Problem: A rocket is launched vertically upward. The rocket has a mass Mr and carries Mo of fuel. The fuel burns at a constant rate (ß) and leaves the rocket at speed Ve relative to the rocket. Assume constant gravity (9.8m/sec^2). There is an air resistance given by F(a) = -kV. Where V is the velocity of the rocket. Take: k = .1 n-sec/m ß = 100 kg/sec Mr = 1000 kg Mo = 10000 kg Ve = 3000 m/sec What is the terminal velocity of the rocket (the velocity when fuel runs out)? ----- The general solution for this problem is an extension of conservation of momentum for a system of variable mass. mdv/dt = ∑F(ext) + Vrel(dM/dt) I have no problem when the external force is solely gravity, however, when this problem presents linear air resistance, I am running into a bit of trouble with the integration. My setup of the differential equation is as follows: Set dm/dt to constant. dM/dt = ß = 100 kg/sec. (=) mdv/dt = 100Ve - mg - kV ---> dv/dt + kV/m = (100Ve/m - g) I am attempting to integrate the velocity from V(0) to V(Terminal), and the time from t=0 to t=t(final)= Mo/ß = 10000/100 = 100 seconds. My question is regarding the velocity portion of the differential equation. I can't seem to get that V over to the left side of the integral by itself... and I'm wondering if I could get some sort of example on how to do that type of integration, or if I've gone astray somewhere. Does this integration qualify as a first order differential? i.e. dy/dx + P(x)y = Q(x) ? Addendum: Okay, so I've set up the equation: m/k(100Ve/m-g) - (m/k(100Ve/m-g)e^(-kt/m)) Thanks! Sean
Tips for making this easier: Remember that ultimately the mass is constant (since the M comes from the force being applied) and don't try and trick yourself into thinking that it is a variable in the end. In a rocket formula with linear air resistance, it is helpful to compare your answer to something you'd expect without air resistance.