Rocket with Variable Mass and Air Resistance.

[holdenthefuries]

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

 

[chaoseverlasting]

Yeah, it does. The solution's ye^{\int p(x)}=\int Q(x)e^{\int P(x)}dx

 

[holdenthefuries]

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.