Physics+III

Rockets and Propulsion

This link explains the mathematics of rockets and propulsion.

Explosion (things flying out) is opposite of collision (things flying together), so we can use momentum. In the following, let's ignore  air resistance   and look only at the  rocket engine  and gravity in one dimension. We will first start out with a finite amount of fuel, D M, being thrown out the back of the rocket with a speed of vexh (relative to the rocket) during a small but finite interval of time, D t. This will cause the initial speed of the rocket, V, to increase by a small but finite amount, D V. Note that the mass of the rocket, including its fuel, changes since part of the fuel is thrown out by the rocket engine, so the final mass of the rocket will be M- D M. Note that we subtract vexh to indicate that this speed is directed opposite to the rocket, and this makes vexh a positive number in all that follows. ** Fexternal = dptotal/dt ** ** -M g = (pf - pi) / **** D **** t ** ** pf-rocket = [M- **** D **** M]*[V+ **** D **** V] ; pf-fuel = **** D **** M * [V-vexh] ; ** ** pf-total = = [M- **** D **** M]*[V+ **** D **** V] + **** D **** M * [V-vexh] ** ** pi-rocket = [M- **** D **** M]*V ; pi-fuel = **** D **** M*V ; ** ** pi-total = MV ** Therefore, putting in pf and pi we get: ** -M g = {[M- **** D **** M]*[V+ **** D **** V] + **** D **** M * [V-vexh] - MV } / **** D **** t ** Writing out all the terms, we get: ** -Mg = {MV - **** D **** M*V + M* **** D **** V - **** D **** M* **** D **** V + **** D **** M*V - **** D **** M*vexh - MV } / **** D **** t ** Cancelling  out terms gives: ** -Mg = {M* **** D **** V - **** D **** M*vexh - **** D **** M* **** D **** V} / **** D **** t ** ** -Mg = M*( **** D **** V/ **** D **** t) - ( **** D **** M/ **** D **** t)*vexh - ( **** D **** M/ **** D **** t)* **** D **** V ** If we now take the limit as D t goes to zero, we see that **(** ** D **** V/ **** D **** t) ** becomes dV/dt, that **(** ** D **** M/ **** D **** t) ** becomes -dM/dt [the negative sign is introduced, since D M/ D t was a positive quantity, but dM/dt is negative since the mass is  decreasing   with time as we burn the fuel], and that the factor ** D **** V ** becomes zero. Thus we have: ** -Mg = M*dV/dt + vexh*dM/dt **, or    ** M*dV/dt = -Mg - vexh*dM/dt **. **//(Eq. 1)//** We see that the acceleration depends on the force of gravity and the thrust of the rocket, where **Fthrust = -vexh*dM/dt**. Note that dM/dt is a negative number since the mass, M, is decreasing with time, so Fthrust is positive. Note that Eq. 1 is a differential equation where both M and V depend on time.


 * Special case: ** If the rocket just hovers, the dV/dt = 0, and we can then try to solve Eq. 1 to see how long the rocket can hover. This is the situation in homework problem 25 (Ch 4, #7).

** dV/dt = -g + vexh*R/(Mo-Rt) ** Bringing the dt to the other side, and integrating gives: ** V - Vo = -gt + vexhR 0 **** ò **** t **** dt / (Mo-Rt) = -gt + vexh ln[1/{1-(R/Mo)t}] **, or    ** V(t) = Vo - gt - vexh ln[1-(R/Mo)t] ** Check: as t goes to zero, V goes to Vo; units work out; as Rt approaches Mo, V(t) approaches positive infinity. Note that Rt can’t actually go to zero since the rocket can’t be made up entirely of fuel – it needs at least the engine and  fuel tank. = Chapter 26 = = Chapter 29 = Binding Energy 1 Binding Energy 2 Binding Energy 3
 * Special case: ** If we burn the fuel at a  constant   rate, R, in the engines, then the quantity dM/dt = -R (since the mass decreases with time, dM/dt is negative), and the mass, M, is then M(t) = Mo-Rt . Equation 1 then becomes:

The above information is from this  webpage. Feel free to  seek   extra help here. Advanced Physics