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)



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?

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    $\begingroup$ I'm not voting to close because this one is extremely easily refuted in a hopefully informative way. $\endgroup$ – Russell Borogove Jun 10 '18 at 3:10
  • $\begingroup$ @RussellBorogove okay $\endgroup$ – uhoh Jun 10 '18 at 5:46
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    $\begingroup$ 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. $\endgroup$ – uhoh Jun 10 '18 at 6:17
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    $\begingroup$ 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. $\endgroup$ – 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:

Plot of pitch angle versus flight time. From 0 degrees (vertical) for the first 20 seconds of flight, the plot slopes down to approximately -70 degrees at 160 seconds (first stage cutoff), holds that pitch angle until about 210 seconds, then continues sloping down generally to almost -110 degrees at Earth parking orbit insertion at around 710 seconds. 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:

Table of nominal and actual values for various flight parameters at first stage separation for Apollo 11. Nominal altitude is 67.4km and nominal surface range (i.e. horizontal travel) is 93.7 km.

If you compute the expected altitude for a pure vertical ascent, you'll likely find a figure much higher than 67km as well.

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  • $\begingroup$ 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. $\endgroup$ – asdfex Jun 10 '18 at 12:37
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    $\begingroup$ 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. $\endgroup$ – 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:

Vector losses

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.

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