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Reposting as a question.

Uhoh's answer allows us to compute the velocity at a given r.

Let's see what velocity is at $r=100000 \text{ km}$:

\begin{align}\mathscr{E}_{tot} &= {1\over 2}v^2 - {GM\over r}\\ -0.7\text{ kg}\cdot \text{m}^2 /\text{s}^2 /\text{kg} &= {1\over 2}v^2 - {4e8\text{ m}^3 /\text{s}^2\over 1e8\text{ m}}\\ -0.7\text{ m}^2/\text{s}^2 &= {1\over 2}v^2 - 4\text{ m}^2/\text{s}^2\\ 3.3\text{ m}^2/\text{s}^2 &= {1\over 2}v^2\\ 6.6\text{ m}^2/\text{s}^2 &= v^2\\ v &= 2.6\text{ m}/s\end{align}

At 100,000 km from Earth the CSM would be traveling at 2.6 m/s... And that's not even halfway to the moon.

I think somebody said the math was wrong, but... how?

So is this bad news?

user26140
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    If you have 10 cubic kilometers, and you divide by 5 kilometers, you do not suddenly get 2 square kilometers. That doesn't even make any sense. (You would arguably just get 2 cubic kilometers, since if e.g. you have a 1x1x10 km rectangular volume, then divide by 5 km, it's now a 1x1x2 km volume, or 2 cubic km.) – Nathan Tuggy Jun 10 '18 at 21:01
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    @NathanTuggy your statement's completely incorrect. A 2 square kilometer square, times 5 kilometers in an orthogonal direction, absolutely is 10 cubic kilometers, and the math works both ways. Let's not pile incorrectness on whatever this is. – Erin Anne Jun 10 '18 at 21:04
  • The value of GM is totally wrong, it is 4E14. Wrong by a factor of 10E6.

    3.986E+5 km^3/s^2 is 3.986E+14 m^3/s^2

     – Uwe Jun 10 '18 at 21:09
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    @ErinAnne: Huh. I think you're actually correct, so now I'm puzzled why this was so hard to verify. – Nathan Tuggy Jun 10 '18 at 21:11
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    @Uwe that's the answer I was just about to write. Just put it in an answer. That's the answer. – Erin Anne Jun 10 '18 at 21:11
  • +1 This is a completely valid question and I don't think it should be down voted at all! "Here is my orbital mechanical calculation but the result isn't generating the expected result; can someone help me resolve this discrepancy?" – uhoh Jun 11 '18 at 11:37

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Uwe's absolutely right. Your unit conversion from the GM value given by uhoh (which I believe is correct but haven't checked) assumes that $1 km^3 = 1000 m^3$. $1 km^3$ is actually $$(1000 m)^3 = 1E9 m^3$$

You'll find that substantially changes the value of $v$.

Erin Anne
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    Thanks for looking into this! When doing physics all by myself in my snow white cube I stick strictly to MKS units, but I frequently see orbital mechanics with mixed kilometers and seconds (e.g. JPL's Horizons, this stuff, etc.) and meters has too many zeros, and I don't want to start adding comma separators. – uhoh Jun 11 '18 at 11:34
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    @uhoh: it is absolutely ok to use km instead of m for large distances. Mm (megameter) and Gm and (gigameter) should be used more often. A light year is 9.46 Pm (petameter). – Uwe Jun 11 '18 at 12:08