In this answer to How close would the Tesla Roadster with Starman have to get to Earth in order to become attracted and fall on Earth? I just wrote:

When Roadster was launched from Earth Elon Musk mis-tweeted that it had achieved 12 km^2/sec^2 excess C3 ("energy") relative to Earth after launch. He got the number wrong but the point is that once Roadster had this positive energy relative to Earth it could never "fall back" to Earth.

See Starman/Roadster in a=1.795 AU orbit, now what's the method to this madness? and this answer.

Question: How and/or why did Elon Musk mistweet Roadster's C3 so badly? He predicted a heliocentric apoapsis of 2.61 AU; way past Mars and almost to Ceres. Where is Roadster's close approaches and milestones says

Far point from Sun on May 20, 2020 at a distance of 1.664 AU.

Is there any somewhat reliably sourced-information about how or why this mistweeted announcement was off by so incredibly much? Surely they would have known roughly how much propellant remained while in the 6 hour high parking orbit around Earth, it's hard to imagine believing a number that was off by 1 AU.

Perhaps it had enough to go to 2.61 but its engine was pointed in the wrong direction because it lost attitude control?


This plot in Musk's tweet is now known to be incorrect! See this answer for further clarification.

Elon Musk mistweet https://twitter.com/elonmusk/status/961083704230674438

  • 3
    $\begingroup$ Maybe he tweeted before he'd had enough covfefe that morning $\endgroup$ Apr 20 '20 at 12:11
  • $\begingroup$ @CarlWitthoft that's always my excuse! Perhaps he wasn't on a sufficient "mix of adrenaline, oxytocin, '...caffeine and a desire to help humanity colonize Mars'" (quoting myself here). $\endgroup$
    – uhoh
    Apr 20 '20 at 12:17
  • 2
    $\begingroup$ This is a three-body, three-dimensional trajectory problem with C3 being a metric of an idealized trajectory around Earth and the plot being of an idealized trajectory around the sun, I have no trouble seeing a BOTE 2D approximation giving that kind of error. Where are you getting the information that it lost attitude control? I'm not aware of any official statements that even hint at such a thing. $\endgroup$ Apr 20 '20 at 13:09
  • 3
    $\begingroup$ Given that the actual apohelion is about the tweeted semi-major axis maybe someone just confused the two values.... $\endgroup$
    – asdfex
    Apr 20 '20 at 14:37
  • $\begingroup$ @ChristopherJamesHuff I think it's far more likely that there was a mixup with the numbers as asdfex theorizes than the rocket went off course. The question mark at the end of my sentence there makes it clear that it's speculation not information! If you have no problem seeing this amount of error in 3D I'd love to see that written up in a convincing way using math in an answer! $\endgroup$
    – uhoh
    Apr 20 '20 at 16:51

The C3 was actually pretty accurate, what was not accurate was the plot that followed. The first version of http://www.whereisroadster.com used a trajectory I calculated from the C3 measurement and the time that it left orbit, which lead to something fairly close to the actual trajectory.

My guess is that the image provided was wrong, probably the best case estimate of what the C3 could have been. They must have had slightly less performance from the upper stage than they had hoped, which lead to the lower performance. I believe (From memory) that the final thrust did not occur directly at the perigee, probably because of the very long delay in launch (It launched at the very end of a several hour long launch window), which might have lead to a less than optimal trajectory for reaching towards the asteroid belt.

Bottom line, the C3 value was basically right, but the image was way off.

  • $\begingroup$ Perhaps they had a previous estimate of a model sent to Elon, and he didn't get an updated one once data was gathered. This has happened with my manager a number of times, where we send a preliminary estimate then they never asked for a new estimate once the project was completed. Often times they also communicated out the preliminary causing confusion among those involved. $\endgroup$ Apr 21 '20 at 12:25
  • 2
    $\begingroup$ I'm sure it was something like that. The bottom line, the C3 given seems to be pretty close to correct, but everything else provided was way off. $\endgroup$
    – PearsonArtPhoto
    Apr 21 '20 at 12:33
  • 1
    $\begingroup$ @PearsonArtPhoto because Stack Exchange provides me a way to not do what I'm supposed to be doing each day, I've gone ahead and re-confirmed that. Thanks for the tip! $\endgroup$
    – uhoh
    Apr 21 '20 at 12:39
  • 1
    $\begingroup$ Sure, no problem! I've said it before, I was really nervous that I messed something up when my image looked very different than Elon's for the same C3. It took me a fair bit of work to confirm that I was right, and Elon was wrong... $\endgroup$
    – PearsonArtPhoto
    Apr 21 '20 at 12:52

To stay in practice I'll do the exercise that confirms what @PearsonArtPhoto's answer already says.

$C_3$ the characteristic energy is twice the total energy (kinetic plus potential) $E$ of a body with respect to a larger gravitational body

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

$$C_3 = v^2 - 2\frac{GM}{r}$$

So for example, Earth is bound to the Sun so we'd expect it to have a negative heliocentric $C_3$. The vis-viva equation tells us that Earth's orbital velocity is

$v = \sqrt{GM_{Sun} / 1 \text{AU}}$

or 29.7 km/s and its heliocentric $C_3$ is

$$C_3 = 2 \left(\frac{1}{2} - 1 \right) \frac{GM_{Sun}}{1 \text{AU}}$$.

Okay, now on to Starman...

With Starman's known semimajor axis of about 1.325 AU we can calculate the velocity at Earth's orbital distance of 1 AU as 33.2 km/sec:

$$v = \sqrt{GM_{Sun} \left(\frac{2}{1 \text{AU}} - \frac{1}{1.325 \text{AU}} \right)}$$

or 3.5 km/s faster than Earth assuming that they are not near each other, i.e. $r$ is large. In that case $C_3 = v^2$ or 12.25 km^2/s^s, which is very close to the value of 12.0 shown in the tweet!

  • 1
    $\begingroup$ Thanks for the link to "characteristic energy"- have not heard that term before. That's a useful quantity, and that Maven example was neat as I didn't consider it could be negative! $\endgroup$ Apr 21 '20 at 12:44
  • $\begingroup$ @MagicOctopusUrn oh, that's a very nice example! $\endgroup$
    – uhoh
    Apr 21 '20 at 12:54
  • 1
    $\begingroup$ Do you happen to know why it is called "C_3"? The 3 seems specific. $\endgroup$ Apr 21 '20 at 12:59
  • $\begingroup$ @MagicOctopusUrn no I don't but I think somewhere I'd seen that explained, possibly in this site. I think it's an excellent new question! $\endgroup$
    – uhoh
    Apr 21 '20 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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