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