Technical Appendix A number of theoretical claims made in Chap. 3 are derived here. The differ- ence between rocket physics and rocket engineering is that the physics calcu- lations are tractable—we can proceed quite a long way down the road to understanding with some high school or undergraduate physics and mathe- matics. The level of physics and math required to make sense of the calcula- tions to follow is higher than that required for the main text, but not by much—you don’t need to be a rocket scientist. If this is your cup of tea, read on; our derivations are condensed to save space, but can be readily unpacked by that mythical creature dubbed by teachers “the interested student”. Engineering calculations are much harder and involve considerable numerical simulation; we won’t attempt any of those. Tsiolkovsky equation From momentum conservation we can say that, for a rocket of mass M (including propellant), where dv is a small change in rocket speed, and u is the speed of a mass dM of exhaust gas: M dv = −u dM (the minus sign is there because rocket mass decreases as fuel is ejected.) Differentiating this equation with respect to time yields the relevant forces: dv dM M =−Fuext (A1.1) dt dt © Springer Nature Switzerland AG 2019 259 M. Denny, A. McFadzean, Rocket Science, https://doi.org/10.1007/978-3-030-28080-2 260 Technical Appendix We have allowed for some external forces here, such as gravity and aerody- namic drag. In the absence of such forces, the rocket acceleration is due to thrust from ejected propellant (2nd term on right). Integration leads to M t ∆vt()≡ vt()−=vuln 0 + dtF (A1.2) 0 Mt() ∫ ext 0 The left side is the famous delta-v. M0 is the initial mass of the rocket, con- sisting of rocket shell mass Ms plus propellant mass Mp plus payload mass m, so that M0 = Ms+Mp+m. The rocket mass at time t = 0 is M0; once all the propellant is burned then the mass is M(τ) = Ms+m, where τ is the burn time. It is common practice to define the rocket mass fraction taken up by propellant: M ε ≡ p (A1.3) M 0 In terms of mass fraction, the total delta-v is given by ∆vuτ =− ln 1−ε (A1.4) () () in the absence of external forces. Equation (A1.4) is the Tsiolkovsky equation.1 If the rocket starts from rest, then its mass is the following exponential function of speed, found by integrating Eq. (A1.2) assuming zero external force, and solving for M: v Mv()=−M0 exp. (A1.5) u The plot shown in Fig. 3.3a in the main text is M(v) obtained from Eq. (A1.5). Figure 3.3b is obtained by plotting v(t) from Eq. (A1.2), again assum- ing that Fext = 0 and also that the mass decreases at a constant rate: M(t) = M0 − μt. The constant mass flow rate (a reasonable approximation, 1 There are many derivations and descriptions of the Tsiolkovsky rocket equation online. See for example www.real-world-physics-problems.com/rocket-physics.html, www.grc.nasa.gov/www/k-12/airplane/ rockth.html, and https://en.m.wikipedia.org/wiki/Tsiolkovsky_rocket_equation. For more on Tsiolkovsky than we provided in Chap. 2 see the Wikipedia article Konstantin Tsiolkovsky, and Andrews (2009). Technical Appendix 261 commonly made in rocketry calculations) is related to other rocket parame- ters via μτ = Mp = ϵM0. In fact there is an extra term that should be added to the right side of Eq. (A1.1). We have derived the rocket thrust as T = udM/dt (it is the same mag- nitude but of opposite sign to the force exerted by the exhaust gas, in Eq. (A1.1)—this is N3 in action). Our derivation is based on momentum flow, but for real rockets there is an extra term due to pressure, so that dM Tu=+()PPea− Ae (A1.6) dt where Pe is the pressure of the exhaust gases at the nozzle exit, Ae is the area of the nozzle exit, and Pa is ambient atmospheric pressure. In practice the pres- sure contribution to thrust is usually negligible, except for large rockets at low altitudes. We note that in the technical literature you will find the curious statement that delta-v is not necessarily the actual change in speed of the rocket, even for motion along a straight line. This is true because many authors absorb the effects of external forces and/or the pressure term into the definition of delta-v (i.e. they move the integral on the right side of Eq. (A1.2) to the other side of the equation). Strictly speaking, delta-v is the true change in rocket speed only if all three of the following conditions apply: • There are no external forces, • Propellant is expelled in the opposite direction to the direction of travel, • Exhaust gas speed u is constant. Finally, for our nerdier readers, we note a subtle technicality of classical mechanics that has led to confusion and error even among professional physi- cists. Newton’s Second Law (N2) applies only to fixed mass systems, not to variable mass systems. If we tried to apply N2 directly we would obtain for the rocket thrust T = d(Mv)/dt = (dM/dt)v + M(dv/dt), which is (wrong and) not compatible with equation (A1.1) because the exhaust speed is not present. The modification of N2 to account properly for variable mass was sorted out, interestingly, over exactly the period that Tsiolkovsky developed the rocket equation (1897–1903) by a fellow Russian I.V. Meshchersky.2 2 Eke (1998), p3. 262 Technical Appendix Drag The force due to aerodynamic drag, i.e. the force that opposes the motion of a body through the air, is 1 2 FCdd= ρ Av (A2.1) 2 where v is the speed of the body through the air, A is its cross-sectional area, and ρ is the air density. Cd is the dimensionless drag coefficient; it is a slowly varying (for the most part) function of a fluid dynamic parameter called the Reynold’s number. Fluid dynamics is very complicated, and we don’t want or need to get into it too much; suffice it to say that the drag coefficient captures all the fluid dynamics of a body as it passes through the air, and that drag coefficient is usually pretty much constant for large values of Reynold’s num- ber, which is the case for our rockets. If we can regard Cd as constant, then the drag force acting on a rocket is seen to be a constant (let us define it as b = ½ ρCdA; it will differ from rocket to rocket) multiplied by the square of rocket speed.3 If the rate at which propellant is expelled from the rocket is a constant μ and if drag is the only external force acting on a rocket,4 then Eq. (A1.1) becomes dv 2 ()Mt0 − µµ=−bv + u . (A2.2) dt Note that there is an equilibrium (terminal) speed c for which no force acts on the rocket: the rocket increases in speed until it reaches uµ c = (A2.3) b 3 A detailed discussion of drag can be found in any fluid dynamics textbook (as well as in many online sources). See e.g. Massey (1997) Chapter 8 For a shorter discussion, see Denny (2011) pp197-200. In the context of the rocket equation, see Denny (2011) pp203-4. 4 This circumstance covers missiles in horizontal flight, for example. If the rocket is moving vertically upward then the force of gravity must also be included, but can be neglected for small model rockets which have very large initial accelerations, so this calculation applies approximately to that case as well. Technical Appendix 263 and then it can go no faster, unless the air thins, or the rocket becomes smaller by dropping a stage, or some other reason for reducing b. This equation inte- grates to give us the speed of the rocket as a function of time: 2uc/ 11−−()ε t /τ vt()= 2uc/ c. (A2.4) 11+−ε t /τ () We can express the rocket mass in terms of speed by eliminating the vari- able t to obtain: cu/2 cv− Mv()= M0 . (A2.5) cv+ Figures 3.4a, b are plotted from Eqs. (A2.4) and (A2.5) respectively. (For the section of Fig. (3.4a) corresponding to t > τ we obtained the curve by integrating Eq. (A2.2) with μ = 0 and with M0 replaced by (1−ε)M0.) Efficiency As with much of rocketry, the physics of rocket engine efficiency is simple enough whereas the engineering is complicated. Here we will stick mostly to calculating the physical propulsive power efficiency,η p. This measure of effi- ciency is defined as the ratio of propulsive power to the rate of production of propulsive kinetic energy. In other words, it is the fraction of power generated by the propellant that is converted into rocket power: Fv η = (A3.1) p 1 dM 1 dM u22+ v 2 dt 2 dt We know from Eq. (A1.1) that the magnitude of the thrust is F = (dM/dt)u and so 2vu/ η p = 2 . (A3.2) 1+ vu/ () 264 Technical Appendix Equation (A3.2) is plotted in Fig. 3.5. This expression for propulsive effi- ciency is the common one5—let us here go off the rails a little by defining another measure of propulsion efficiency based on energy: for a rocket with negligible shell mass (that is, Ms is small so that M0 ≈ Mp + m) the mechanical efficiency might be defined as payload kinetic energy divided by the sum of payload and propellant kinetic energy: 1 2 mv vv2 exp/()− u η ≡ 2 = .