# Was the Saturn V only going at 1.1km/s after the first stage?

Using data from Wikipedia and NASA, I compute the Saturn V velocity after jettisoning the first stage (S-IC)

\begin{align}V &= \ln(M_{initial}/M_{final}) \times V_e - g \times T_{burn}\\ \\&=\ln(2970t/819t) \times 3kms - 9.81m/s^2 \times 304s\end{align}

Left term

1.3x3kms = 4kms

Right term

-9.81m/s2 (304s)

=-2.9kms

Together

4 - 2.9 = 1.1 km/s

This is a lot less than NASA documents that imply 2kms or more If the S-IC was flying perpendicular to earth at this point how do I account for that?

• I'm not voting to close because this one is extremely easily refuted in a hopefully informative way. – Russell Borogove Jun 10 '18 at 3:10
• @RussellBorogove okay – uhoh Jun 10 '18 at 5:46
• related: space.stackexchange.com/q/27749/12102 and also space.stackexchange.com/q/27743/12102 and also space.stackexchange.com/q/27703/12102 similar MO but each time a new unregistered user. – uhoh Jun 10 '18 at 6:17
• I hadn't realized OP was using a different generic account on every post including responses to self; I'll probably start VTC similar ones in future, unless I get baited into answering again. – Russell Borogove Jun 10 '18 at 12:51

If the S IC was flying perpendicular to earth at this point how do I account for that?

The S-IC was not flying vertically. At first-stage cutoff, at about 161 seconds into the flight, it had pitched over 70 degrees from the vertical -- it was accelerating almost horizontally. Here's a plot of time versus pitch angle from the SA-507 (Apollo 12) Saturn V Flight Manual: At first stage cutoff the rocket had traveled 95 km downrange -- difficult to achieve in a vertical ascent. This table is from the Apollo 11 flight report: If you compute the expected altitude for a pure vertical ascent, you'll likely find a figure much higher than 67km as well.

• Could you elaborate on the definition of "pitch angle" here? -120° at 12 minutes into the flight sounds like it is pointing downwards. If it's due to curvature of Earth and speed increasing, I would expect the curve to get steeper over time. – asdfex Jun 10 '18 at 12:37
• It must be angle relative to the launch pad vertical rather than the local vertical. Occasional downward excursions at orbital insertion aren't unusual, but this chart shows a lot of time spent pointing downwards! The pitch angle would be certainly affected by the curvature of the Earth but not completely determined by it. A person with more spare time than I could cross-reference a table of downrange distance to produce a diagram of the pitch angle relative to the local vertical for a different view on things. – Russell Borogove Jun 10 '18 at 12:47

You significantly overestimate the losses of gravity drag and get some numbers wrong.

The first stage burned for about 2 minutes and 41 seconds or 161 seconds, not 304 seconds which you used for the burn time. Adjusting for this alone brings your solution to about 2284 $\frac m s$. But the Saturn V did about 2756 $\frac m s$ at first stage separation, so where does the rest come from?

$$\Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}} \\ = 3000 \frac m s \cdot \ln \left ( \frac{2970t}{819t} \right ) \\ \approx 3864 \frac m s$$

So we get about 3864 $\frac m s$ of $\Delta v$ from the first stage, and then we sub $161s \cdot \frac {9.81m}{s^2} \approx 1579$ to get $\approx 2284 \frac m s$

But this assumes the rocket went completely vertical, which is wrong.

Take a look at this graphic: The point is you need to do vector addition of both the downward component (gravity) as well as the acceleration vector, which quickly becomes non-vertical. @RussellBorogove has given the graphic, from which you can see that the rocket is at abut 45° pitch angle at about 90s into the flight. Thus, gravity losses are minimized since the rocket actually doesn't point upwards.