6
$\begingroup$

This comment mentions:

This relation between drag and mass is taken to a relative (for satellites) extreme in GOCE which I think needed to be close to Earth to accurately sense the changes in gravity, but "streamlined" (have a low frontal surface area) to not decelerate too much in order to reach a mission duration of at least 20 months. Some called it "the Ferarri of Space" (Phys.org, 12-Sep-2013: 'Ferrari of space' set to fall to Earth)

And that Phys.org article says:

Circulation Explorer (GOCE) orbits at an extremely low altitude of just 260 kilometres (160 miles), where there are lingering molecules of atmosphere.

To reduce drag, it has an arrow-like octagonal shape and two fins to provide extra aerodynamic stability, a departure from the box-like form of satellites that operate in the complete vacuum of space.

It stays aloft thanks to an ion engine that began with a stock of 41 kilos (90.2 pounds) of fuel and is now down to about two kilos (4.4 pounds), Rune Floberghagen said from an ESA symposium in Edinburgh, Scotland.

Question(s):

  1. Did the spacecraft orbit so low that the engines had to run continuously to counteract drag, or were they just run intermittently? If intermittent, roughly what was the duty cycle.
  2. Roughly how much lower was GOCE's drag compared to a typical spacecraft, or to a sphere of the same mass. Did it have a drag coefficient as low as a real Ferrari?
Improve this question
$\endgroup$
1
  • 8
    $\begingroup$ FWIW, Ferraris actually don't have particularly low drag. They're designed mainly to look cool, and perhaps to produce downforce at higher speeds. The lowest drag Ferrari I could find in a quick search is the California, at 0.32: en.wikipedia.org/wiki/Ferrari_California Many ordinary sedans have much lower drag these days: carbuzz.com/features/… $\endgroup$
    – jamesqf
    Aug 24, 2020 at 3:31

2 Answers 2

Reset to default
10
$\begingroup$

  1. The ion propulsion was run continuously to compensate for drag immediately. The drag force varied strongly during each orbit (i.e. changing from night to day), typically between 4 and 12 mN on its 1 m² surface.

  2. Absolutely not. The drag coefficient was about 10 times higher. The low $c_W$ that can be reached in a dense atmosphere mostly comes from flow effects - the flow of air creates a cushion around the object. This forces the bulk of air to move gently around and not to hit the object directly.

    In space the air is so sparse, there is no such effect and each particle hits the satellite directly. Hence, what matters is not the shape of the object, but just its cross section area (to be precise, $A\cdot \cos^2(\alpha)$ for angled surfaces).

    GOCE had a cylindrical shape with a 1 m² bottom side and a length of 5 m. A cube with the same volume would have a 3 times larger cross section and therefore 3 times more drag.

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ Re A cube with the same volume would have a 3 times larger cross section and therefore 3 times more drag. That assumes the coefficient of drag of a cube is the same as was GOCE's coefficient of drag. That is not the case. GOCE had a higher than average coefficient of drag, $\endgroup$
    – David Hammen
    Aug 23, 2020 at 13:43
  • $\begingroup$ @DavidHammen - the cube has almost the same coefficient of drag, if it moves face-on into the wind. The average satellite won't do that and therefore has a lower $c_w$. $\endgroup$
    – asdfex
    Aug 23, 2020 at 14:32
  • $\begingroup$ Is a pointy cone not advantageous at all? $\endgroup$
    – ikrase
    Aug 23, 2020 at 20:43
  • $\begingroup$ @ikrase Not so much. In the case of perfect elastic collisions you could gain a bit, but as uhoh wrote in another comment: space.stackexchange.com/questions/46144/… $\endgroup$
    – asdfex
    Aug 24, 2020 at 8:54
6
$\begingroup$

Roughly how much lower was GOCE's drag compared to a typical spacecraft, or to a sphere of the same mass. Did it have a drag coefficient as low as a real Ferrari?

GOCE's drag coefficient was higher than that of a typical spacecraft. From Geul, J., E. Mooij, and R. Noomen. "GOCE statistical re-entry predictions." Proceedings of 7th European Conference on Space Debris. 2017,

The drag coefficient for α = β = 0◦ is 3.15, compared to CD = 13.24 for α = 90.

Similar, if not worse, numbers can be found elsewhere. The canonical value for a spacecraft's coefficient of drag is 2.2, regardless of shape. This canonical value dates back to the 1960s.

GOCE had a small cross section to the wind, but it was also longer than typical spacecraft. Long, skinny spacecraft suffer drag across the portion of the spacecraft that is nominally parallel to the wind vector. Ferraris and well-designed jet airplanes also suffer this sort of drag, but to a much lesser extent. An aerodynamic shape tends to push the air aside in a dense atmosphere. This does not apply in low Earth orbit.

The reason GOCE's drag coefficient is higher than average is because the definition of a spacecraft's coefficient of drag depends only on its cross section area. Any drag that results portions of the spacecraft that are nominally hidden by the spacecraft's cross section add to the drag coefficient.

That said, GOCE would have needed a good deal more fuel for orbit maintenance if it had had a cube-like or spheroidal shape with the same volume. GOCE's high drag coefficient is in part an artifact of how the drag coefficient is defined. But GOCE is not a Ferrari. No vehicle in low Earth orbit is Ferrari-like in the sense of how the shape of a Ferrari drastically reduces drag compared to a car or truck that is shaped like a toaster.

Did it have a drag coefficient as low as a real Ferrari?

A real Ferrari has a drag coefficient of about 0.33. A well designed parachute has a drag coefficient of about 1.5. GOCE's drag coefficient is over twice that of a well designed parachute. So, no, GOCE was not a Ferrari. It was however "better" than a parachute -- as a parachute. (Designed right, a parachute should have a high drag coefficient.)

Improve this answer
$\endgroup$
8
  • $\begingroup$ Okay this is a thought provoking answer! Starting from the form $F_D = 0.5 \rho v^2 C_D A$, the object is a flat 2D surface, the air molecules have zero transverse velocity and they are elastically scattered backwards, then I can see how $C_D$ can be conceivably as high as 4. Of course that's not what actually happens in collisions. But the only way I can see that the drag coefficient could depend on the length of the sides if it''s pointing straight "into the wind" is if the air has a substantial temperature and there is a lot of random transverse velocity. $\endgroup$
    – uhoh
    Aug 23, 2020 at 14:16
  • $\begingroup$ I'll go think about this a day or two and then may formulate another question. Thank you for this thorough answer! $\endgroup$
    – uhoh
    Aug 23, 2020 at 14:17
  • 3
    $\begingroup$ @david-hammen Could you provide a citation for the contribution of areas that are aligned with the wind? I don't see how these can have any significant influence in orbit with very low molecular densities. $\endgroup$
    – asdfex
    Aug 23, 2020 at 14:24
  • $\begingroup$ @uhoh: "Of course that's not what actually happens in collisions" - that's exactly what happens - apart from those molecules that stick to the satellite and are released later at lower relative velocity. But these still contribute 50% as much as those that are 'reflected'. $\endgroup$
    – asdfex
    Aug 23, 2020 at 14:30
  • 2
    $\begingroup$ @DavidHammen The "CD = 13.24" you cite seems to use a strange definition of $c_D$ that doesn't take the changed cross-section into account, so which definition of $c_D$ do they use? $\endgroup$
    – asdfex
    Aug 23, 2020 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.