SpaceX Demo-2

Kinematic Study of the Falcon 9 ๐Ÿš€

Crรฉdit: NASA/Bill Ingalls

After having to be postponed due to bad weather1, last Saturday May 30, at 21:22 peninsular time, the Falcon 9 rocket from SpaceX launched the second demonstration mission (Demo-2) of Crew Dragon from the Launch Complex 39A (LC-39A) at the John F. Kennedy Space Center of NASA in Florida.


Approximately 19 hours later, the Crew Dragon ship docked autonomously with the International Space Station, with astronauts Bob Behnken and Doug Hurley on board, which has meant that the United States is putting humans back into space for the first time since 20112.

Demo-2 is the last major test of the SpaceX manned space flight system to be certified by NASA for manned missions to and from the International Space Station.

In this video of almost 5 hours of duration, published by SpaceX, you can know many more details about the mission (if you only want to see the launch jump to 4:22:46):

If you wish to read more about this historic mission you can do so at NASA’s official website.

Kinematic Study of the Falcon 9 ๐Ÿš€

The post could have ended with the previous paragraph, but then you wouldn’t know if you were in Hello! magazine or in Physichemically with Rodri’s blog ๐Ÿ˜.

If you look at the video of the launch, in the lower left corner you can see the module of the velocity (speed from here on), in km/h, and the altitude, in km, of the rocket in real time as it ascends to approximately 200$\thinspace$km. What did I think when I saw that data? Well, write the values down3, represent them and make a small empirical study about the cinematics of the Falcon 9.


The following plot shows the altitude of the Falcon 9, in km, as a function of the elapsed time, in minutes 4:

The altitude rises rapidly during the first two minutes approximately (until minute 2.6), surpassing the 75$\thinspace$km of altitude, when the nine Merlin engines of the Falcon 9 shut down, instant that is known as MECO (Main Engine Cutoff) 5.

From that moment the altitude continues to increase reaching 200$\thinspace$km after approximately 5 minutes of flight and remaining constant.

SECO means Second-Stage Engine Cutoff and represents the moment when the Merlin Vacuum engine, the only one that was driving the second stage of the rocket (to which the Dragon ship itself is attached, where the astronauts were going), stops, which doesn’t seem to affect too much the altitude of the Dragon.


The following plot shows the speed of the Falcon 9, in km/h, as a function of the elapsed time, in minutes 6:

The speed increases in a non-linear way, reaching 6724$\thinspace$km/h, more than 5 times the speed of sound in air 7, in the MECO. It is observed that at that moment the speed is even reduced, until the Merlin Vacuum engine of the second stage is started and accelerates the Dragon with a similar tendency as it had done during the first stage.

It’s nice to see how in SECO the Dragon stops accelerating, because it doesn’t have any engine driving it anymore, keeping its constant speed from then on (describing an uniform circular motion).

Orbital Speed

The maximum value of the speed is approximately 27000$\thinspace$km/h. Can we understand this value? Indeed, as of approximately minute 9, the Dragon ship is in an orbit at an altitude of about 200$\thinspace$km. Assuming a circular orbit, the orbital speed is given by the expression:

$$ v_\text{orbital} = \sqrt{\frac{GM_\mathrm T}{r}}, $$ where $G = 6.67\times 10^{-11}\thinspace\mathrm{m^3\thinspace kg^{-1}\thinspace s^{-2}}$, $M_\mathrm T = 5.97\times 10^{24}\thinspace\mathrm{kg}$ is the mass of the Earth y $r = R_\mathrm T + h$ is the distance the ship is measured from the center of the Earth, with $R_\mathrm T = 6371\thinspace\mathrm{km}$. For an altitude $h = 200\thinspace$km, we have:

\begin{align*} v_\text{orbital} = \sqrt{\frac{GM_\mathrm T}{r}} &= \sqrt{\frac{6.67\times 10^{-11}\cdot 5.97\times 10^{24}}{(6371+200)\times 10^3}} \\
&= 7784.6\thinspace\mathrm{m/s} \approx 28000\thinspace\mathrm{km/h} \end{align*}

which is a relative error of about 3.7$\thinspace$%.


From the speed values it is possible to obtain the tangential acceleration of the rocket by means of a numerical derivation8.

The following plot shows the acceleration of the Falcon 9, in m/s2, as a function of the elapsed time, in minutes:

It is clear that the acceleration is not constant, increasing until MECO, when it even takes negative values (remember that speed is reduced). Then it increases again to values above 30$\thinspace$m/s2 (more than three times the acceleration of gravity on the surface of the Earth), until SECO, when the tangential acceleration vanishes because there is no longer any engine powering the ship.

What if we assume that the acceleration is constant?

If the acceleration of the rocket was constant, then its ascent could be modelled by an uniformly-varied linear motion (UVLM). Looking at the previous plot it seems crazy to think that it could be like that, but it’s worth trying as a mental exercise.

The following plot shows again the empirical acceleration of the rocket, obtained by numerical derivation from its speed, and the constant acceleration that it would have assuming a UVLM, obtained as the arithmetic mean9:

The resulting average value of the acceleration before SECO is 14.1$\thinspace$m/s2, $\approx 1.4$ times higher (in module) than the acceleration of gravity on the Earth’s surface (9.8$\thinspace$m/s2). This can be interpreted as that, on average, astronauts have spent almost 9 minutes experiencing something worse than a free fall, and on top of that, upwards ๐Ÿ™ƒ 10.

Once we have our constant acceleration value, we can compare the empirical altitude and speed with those obtained from the UVLM expressions (taking into account that after SECO the acceleration is zero and therefore the ship will move with an uniform linear motion or ULM).


After four minutes, the rocket is maintained at a practically constant altitude, so the UVLM or ULM expressions are not valid, as they imply an indefinite increase.

Up to approximately 4 minutes, the theoretical altitude is calculated from the expression:

$$ h(t) = h_0 + v_0 t +\frac{1}{2} a t^2, $$ where $h_0 = 0$, $v_0 = 0$ and $a = 14.1\thinspace$m/s2.

In the following plot both the empirical altitude and the one calculated assuming an UVLM, during the first four minutes of the Falcon 9’s ascent, are shown:

The theoretical expression is only able to model the movement of the rocket during the first instants of time (already in the first minute of the ascent the theoretical expression has a relative error of almost 130$\thinspace$%).


The theoretical speed is calculated from the expression 11:

$$ v(t) = \begin{cases} v_0 + a t & \text{before SECO (UVLM)} \\
26734.6 & \text{after SECO (ULM)} \end{cases} $$ where $v_0 = 0$ and $a = 14.1\thinspace$m/s2.

The following plot shows both the empirical and calculated speed assuming an UVLM and subsequent ULM:

It is observed that the theoretical expression overestimates the speed of the ship before SECO (with a maximum relative error of more than 300$\thinspace$%, for $t = 0.2\overline{6}\thinspace$min), and underestimates it slightly afterwards.

Still, it seems that the theoretical expression does not deviate so much from the empirical values, indicating that, at least for estimating speed, it does not seem so farfetched to model the rocket ascent by an UVLM (and subsequent ULM after SECO).

  1. The launch was originally scheduled for Wednesday, May 27, but had to be cancelled with only 17 minutes left because of Tropical Storm Bertha. ↩︎

  2. On July 8, 2011, the 135th and final mission of NASA’s Space Shuttle Program took place. ↩︎

  3. I’d love to tell you that I used a fully automated algorithm with optical character recognition (OCR) to read the values in the video, like some other freaks and more capable than me have managed to do. But no, I’m afraid all I did was play the video in 10 second jumps, manually entering the speed and altitude values ๐Ÿคทโ€โ™‚๏ธ. ↩︎

  4. To put this data into perspective, it takes a commercial plane about 10 minutes to reach its cruising altitude, which is about 10$\thinspace$km. In other words, in half the time, the Falcon 9 is capable of reaching an altitude about 20 times higher than the cruising altitude of a commercial plane. ↩︎

  5. One of the distinguishing features of SpaceX’s Falcon 9 is that the first stage of the rocket, once separated, is capable of returning to Earth and landing on its own, as shown in this gif ๐Ÿ˜ฒ:

    via GIPHY

  6. Again to put this data into perspective, it takes a commercial plane about 10 minutes to reach its cruising speed, which is about 900$\thinspace$km/h. In other words, in the same time, the Falcon 9 is capable of reaching a speed about 30 times higher than the cruising speed of a commercial plane. ↩︎

  7. At 20$\thinspace^\circ$C temperature, 50$\thinspace$% humidity and sea level ( ↩︎

  8. Specifically, the acceleration has been obtained using the diff function of MATLABยฎ. ↩︎

  9. Actually, two different averages have been taken, before and after the SECO, due to the importance and influence that moment has on the ship’s movement. ↩︎

  10. Actually it would have been much worse than this ๐Ÿคฆโ€โ™‚๏ธ, but as a matter of fact, a skydiver usually reaches the terminal velocity (around 180$\thinspace$km/h) in only 12 seconds, after which he stops experiencing the feeling of falling. ↩︎

  11. The value of 26734.6$\thinspace$km/h is the speed the ship has, according to the theoretical expression of the UVLM, right in the SECO. ↩︎

Rodrigo Alcaraz de la Osa
Rodrigo Alcaraz de la Osa
PhD in Physics and Physics and Chemistry Teacher

I teach Physics and Chemistry at IES Rรญa San Martรญn in Cantabria (Spain).