7
$\begingroup$

I would like to know the performance for escape mission of the Ariane 5 launcher. More precisely, I would like "LV Performance vs. C 3" curve. My problem is, I can find this info for a lot of launchers but not Ariane. Does anybody have such plot or, at least the value for low C3 (around 0-4)?

Soyuz curve Here the kind of plot I would like. This one is for Soyuz and I found it in the Soyuz user's manual but this is not in the Ariane one's.

$\endgroup$
5
$\begingroup$

The Ariane 5 user manual has the following data:

Using a storable propellant upper stage, through a delayed ignition of this upper stage, Ariane 5, in the A5G version, has demonstrated its ability to carry a satellite weighing 3065 kg, leading to a total required performance of 3190 kg, towards the following earth escape orbit:
- infinite velocity V∞ = 3545 m/s
- declination δ = - 2°

The typical Ariane 5ECA performance on a similar orbit is 4.3 t.

No graph though.

C$_3$ = (V∞)$^2$. And V∞ is related to payload mass via the rocket equation, so you should be able to draw a plot using this one data point. I've no time to do that today though.

$\endgroup$
  • $\begingroup$ Yes, I found only information like that. Is it possibilities to extrapolate these data for interplanetary mission? $\endgroup$ – Tonio Oct 30 '15 at 14:14
  • $\begingroup$ I think so, yes. See the last paragraph of my answer. $\endgroup$ – Hobbes Oct 30 '15 at 20:09
  • $\begingroup$ The rocket equation assume there is no extern forces. I do you take account of drag, gravity and others forces due to the Earth and atmosphere? Do you know a good model for this? $\endgroup$ – Tonio Oct 31 '15 at 1:39
  • 1
    $\begingroup$ Well, you have one data point that takes those forces into account. I expect you can plug that into the rocket equation and get reasonably close. The rocket equation will set the shape of the V∞/m curve, your data point will set the starting point. But that's just an assumption. I recommend you ask a new question about how to calculate these curves, that might trigger an answer from someone who knows more than me. $\endgroup$ – Hobbes Oct 31 '15 at 7:03
  • $\begingroup$ I didn't think about that! I'll try it and eventually ask another question. Thanks you very much! $\endgroup$ – Tonio Oct 31 '15 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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