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?

  • 1
    $\begingroup$ 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.) $\endgroup$ Commented Jun 10, 2018 at 21:01
  • 3
    $\begingroup$ @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. $\endgroup$
    – Erin Anne
    Commented Jun 10, 2018 at 21:04
  • $\begingroup$ 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 $\endgroup$
    – Uwe
    Commented Jun 10, 2018 at 21:09
  • 1
    $\begingroup$ @ErinAnne: Huh. I think you're actually correct, so now I'm puzzled why this was so hard to verify. $\endgroup$ Commented Jun 10, 2018 at 21:11
  • 1
    $\begingroup$ @Uwe that's the answer I was just about to write. Just put it in an answer. That's the answer. $\endgroup$
    – Erin Anne
    Commented Jun 10, 2018 at 21:11

1 Answer 1


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 $v$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – uhoh
    Commented Jun 11, 2018 at 11:34
  • 1
    $\begingroup$ @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). $\endgroup$
    – Uwe
    Commented Jun 11, 2018 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.