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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 310 17 "Rocket propulsion" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 242 "The prob
lem of mechanical systems in one dimension with varying mass represent
s an interesting application of Newton's law. It comes up in the rocke
t problem (variable mass and velocity), the conveyor belt problem, and
 others. (cf. K. Symon: " }{TEXT 317 9 "Mechanics" }{TEXT -1 33 ", 3rd
 ed. p. 172ff, L.S. Lerner: " }{TEXT 316 7 "Physics" }{TEXT -1 52 ", p
. 226 ff, or almost any first-year physics text)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "We consider a rocket wit
hout external forces (no gravity). The rocket's momentum for motion in
 one dimension is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "P_R:=M(t)*v(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$P_RG*&-%\"MG
6#%\"tG\"\"\"-%\"vGF(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Here " 
}{TEXT 257 1 "v" }{TEXT -1 1 "(" }{TEXT 258 1 "t" }{TEXT -1 141 ") is \+
the velocity of the rocket with respect to a stationary frame. The roc
ket is propelled by the exhaust of gases with a constant velocity " }
{TEXT 256 1 "u" }{TEXT -1 110 " relative to the rocket. With respect t
o the stationary (laboratory) frame the exhaust gas has a velocity of \+
 " }{TEXT 259 1 "v" }{TEXT -1 1 "+" }{TEXT 260 1 "u" }{TEXT -1 292 ". \+
The total rate of change of momentum has to vanish, as there are no ex
ternal forces. Thus, the total rate of change of the rocket's momentum
 is balanced by the rate of change of the exhaust gas momentum. The la
tter is given by the change in mass times the exhaust velocity in the \+
lab frame." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "NEq:=diff(P_R
,t)-diff(M(t),t)*(v(t)+u)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NEq
G/,(*&-%%diffG6$-%\"MG6#%\"tGF.\"\"\"-%\"vGF-F/F/*&F+F/-F)6$F0F.F/F/*&
F(\"\"\",&F0F/%\"uGF/F/!\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "NEq:=simplify(NEq);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%$NEqG/,&*&-%\"MG6#%\"tG\"\"\"-%%diffG6$-%\"vGF*F+F,F,*&-F.6$F(F+F,
%\"uGF,!\"\"\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "It appears, \+
as if there is an asymmetry in how the velocity " }{TEXT 276 1 "v" }
{TEXT -1 1 "(" }{TEXT 277 1 "t" }{TEXT -1 372 ") is treated: in the ra
te of change of the rocket momentum derivatives are considered both fo
r the mass and for the velocity. They are both important as they are m
ultiplied by the other respective variable (as dictated by the product
 rule). In the rate of change of the exhaust fuel momentum it appears \+
as if the change in part of the velocity in the lab frame (namely in \+
" }{TEXT 279 1 "v" }{TEXT -1 1 "(" }{TEXT 278 1 "t" }{TEXT -1 164 ")) \+
is ignored. This is indeed the case, but it is justified since the vel
ocity is multiplied at each instant in time by a small amount of mass:
 between time layers " }{TEXT 281 1 "t" }{TEXT -1 5 " and " }{TEXT 
282 1 "t" }{TEXT -1 2 "+d" }{TEXT 283 1 "t" }{TEXT -1 55 " the only ma
ss that counts on the exhaust gas side is d" }{TEXT 280 1 "M" }{TEXT 
-1 53 ". One is justified to ignore the change in velocity d" }{TEXT 
318 1 "v" }{TEXT -1 17 " as the product d" }{TEXT 284 1 "m" }{TEXT -1 
1 "d" }{TEXT 285 1 "v" }{TEXT -1 125 " is of higher order and does not
 come into play. It is in fact cancelled by a similar term arising in \+
the rocket momentum at " }{TEXT 291 1 "t" }{TEXT -1 2 "+d" }{TEXT 290 
1 "t" }{TEXT -1 45 " which expressed with differentials reads as " }}
{PARA 0 "" 0 "" {TEXT -1 1 "(" }{TEXT 288 1 "M" }{TEXT -1 2 "+d" }
{TEXT 289 1 "M" }{TEXT -1 3 ")*(" }{TEXT 286 1 "v" }{TEXT -1 2 "+d" }
{TEXT 287 1 "v" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 101 "We can solve the differential equation e
ither for the mass or for the velocity as a function of time:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "solM:=dsolve(NEq,M(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%so
lMG/-%\"MG6#%\"tG*&%$_C1G\"\"\"-%$expG6#*&-%\"vGF(\"\"\"%\"uG!\"\"F," 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solv:=dsolve(NEq,v(t));" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solvG/-%\"vG6#%\"tG,&*&%\"uG\"\"
\"-%#lnG6#-%\"MGF(F-F-%$_C1GF-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 
"In the solution for the mass " }{TEXT 269 1 "M" }{TEXT -1 1 "(" }
{TEXT 270 1 "t" }{TEXT -1 57 ") the integration constant _C1 is clearl
y a mass, namely " }}{PARA 0 "" 0 "" {TEXT 271 1 "M" }{TEXT -1 1 "(" }
{TEXT 273 1 "t" }{TEXT -1 4 "=0)=" }{TEXT 272 1 "M" }{TEXT -1 4 "(0).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 218 "In t
he second solution for v(t) some further thinking is required. It make
s no sense to take the logarithm of a dimensionful quantity (e.g. a ma
ss). Thus _C1 should really be reexpressed as a constant that represen
ts -" }{TEXT 266 1 "u" }{TEXT -1 4 " ln(" }{TEXT 265 1 "M" }{TEXT -1 
1 "(" }{TEXT 268 1 "t" }{TEXT -1 53 "=0)), in which case the solution \+
can be expressed as " }}{PARA 0 "" 0 "" {TEXT 264 1 "u" }{TEXT -1 4 " \+
ln(" }{TEXT 263 1 "M" }{TEXT -1 1 "(" }{TEXT 267 1 "t" }{TEXT -1 2 ")/
" }{TEXT 262 1 "M" }{TEXT -1 80 "(0)). Now the logarithm is taken of a
 dimensionless quantity, which makes sense." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Note that " }{TEXT 261 1 "u" }
{TEXT -1 102 " is the exhaust velocity (not just the speed), i.e., the
 direction is opposite to the rocket velocity " }{TEXT 274 1 "v" }
{TEXT -1 1 "(" }{TEXT 275 1 "t" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "With specified initial co
nditions the solution for the velocity is obtained as:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "solv:=dsolve(\{NEq,v(0)=v0\},v(t));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solvG/-%\"vG6#%\"tG,(*&%\"uG\"
\"\"-%#lnG6#-%\"MGF(F-F-*&-F/6#-F26#\"\"!F-F,\"\"\"!\"\"%#v0GF-" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "The fact that the solution can be \+
found either for " }{TEXT 292 1 "M" }{TEXT -1 1 "(" }{TEXT 293 1 "t" }
{TEXT -1 9 ") or for " }{TEXT 295 1 "v" }{TEXT -1 1 "(" }{TEXT 294 1 "
t" }{TEXT -1 51 ") should have made us suspicious. The solution for " 
}{TEXT 299 1 "v" }{TEXT -1 1 "(" }{TEXT 298 1 "t" }{TEXT -1 35 ") depe
nds on the time evolution of " }{TEXT 297 1 "M" }{TEXT -1 1 "(" }
{TEXT 296 1 "t" }{TEXT -1 317 ") and vice versa. This makes sense, as \+
we have specified the exhaust velocity (or speed) of the gas, but not \+
the rate at which the fuel is being burned. The rocket velocity depend
s clearly on how much mass is being sent off in the opposite direction
 at a given exhaust speed. Let us assume a linear relationship with " 
}{TEXT 311 1 "m" }{TEXT -1 74 " being the mass of the rocket without f
uel, and a valid time range of 0 < " }{TEXT 312 1 "t" }{TEXT -1 42 " <
 10 (all quantities in MKSA (SI) units):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "m:=500;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\"$+&
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "M:=t->m+1000-100*t;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MGR6#%\"tG6\"6$%)operatorG%&arrowG
F(,(%\"mG\"\"\"\"%+5F.9$!$+\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "v_R:=unapply(rhs(solv),u,v0,t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$v_RGR6%%\"uG%#v0G%\"tG6\"6$%)operatorG%&arrowGF*,(*&
9$\"\"\"-%#lnG6#,&\"%+:F19&!$+\"F1F1*&-F36#F6F1F0\"\"\"!\"\"9%F1F*F*F*
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "plot(v_R(-100,0,t),t=0.
.10);" }}{PARA 13 "" 1 "" {GLPLOT2D 648 200 200 {PLOTDATA 2 "6%-%'CURV
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" 0 "" {TEXT -1 4 "For " }{TEXT 319 2 "t " }{TEXT -1 73 "> t_f = 10 th
e solution is not meaningful since the fuel contribution to " }{TEXT 
320 1 "M" }{TEXT -1 1 "(" }{TEXT 321 1 "t" }{TEXT -1 56 ") becomes neg
ative according to the mapping given above." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "The acceleration of the rocket
 changes as a function of time. It is calculated by differentiating th
e velocity:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "v_R(-100,0,t
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%#lnG6#,&\"%+:\"\"\"%\"tG!$+
\"F+-F%6#F(\"$+\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "a_R:=d
iff(%,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$a_RG,$*&\"\"\"F',&\"%+
:\"\"\"%\"tG!$+\"!\"\"\"&++\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "plot(a_R,t=0..10);" }}{PARA 13 "" 1 "" {GLPLOT2D 646 230 230 
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-1 166 "The strong rise in the rocket's acceleration near the end of t
he fuel burn cycle is associated with the fact that according to the s
olution the increase in velocity d" }{TEXT 305 1 "v" }{TEXT -1 77 " is
 given by the product of the exhaust speed and the log of the mass rat
io (" }{TEXT 300 1 "M" }{TEXT -1 1 "(" }{TEXT 304 1 "t" }{TEXT -1 2 "+
d" }{TEXT 303 1 "t" }{TEXT -1 2 ")/" }{TEXT 301 1 "M" }{TEXT -1 1 "(" 
}{TEXT 302 1 "t" }{TEXT -1 140 ")). The mass ratio is close to unity w
hile a lot of heavy fuel is on board, but  deviates strongly from that
 value as the fuel is exhausted." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 427 "Another interesting observation can be m
ade from the graph of the acceleration at early times The rocket start
s with a positive non-zero acceleration. If we add the gravitational a
cceleration that acts in the opposite direction we may not have enough
 thrust for lift-off. We discuss thrust further below: it is the magni
tude of the equivalent force that results as a consequence of hurling \+
out particles at a given flow rate (d" }{TEXT 313 1 "M" }{TEXT -1 2 "/
d" }{TEXT 314 1 "t" }{TEXT -1 27 ") with some exhaust speed |" }{TEXT 
315 1 "u" }{TEXT -1 604 "|. The product of these two quantities has th
e dimensions of a force and it is responsible for the rocket's acceler
ation. For a constant rate of fuel burn-up and for constant exhaust sp
eed this force is constant. Thus, the increase in the rockets accelera
tion during its fuel-burning phase is a result of the decreasing rocke
t mass. Multi-stage rockets have the advantage that the mass associate
d with a big motor (or entire booster rockets on the Space Shuttle) ca
n be jettisoned as the contained fuel is burned up. For the subsequent
 stage(s) a more convenient fuel/gross payload ratio can be achieved.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 523 "The \+
design of rockets calls for a thrust that is several times larger than
 the initial weight in order to gain speed quickly. When a rocket is i
gnited, however, it takes some time for the build-up of full thrust: t
his is repsonsible for a short delay between firing of engines and lif
t-off. For the example given above the product of fuel burning rate an
d exhaust velocity, i.e., the thrust is not enough to overcome gravity
 (the acceleration does not exceed 9.8 m/s^2). In free space, however,
 this issue is not important." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 386 "A similar phenomenon can be experienced \+
during take-off of jet airplanes, i.e., a time delay between the revvi
ng up of engines and actual forward acceleration (here one needs to ov
ercome friction to set the plane in motion). In a jet engine thrust is
 not achieved by just hurling burned fuel out of a nozzle: air is take
n in and compressed by burning kerosene using air from the intake." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 211 "If we do
 not like to rely on Maple's dsolve engine, we can perform the require
d steps to find the solution ourselves. We restart the session and red
efine Newton's equation as it is safer than to simply unassign " }
{TEXT 322 1 "M" }{TEXT -1 1 "(" }{TEXT 323 1 "t" }{TEXT -1 64 ") (we f
ound this to be unreliable in early releases of Maple V):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 57 "NEq:=simplify(diff(M(t)*v(t),t)-diff(M(t),t)*(v(t)+
u)=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NEqG/,&*&-%\"MG6#%\"tG\"
\"\"-%%diffG6$-%\"vGF*F+F,F,*&-F.6$F(F+F,%\"uGF,!\"\"\"\"!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "accel:=solve(NEq,diff(v(t),t));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&accelG*&*&-%%diffG6$-%\"MG6#%\"tGF-
\"\"\"%\"uGF.\"\"\"F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "sol:=v(s)-v(0)=int(accel,t=0..s);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%$solG/,&-%\"vG6#%\"sG\"\"\"-F(6#\"\"!!\"\",&*&-%#lnG6#-%\"MGF)F+%
\"uGF+F+*&-F36#-F6F-F+F7\"\"\"F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
258 "To combine the solution to the Tsiolkovski formula (named after a
 rocket pioneer who found it in 1897) we need some tricks: it is neces
sary to assume the arguments of the ln function to be positive for the
 combine command to work; the assumption of positive " }{TEXT 324 1 "M
" }{TEXT -1 156 " doesn't work properly as it is a subscripted variabl
e, so we substitute two symbols for the initial and final mass, factor
 the expression and then combine:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "assume(M_f>0,M_i>0); interface(showassumed=0);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "Delta:=combine(factor(subs(M
(s)=M_f,M(0)=M_i,rhs(sol))),ln);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
&DeltaG*&%\"uG\"\"\"-%#lnG6#*&%%M_f|irG\"\"\"%%M_i|irG!\"\"F'" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "The interpretation of this expres
sion for the change in the rocket velocity Delta for given exhaust vel
ocity " }{TEXT 306 1 "u" }{TEXT -1 15 " is as follows:" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 254 "The mass ratio that
 appears as the argument of the logarithm is less then unity, and thus
 the logarithm yields a negative number. This is consistent with the f
act that the rocket acquires a change in velocity of opposite sign to \+
the exhaust velocity. If " }{TEXT 307 1 "u" }{TEXT -1 275 " were to st
and for the exhaust speed, the mass ratio would need to be inverted. N
ote that the formula is valid for any finite time interval, as it repr
esents the solution to the differential equation. In particular, it is
 not important how the fuel was burned, i.e., whether " }{TEXT 308 1 "
M" }{TEXT -1 1 "(" }{TEXT 309 1 "t" }{TEXT -1 68 ") was linear in time
 as assumed for the graph of the solution above." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "29" 0 }{VIEWOPTS 1 1 0 1 1 1803 }