How and/or why did Elon Musk mistweet Roadster's C3 so badly?

Question

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?

WARNING:

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

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Answer 1

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.

Answer 2

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!