Question:

Answer:

This question can be answered in different ways, at different levels of precision. Even when we want an accurate answer, it is worthwhile to do rough estimates first, both as a check on later answers and because the insight gained in the estimation process is often valuable.

1. At the simplest level, the Saturn V was required to launch the lunar exploration spacecraft (the Command Module, CM, Service Module, SM, and Lunar Module, LM) on a trajectory to the Moon. In the approximation that the Moon is far from the Earth (compared to the size of the Earth), it must provide the payload with a mission velocity close to the earth escape velocity, 11.2 km/s, or roughly 7 miles per second. While there are corrections to this figure for several reasons (see below), they are mostly at the few percent level for a large rocket like the Saturn V, so the final answer must be not too different from 11.2 km/s, but of course somewhat larger, to take account of the various inefficiencies discussed below.

2. Because the question specified the velocity reached in space when beginning fully fueled (sometimes called the ideal velocity), now let us guesstimate the major corrections to the above 11.2 km/s.

   These are as follows, roughly in order of magnitude.

   Gravity:

   While the engines must obviously do work against gravity to lift the vehicle, here this term refers to another sort of gravitational loss that may be understood by imagining the extreme case in which the rocket's thrust was exactly equal to its weight. Then the vehicle would not move at all (at least until it became lighter), yet fuel would be burned: it would simply sit on its exhaust, being supported against gravity but making no progress.

   If, on the other hand, the thrust were extremely high, so the propellant were used almost instantly, the rocket would immediately reach its final velocity, and no such loss would be incurred. Because of the large heavy engines required, atmospheric drag, and stress on the vehicle and its occupants, this extreme is not a feasible choice.

   The optimal compromise is somewhere in between these two extremes. Alternatively, one could avoid gravitational losses entirely by accelerating only horizontally to escape velocity. This is also not possible due to the effects of atmospheric drag. The best trajectory will thus go up to get above the effective limits of the atmosphere as fast as possible, but also bend over to do the major acceleration somewhat above about 60 km. The losses due to gravity near the optimum trajectory are typically about 10% of the mission velocity for launches from the surface of the Earth.

   Earth rotation:

   Because the Earth rotates, whenever possible one attempts to take advantage of this free velocity increment by launching towards the east. At the latitude of Cape Kennedy, this gave about 410 m/s, or -3.6% of the mission velocity (minus meaning reduces necessary velocity requirement from the launch vehicle).

   Atmospheric drag:

   The trajectory is designed to get above the denser atmosphere as quickly as possible, then bend over to the east to avoid unnecessary gravitational losses. Here the large size of the Saturn V is a big help; let us guess +5% of the mission velocity.

   Finite distance to the Moon:

   Since the Moon is about 60 Earth radii away, the potential energy (proportional to 1/R) is about 1/60th that at the Earth's surface. Thus the energy to reach the Moon is (59/60)*(escape to infinity energy) and the lunar mission velocity is about -1% (less) than escape velocity, or 11.1 km/s.

   Collinearity:

   Ideally the rocket would accelerate in a straight line. The same effect may be largely achieved by keeping the thrust direction parallel to the velocity vector, but some steering accuracy errors must be expected. They should be quite small, let us guess 1%.

   Contingency reserve.

   This is a matter of policy as much as physics, although the uncertainties in the above factors from one flight to the next play a role. In any case, it will not do to come out 0.5% short, and some margin must be allowed. For the space shuttle I believe I recall 3%, so I adopt that here.

   Adding everything together, we get 14.4%, (1.00+0.10-0.036+0.05-0.01+0.01+0.03 = 1.144), which then gives about 12.8 km/s as the necessary vehicle velocity capability. While the approximations and guesses above are rough at best, they give us a tighter bound on the answer to expect. I will be surprised if the answer computed in part 3 below comes out above 14.4 km/s or less than 12.0 km/s, in particular.

3. At the next level of precision, we can compute the ideal answer based on the various masses and the performance of the engines.

   The essential fact one needs in computing the velocity is the rocket equation, which gives the ideal velocity v of a single-stage rocket in terms of the exhaust velocity c and the masses via

   v = c*ln(M1/M0).

   Here M1 is the initial mass of the rocket, M0 is the final mass (empty tanks, engines, payload, any residual propellants, etc) at engine cutoff, and ln is the natural logarithm. ( The natural logarithm is the inverse of the exponential function. That is, if y=ln(x), then x=e^y = exp(x), and vice versa, where the number e=2.71828... is a universal irrational mathematical constant, rather like pi. ) The equation is valid for a rocket accelerating in a straight line, in empty space, in the absence of any complications due to gravity, atmospheric drag, or other factors.

   The quantity M1/M0 = R is known as the mass ratio of the rocket. Hence, if R = e, v = c, and the rocket reaches a speed just equal to its own exhaust velocity; if R = e^2 (= 7.389... approx), v = 2c, and if R = e^3 (= 20.086... approx), v = 3c, etc. In practice engineering factors limit R to fairly small values, generally less than 10 for vehicles taking off from the Earth and carrying any significant payload.

   For a multi-stage rocket like Saturn V, we apply the formula to each stage in turn, adding up the velocities as we go. Thus we need, for each of the three stages,

   1. the total mass, M1, at ignition
   2. the remaining mass, M0 at engine cut-off
   3. the exhaust velocity, c.

   The Saturn V used five large Rocketdyne F1 engines, powered by liquid oxygen (LOX) and RP-1 (roughly kerosine) for the first stage, five Rocketdyne J2 liquid hydrogen (LH2) and LOX engines for the second stage, and a single J2 for the third stage.

   Sutton & Ross (1975 Rocket Propulsion Elements, Wiley) quote a specific impulse Isp of 304.8 s for the F1 and 426 s for the J2, both in vacuum. Isp is defined as the number of pound-s of impulse (thrust times duration) given by 1 pound (mass) of propellant. It is related to the exhaust velocity c [m/s] by c = g*Isp, where g = 9.80665 m/s^2 is the standard acceleration of gravity.

   The most comprehensive information on the masses that I have been able to find is in The Saturn V Launch Vehicle Home Page:

    #    Stage      m0       mp       m1   
   1.    S1C     300.0   4492.0   4792.0
   2.    S2       95.0    942.0   1037.0
   3.    SIVB     34.0    228.0    262.0
   4.    IU        4.5      0.0      4.5
    5.   payload     -        -     109.6 
   

   where the payload is for Apollo 11, IU is the Instrument Unit (logically, part of the dry mass of stage 3), and SIC, SII, and SIVB are the three rocket stages.

   Here m0 is the stage empty weight, mp is the propellant, and m1 is the stage total, in thousands of pounds. Note that M0 differs from m0 (and M1 from m1) in that the lower case letters denote the masses just for each stage alone, without regard to the mass of any stages higher up, where the upper case letters, which we need to apply the rocket equation, are cumulative, so that, eg, for the Saturn V, M0 for the second stage includes the loaded mass m1 of the SIVB, and the masses of the IU and the payload.

   It is important to realize that the above figures are not absolute; the launcher improved steadily even during the few years it was in use. For example, Von Braun (1975, NASA SP-340) quotes the payload mass above the IU as rising from less than 80,000 lb on Apollo 8 (which carried a structural test article instead of a live LM) to 116,000 lb on Apollo 17 -- a 45% improvement.

   Then converting Isp to c, m's to cumulative M's and expressing everything in metric units, we obtain for the exhaust velocities & cumulative masses, [kgm]:

   Stage  c [m/s]       M0          M1
      1    2989.1    777065.8   2814637.0
      2    4177.6    213694.6    640985.8
      3    4177.6     67181.8    170602.6
   

   With these numbers in hand, it is easy to compute the mass ratios and the natural logarithms ln(R) of each stage, the stage velocities and finally the total velocity:

         Mass Ratio (R) and Velocities [m/s]
   
   Stage  mass ratio R    ln(R)   Stage v   Total v
      1       3.622       1.287    3847.1    3847.1
      2       3.000       1.098    4589.0    8436.1
      3       2.539       0.932    3893.3   12329.4
   

   The answer to the problem is thus 12.33 km/s.

   Looking back at our guesstimates in Section 2 above, we see that the value we obtain is close to, but a little less than, our best guess. Considering that the guesses were really just off the top of my head, this seems okay. Taking 10.7 km/s as the absolute minimum (considering the rotation of the Earth and the finite distance of the Moon), the ratio 12.3/10.7 is 1.1495, so the ideal vehicle capability exceeds the bare minimum mission velocity by about 15%.

Afterword:

After submission of the above I realized that in Part 2 I had neglected to include the effect of reduced engine efficiency at sea-level pressure. This does not affect the final answer of 12.33 km/s, but does come into the efficiency estimates. Sutton & Ross (1975) give the sea-level Isp of the F1 engine as only about 265 s, 15% less than the vacuum value. The J2 engines all operate at high altitude and should be unaffectd. The effect for the F1 engines on the SIC 1st stage can be roughly estimated as a 15% loss during the time it takes to traverse the first scale height of the atmosphere (about 1 min), and 0 thereafter. Given the first-stage burn duration of 170 s, this means the loss is about 5% of the SIC ideal velocity of 3847 m/s, which is 193 m/s, or about 1.7% overall. In particular, if considered above it would have raised my lower "surprise threshold" from 12.0 km/s to 12.2 km/s, nearly the 12.3 obtained. While the discrepancy can hardly be called serious, the obvious places to look for its origin would be my guesses of the gravitational effect and atmospheric drag, as these are both relatively large and uncertain. If they were reduced from 10% and 5% to 9% and 4.5%, respectively, good agreement would be restored. At the least, this suggests that the losses due to the factors in Section 2 may be somewhat less than my guesses.