I don't understand the speed readings for stage 1 on CRS-12

Question

I don't understand the speed reading for the CRS-12 misson's stage 1.

While being around an altitude of ~100km the minimum speed that is shown in the webcast for stage 1 is ~1600km/h.

If the speed shown were the horizontal speed measured relative to earth it should reach zero at some point, as it turns around and flies back to the Landing Zone in Cape Caneveral.

It can't be the vertical speed as it is still a couple of thousand km/h while being on a constant altitude.

A combination of the two couldn't be much more than 100km/h at some point during the turnaround / minimal climb period.

So how am I to read that velocity data, does anyone know?

There is a nice plot of the trajectory in the image shown below, which seems plausible because of the fact that the stage 1 actually lands only a few km away from the lift-off site, but the speed data shown in the video simply does not match with that plotted trajectory.

So what am I missing here?

UPDATE:

After looking at the great answers below i now realize that even with the 'not to scale' image above you can understand whats happening (once the nice ppl on stackoverflow explain it to you ;) ), the turnaround is 'roughly' circular, so imagening it as almost circular explains why the speed does not drop to zero, the combination of vertical + horizontal movement keeps the total speed high.

Thanks guys :)

Answer 1

Your intuition about the combination of the two is leading you astray. In fact, the speed is a combination of the two. 1600km/h is 450m/s, which is practically stationary in terms of orbital mechanics. Also, remember that any graphics you see are not to scale.

For the more detailed answer: The boostback burn has to negate all the horizontal velocity and add reverse horizontal velocity to come back. Given that it has probably flown downrange on the order of 50km by stage separation, it needs to supply enough horizonal velocity to come back all of those 50km. It can be done quickly or slowly. If it is done quickly, a lot of fuel will be used in adding horizontal reverse velocity and then cancelling it out again. If it is done slowly, then the rocket might fall back to earth before it returns close enough to the pad. But the magic of the boostback burn is that it preserves vertical upwards velocity, and so a large amount of time is gained for the horizontal trip to the pad. But this time gained is on the order of a minute or two, and so the horizontal velocity reflects that.

Just some rough numbers: $$50\mathrm{km} / 120\mathrm{sec} \approx 420\mathrm{m/s} \approx 1500 \mathrm{km/h}$$

So the speeds shown in the webcast at highest point are in fact spot on.

Edit: uhoh has put together some awesome graphics, showing just how aggressive the horizontal trajectory is. Check uhoh's answer out!

Answer 2

I like the approach and explanation in Quietghost's answer and just want to add a screen shot and link to some helpful resources.

I took a screen shot at around T+ 00:04:10 or about 250 seconds after launch. At an altitude of 118 km the speed relative to the Earth's rotating frame is 1658 km/h or 461 m/s.

Remember that these numbers are in the rotating frame of the Earth, and that you have to tack on roughly 70 m/s to translate that back into an Earth-centered inertial frame.

Flightclub.io is a really fun site to run simulations of particular launches. I re-ran the the CRS-12 simulation that is already set up on the website and plotted velocity and the altitude/range profile below.

Also don't forget that the Downrange distance plot is a 2D projection of a three-dimensional orbit in the Earth's rotation frame. The x-axis is distance from the launch site to the ground-track of the orbit, the point on the Earth below each point in the orbit. But that distance along the surface can be in any direction. While it is mostly East in the beginning, the final landing site is more south-southwest (shown below in a screnshot from Google maps.