[nkoren]

With regards to re-entry: the rocket was going much slower than orbital velocity, and the re-entry was therefore much more moderate.

Some ballpark math: The 1st stage was travelling at 8,300 kph at MECO, and almost of that velocity would have been horizontal. There was no boost-back burn, so it would have kept that velocity until re-entry. Furthermore, it would be re-entering with some vertical velocity. They mentioned that apogee -- when the vertical velocity is zero -- was two minutes prior to the start of the re-entry burn. 2 minutes of freefall would add another 4,200 kph of vertical velocity. So that's 12,500 kph, which is less than 1/2 of a typical orbital re-entry of 26,000 kph. Because the energy grows with the velocity squared, it's more like a quarter of the heat of a typical orbital re-entry.

Finally, as I understand it, much of the point of the re-entry burn is to deflect the re-entry shockwave, effectively creating a shield of "cooler" rocket exhaust around the vehicle.

Still not anything you'd want to take a leisurely stroll in, but nowhere near as stressful as a proper orbital re-entry.

[skrause]

It's actually doing an entry burn to slow down before hitting the atmosphere, you can see it starting here: https://youtu.be/zBYC4f79iXc?t=27m34s

Also your calculation is not correct: You add another vertical 4,200 kph of free fall starting from apogee, but didn't factor in that of course the vehicle also slows down as much in the 2 minutes before reaching apogee. Since MECO is at about 65 km height at 8,300 kph, it will also have 8,300 kph at 65 km height when falling down again (minus a little atmospheric drag).

You also can't just add 8,300 kph horizontal (and it's not actually all horizontal, look how quickly the altitude is rising around MECO) and 4,200 kph vertical speed. Those are not scalar values but vectors, so you need to apply pythagoras for vector addition. 8,3 km/s horizontal plus 4,2 km/s vertical speed is just 9,3 km/s total speed.

[nkoren]

Thanks, I didn't consume enough coffee today.

[bisby]

If it had 8300 horizontal, and 4200 vertical, the total speed would be the hyponetuse of a triangle made by those, not the sum. sqrt(4200^2 + 8300^2) gives me 9,302 kph, which is a bit more than a third of 26000 (.3578). That squared gives you a bit more than an eighth (.1280) of the heat.

and thats not including the fact that someone else pointed out that it would go with velocity cubed.

[LNMNMMMC]

Because the energy grows with the velocity squared, it's more like a quarter of the heat of a typical orbital re-entry.

No, the drag force grows with the square of velocity. https://en.wikipedia.org/wiki/Drag_(physics)#Drag_at_high_ve...

Force * distance / time = power, so the heating will scale with the velocity cubed. Therefore at 1/2 the speed, it would experience 1/8 the heating.

[nkoren]

You are correct!