I was watching SpaceX's launch of the Japanese satellite JCSAT-16 yesterday, and the second stage at a point seemed to decrease in altitude (from 171km to 165km) at about T+00:07:00 in this video. Is that normal or was that a glitch?

  • 2
    $\begingroup$ Are you referring to what starts happening around T+00:07:00, which is about 24 minutes into the video? A wild guess, but as velocity is still increasing (and IIRC, below orbital at that altitude) it could be the effect of an elliptical orbit injection. I'm pretty sure that 5.5 km/s is not sufficient to attain orbit at 170 km. $\endgroup$
    – user
    Commented Aug 15, 2016 at 9:35

2 Answers 2


This is actually pretty common in rocket launch profile design. For the Ariane rocket, we covered it in What is the reason for the Ariane 5 launcher with Intelsat 29e losing altitude?, but the same principal applies in general. When you have a low thrust high ISP upper stage, it is common for it to thrust more horizontally at a certain point than vertically, which might cause the altitude to go down somewhat. The speed eventually will be such that the altitude begins to rise, as the curvature of the Earth drops off more.

  • $\begingroup$ omg - the question is only "it started going down a little - is that wrong/bad?" I saw "glitch" and went deep! $\endgroup$
    – uhoh
    Commented Aug 15, 2016 at 12:21
  • $\begingroup$ That's what I assume, and, well, my answer was accepted, so I guess it was? $\endgroup$
    – PearsonArtPhoto
    Commented Aug 15, 2016 at 12:35
  • 2
    $\begingroup$ Oh, I'm sure now that you are right. I'm poking a bit of fun at myself. $\endgroup$
    – uhoh
    Commented Aug 15, 2016 at 12:40
  • $\begingroup$ This is the correct answer. To add to this, the term of art is a "lofted trajectory." $\endgroup$
    – Tristan
    Commented Aug 15, 2016 at 15:49

There doesn't seem to be any glitch as far as I can tell.

I watched the video. The altitude is expressed in integer kilometers. We don't know for sure if it's rounded or truncated, but a good guess would be truncated, so that 171000 and 171999.99 would both show up as 171 km, 'odometer style'.

Every time I see the altitude change, I hit the spacebar (1/5 second reflexes after coffee, on a good day) and type the T+ time, the altitude, and the speed.

Why speed? for two reasons1

  1. Always take more data than you think you need! For a reason, you can choose between "Just do it" or "Hey, you never know". I'm a Hey, you never know person myself. Data can come in handy for many reasons later on. One example is the next item
  2. See if other data has a glitch - is the frame rate smooth, is the telemetry smooth, is the trajectory smooth... helps differentiate between different plausible explanations if you verify a potential glitch.

Here's the result. I don't see a glitch. There is gap in the altitude plot because the 171 never "rolls over" to 172, so we have no information about the altitude except that it's somewhere between 171 and 172. Remember, the data points represent when the odometer rolls over - the only time when we can even guess at the fractional part.

enter image description here

above: Plot of the data I took from the video, using the Python script below.

Discussion: For a circular low earth orbit (LEO), or any orbit actually, you can get the orbit velocity $v_{orbit}$ at a distance $r$, as long as you know the semi-major axis $a$, the gravitational constant $G$ times the mass of the body you are orbiting, in this case $M_e$ for the Earth by using the vis-viva equation:

$$v_{orbit}^2=GM_e (\frac{2}{r}-\frac{1}{a}) $$

where $GM_e$ is about 3.9860E+14 m${}^3$ / s${}^2$. If choose the radius of the earth $r_e$ to be 6371 km, and calculate a circular orbit ($r = a$) at an altitude $h$ of 171 km so that $a = r_e+h$, I get an orbital speed of about 7805 m/s. If I project that on to a non-rotating Earth 171km below, that's 7601 m/s.

At 28 degrees latitude, the earth is spinning at about 410 m/s, so assuming we're in a 28 degree inclination plane and still nearby the launch site, the ground speed will be about 7192 m/s.

You can see in the plot that throughout this part of the video, the 2nd stage is well below orbital speed, and thus is sub-orbital. However it's picking up substantial speed and getting much closer. Still you see a rising and falling arc that you might expect of something suborbital, and the decrease in the rate of fall is consistent with that - thus the asymmetric shape.

enter image description here

above: Odometer rollover from here.

screendata = """05:39   165  3646
05:45   166  3715
05:50   167  3790
05:57   168  3879
06:05   169  3988
06:14   170  4124
06:28   171  4333
06:45  -171  4623
06:55  -171  4804
07:05  -171  4990
07:12   170  5132
07:28   169  5466
07:40   168  5739
07:50   167  6006 
08:01   166  6299
08:13   165  6651"""

import matplotlib.pyplot as plt
import numpy as np

lines = screendata.split('\n') 

time, alt, speed = zip(*[line.split() for line in lines])

time = [ [int(thing) for thing in t.split(':')]
          for t in time]

time = np.array([ 60*m + s for m, s in time], dtype=float)

alt, speed  = [np.array([int(item) for item in thing], dtype=float)
               for thing in [alt, speed] ]

alt[alt<0] = np.nan       # those marked with a minus didn't actually change



plt.plot(time, alt)
plt.plot(time, alt, 'ok')

plt.ylim(min(alt)-1, max(alt)+1)
plt.title("altitude (final upon each integer km change in video) vs T+ (sec)")


plt.plot(time, speed)
plt.plot(time, speed, 'ok')

plt.ylim(min(speed)-1, max(speed)+1)
plt.title("speed vs T+ (sec)")

  • 1
    $\begingroup$ +1 for teaching a man to fish and handing him the requested fish in the same answer. $\endgroup$
    – Mast
    Commented Aug 15, 2016 at 22:41

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