# Rocket Equation Notes ```The Rocket Equation (Constant Acceleration Version, Zero Gravity)
Assume we have a rocket with:
MP = mass of payload
MF = mass of fuel
M0 = MP + MF = total initial mass
v0 = constant exhaust speed of the fuel
R = dm/dt = fixed rate at which the fuel is burnt
F0 = v0R = constant thrust
The rocket mass then has a time dependence of m(t) = M0 – Rt. (Note that fuel burning must stop
when Rt equals the mass of the fuel, or tF = MF/R). Isaac Newton gives us F0 = v0R = m(t)a, or
a(t) = v0R / (M0 – Rt).
We find the rocket’s velocity by integrating a(t):
v(t) = v0R ln(M0 – Rt) (–R–1) + C = –v0 ln(M0 – Rt) + C
To evaluate the constant of integration C, we note that the rocket’s speed is zero when t = 0:
0 = –v0 ln(M0) + C, or C = v0 ln(M0)
We do a bit of algebra:
v(t) = –v0 ln(M0 – Rt) + v0 ln(M0) = v0[ln(M0) – ln(M0 – Rt)], or v(t) = v0 ln[M0 / (M0 – Rt)].
When t = tF, then vF = v0 ln[M0 /(M0 – RtF)] = v0 ln[M0 /(M0 – MF)] = v0 ln(M0 /MP), which is exactly
the rocket equation (Eqn 8.40) as derived on page 271 of the textbook.
To find how far the rocket will travel as a function of time, we integrate v(t). It is convenient to
rewrite the velocity as v(t) = v0 ln[M0 / (M0 – Rt)] = –v0 ln[(M0 – Rt) / M0] = –v0 ln(1 – Rt/M0)
before doing the integration. We recall that the integral of ln x is x ln x – x, and the algebraic details
are left as an exercise for the reader:
x(t) = (v0 M0/R)[(1 – Rt/M0) ln(1 – Rt/M0) + Rt/M0]
The distance the rocket has travelled at the moment the fuel runs out is given by inserting t = tF, which
means RtF/M0 = MF/M0. We have xF = (v0 M0/R)[(1 – MF/M0) ln(1 – MF/M0) + MF/M0]
= (v0/R)[(M0 – MF) ln(1 – MF/M0) + MF] = (v0/R)[MP ln(1 – MF/M0) + MF].
Rocket Energy
Rocket exhaust of mass m escaping at a speed of v0 will have EK = ½ mv02, which must come from
somewhere. Usually this energy comes from the mass itself: rocket fuel normally has a double role as
both reaction mass and energy source. (This connection is not essential. The little air-pumped rocket I
fired off in class used water as the reaction mass, and air pressure as the energy source.)
The rocket equation (there are many rocket equations, but only this one is called the rocket equation) is
vF = v0 ln(M0 /MP) = v0 ln(1 + MF/MP). This equation is critical because it says that no matter how you
design your rocket, no matter how the engine works, no matter what super-technology you use, the
final velocity vF of the rocket depends only on the fuel’s exhaust speed v0 and the fuel-to-payload mass
ratio. Alas, the fuel-to-payload ratio is inside a logarithm. Even huge increases in the ratio on the
order of a thousand to one only gain you about seven times the final speed. In a nutshell, the more fuel
you pack onto a rocket then the more energy the fuel has to waste pushing the rest of the fuel.
To achieve ultrahigh speeds, one is forced to concentrate on increasing v0. The exhaust speeds of the
hottest chemical fuels are about 4000 m/s, and if we select a typical fuel-to-payload ratio of 10 to 1,
then the rocket equation gives a final velocity of vF = (4000) ln(1 + 10) = 9590 m/s. This might sound
speedy, but the distance to the nearest star, Alpha Centauri, is 4.26 light years, which translates into a
staggering 40,300,000,000,000,000 meters. The time needed for our spaceship to cross that canyon
would be t = 4.03 x 1016 m / 9590 m/s = 3.1 x 1012 s = 133,000 yr.
We could tweak this number by pushing up the fuel-to-payload ratio, or using some exotic chemical
fuel so flammable that it starts burning before you’ve ignited it, but factors of two or three are the best
you can do. If you want your spaceship to really move, then you have to go beyond chemicals.
When the energy of the propellant comes from the propellant, our real interest is in the energy density
of the fuel, or EK/m. This is because v0 = [2(EK/m)]½, so the energy per kilogram of a fuel determines
how “hot” its exhaust will be. For example, take kerosene. (Kerosene was used on the Saturn V.)
Reference books say that kerosene – used as a heating fuel – gives off 43.5 megajoules/kg. This is
about the same energy density as a Hershey’s chocolate bar, so we see that hydrocarbons are
hydrocarbons when it comes to burning them into CO2 and H2O. One might hope that this density
would lead to an exhaust speed of v0 = [2(EK/m)]½ = [2(43.5 MJ/kg)]½ = 9330 m/s, but no.
One needs oxygen to burn kerosene, and a rocket must carry its own oxygen. For every kg of kerosene
you also need 3.47 kg of oxygen for full combustion. This means the energy density for everything in
the rocket’s exhaust is actually 43.5 MJ / (1 kg + 3.47 kg) = 9.7 MJ/kg. If we calculate v0 again in
view of this depressing news, we obtain 4400 m/s. Is this the true exhaust speed?
Well, no. A rocket engine is a heat engine, and no heat engine can convert heat into mechanical
energy with 100% efficiency. Even the most grainy YouTube video of a NASA launch reveals that the
rocket exhaust is extremely hot – literally white hot. A typical exhaust temperature is about 2000 C°.
The temperature inside a liquid rocket combustion chamber is about 3300 C°, so we can estimate an
thermodynamic efficiency e = (1 – TL / TH) = (1 – 2270/3570) = 36%. This reduces the EK/ kg
available to our rocket to (0.36)(9.7 MJ) = 3.5 MJ. Calculating v0 one last time gives [2(3.5 MJ/kg)]½
= 2650 m/s. The actual measured number for the Saturn V was 2580 m/s, so it seems this grand old
paragon of 1970’s rocket technology was doing about as well as could be hoped for, given that it was
using the same fuel as a Coleman camping lantern.
If we want real power, we must turn to nuclear reactions. The details of nuclear physics are outside
this course, so I will simply state that the most likely energy source for a 24th century supership is
deuterium fusion. Nuclear reactions are so powerful that physicists don’t quantify them with reaction
energies – instead we specify how much of the mass has disappeared. Einstein’s equation E = mc2
very literally says that mass is energy and vice versa. (If you heat up a pot of noodles, then you’ve
increased the pot’s mass because you’ve added energy to it.) A thousand joules divided by 9 x 1016
doesn’t give you much of a mass increase, but there is one nonetheless.
When deuterium is fused into helium, 0.635% of the original mass literally disappears. This
corresponds to an energy per kg of ∆E/kg = ∆mc2 = (0.00635 kg)(3 x 108)2 = 5.7 x 1014 J, or an
incredible 59 million times the energy density of kerosene. Putting this number into [2(EK/m)]½ gives
v0 = [2(5.7 x 1014 J/kg)]½ = 3.38 x 107 m/s, or 11.3% of the speed of light. Assuming the same 10:1
fuel ratio as we did previously gives our rocket a final velocity of vF = (11.3%) ln(1 + 10) = 27% of
lightspeed. Now we’re talking! At this speed it would only take 4.26 yr / 0.27 = 15.8 yrs to reach
Alpha Centauri. A little long for a family vacation, perhaps, but not outlandish by the standards of
scientific exploration.
Except. There are problems. For one thing, our deuterium fusion rocket is still a heat engine, and that
means we must contend with thermodynamic efficiency. It is very difficult to estimate the efficiency
of a spaceship engine which doesn’t exist, especially when it is based on nuclear technology which
also doesn’t exist, so I will guess e = 50%. That revises our numbers to: v0 = 8% c, vF = 19% c, and
the time to Alpha Centauri to 22.2 yrs.
Far more of a problem is the energy release of our nuclear spaceship. The Saturn V rocket developed a
thrust of TS = v0(dm/dt) = 3.5 x 107 N, based on v0 = 2580 m/s and dm/dt = R = 13,570 kg/s. It
therefore had an energy release of E = (13,580 kg/s)(9.7 MJ /kg) = 1.32 x 1011 watts. Or, or to put it
another way, since the Saturn V was 36% efficient at turning its fuel into thrust, its engines released a
massive blast of (1 – 0.36)(1.32 x 1011 W) = 8.4 x 1010 W of raw heat in addition to their thrust.
This is a wow number. 8.4 x 1010 W is enough to heat all of New York City in a bad winter. To
contain it, the interior linings of the Saturn V rocket engines were engineered right to the breaking
limit of what any material can take before it vaporizes. Yet, even though the Saturn V is the largest
rocket ever launched, there can be no question that any Battlestar Galactica Space Ark large enough to
carry an entire scientific colony (plus the resources for them to live on) through the depths of
interstellar space for over 20 years will have to be far larger than the puny 130 ton payload of the
Saturn V. This is a problem. If the Saturn V rockets were already at the limit of what materials can
stand, what does this say about our deuterium rocket?
Let us assume that our Space Ark is limited to the same heat loss as the Saturn V, or 8.4 x 1010 W.
Dividing by our assumed efficiency of 50% means the deuterium reactor can release 16.8 x 1010 W.
Dividing this by our energy density of 5.7 x 1014 J /kg gives a burn rate of R = 2.95 x 10–4 kg, or only
0.294 grams! The thrust of the deuterium rocket plunges to F = v0R = (0.08)(3 x 108)(2.95 x 10–4) =
7080 N, a factor of nearly 5000 below that of the Saturn V.
The moral of the story? The rocket equation is still good, and the final velocity of the deuterium rocket
will still be 19% of lightspeed. However, the acceleration of the deuterium rocket is going to be
unbelievably slow. A force of only 7080 N, placed against a mass that one realistically must assume to
be much greater than that of the Saturn V, means that it could well take centuries for the starship to
reach that theoretical 19%.
You Do The Calculation
Let us make some educated guesses about the Battlestar Galactica Space Ark, and see where they lead
us. We will assume that some miracle of nanofiber technology allows the Ark’s engine to emit 10
times the waste heat of the Saturn V, or 8.4 x 1011 W. We will assume that advances in controlling
turbulent gasses has raised the thermodynamic efficiency of the ship’s engines to 70%. The energy
density of deuterium is still 5.7 x 1014 J /kg, since that cannot be changed.
For reasons I don’t have room to detail here, we will assume that the payload of the Ark has the mass
of a small ocean-going vessel, about 13,000 tons = 1.18 x 107 kg. The maximum fuel capacity of the
Ark is twice this, or 2.36 x 107 kg, due to concerns about minimizing its mass.
1) At what rate (dm/dt) is the Ark burning its deuterium?
2) What is the thrust of the Ark’s engine?
3) What is the Ark’s acceleration at launch?
4) How long will it take the Ark to reach full velocity?
5) How far from Earth will the Ark be when it reaches full velocity?
6) Suppose the destination of the Ark is Alpha Centauri, 4.26 ly away. Suppose the Ark’s crew is not
interested in waving at Alpha Centauri as they zoom past, but that they wish to stop and visit. This
means their speed must be zero when they arrive. Calculate fuel quantities (M1 = accelerate,
M2 = deaccelerate) that would allow them to make the journey in the minimum amount of time.
```