# CSM velocity / apollo program

Reposting as a question.

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?

• 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
• @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
• @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
• @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

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 $v$.