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This is from the Apollo 11 mission report. It says the vehicle was going at 7.6 km/s when the CSM separated from the third stage. It was not in orbit at this point, it was thousands of KM from earth.

Is this accurate? Am I missing something? That seems incredibly far below other sites saying that you need 10km/s or more to leave the earth.

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    $\begingroup$ "It was not in orbit at this point, it was thousands of KM from earth." One does not follow from the other. In this answer you can see four spacecraft that orbit as far as 20,000 to 40,000 km from the Earth, yet are still definitely in "Earth orbit". $\endgroup$ – uhoh Jun 10 '18 at 6:15
  • $\begingroup$ @JCRM No the numbers are not from that link. Those altitudes and velocities are in the 7,000's The numbers are from this link about Apollo 11. ibiblio.org/apollo/Documents/lvfea-AS506-Apollo11.pdf $\endgroup$ – uhoh Jun 11 '18 at 1:05
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As @Rikki-Tikki-Tavi points out escape velocity is the velocity you would need at (or near) the surface of Earth to make out out of Earth orbit. Of course, just like anything thrown up into the air, the spacecraft decelerates as it moves away from the Earth.

Having escape velocity means that your total energy (relative to the Earth in this case) is greater than or equal to zero.

Using the equations from this answer

$$\mathscr{E}_{tot} = \frac{1}{2}v^2 - \frac{GM}{r}$$

where $\mathscr{E}_{tot}$ is energy per kilogram. Since both terms use the same $m$ you can just calculate energy per kilogram by dropping it.

Let's do the calculation for the "actual" numbers in the table.

altitude       7066 km
Earth radius   6378 km
r             13444 km
GM         3.986E+5 km^3/s^2
v             7.609 km/s

$\begin{align}\frac{1}{2}v^2 & =28.948\\ -\frac{GM}{r} & = -29.649\\ \mathscr{E}_{tot} & = -0.701 \ km^2/s^2 \end{align}$

So the spacecraft has plenty of energy, but not quite enough to completely escape Earth's gravity. That means if it didn't approach the moon, it would keep going until all of the kinetic energy was used up by being stored as potential energy.

$-\frac{GM}{r_{max}} = -0.701$ gives

$r_{max} = 569,000$ km. or a little past the orbital radius of the Moon if the Moon's gravity wasn't there.

So the numbers work out just fine.

If the spacecraft returned to LEO at say 400 km altitude or $r = 6778$, the velocity would then be given by

$-0.701 = \frac{1}{2}v^2 - \frac{GM}{6778 km}$

$v = \sqrt{2(-0.701 + GM/6778 km)} $

and that gives 10.78 km/s

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    $\begingroup$ I think the point here is that Apollo never required Earth escape velocity because it never left Earth's sphere of influence (after all, the Moon is orbiting the Earth). It would have been undesirable for Apollo to be on an Earth escape trajectory for a couple of reasons: (1) it would have involved needless expenditures of propellant acquiring and then discarding excess velocity; (2) if the burn to enter lunar orbit failed to occur, Apollo would have simply orbited back to Earth, rather than wander off into a solar orbit that may never bring it home. $\endgroup$ – Anthony X Jun 10 '18 at 23:54
  • $\begingroup$ @JCRM "not sure where 7.609 comes from" I used the values that are shown in the "actual" column of the data show right there in the OP's question. i.stack.imgur.com/3SGHo.png I've rolled back as I'm answering the question as asked. Thanks for the formula align though, that looks great! $\endgroup$ – uhoh Jun 11 '18 at 0:33
  • $\begingroup$ From the linked PDF Table 7-11 page 7-9 (117/359) I was unable to find table 4-3 in that document. $\endgroup$ – JCRM Jun 11 '18 at 0:43
  • $\begingroup$ @JCRM That's an interesting, but different problem. The OP may have been reading multiple documents and linked a different pdf, but the question is about making sense of the numbers so it's best to answer using the numbers the OP gives. $\endgroup$ – uhoh Jun 11 '18 at 0:49
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    $\begingroup$ Absolutely agree you should use the OP's figures - I thought I was (reading fuzzy graphics isn't my strong point) $\endgroup$ – JCRM Jun 11 '18 at 1:32

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