Question Launch trajectory, Tianwen-1 Chinese Mars Mission

[rodion_herrera]

Hello, anyone has any information on launch trajectory of the Tianwen-1 Mars Mission? Just initial ascent path to orbit is fine, thank you.


-RODION

 

[BrianJ]

Hi Rodion
JPL Horizons has a "provisional" trajectory (although with disclaimer regarding accuracy). It might be some help.
Cheers,
Brian

 

[N_Molson]

I just found this, data from NORAD :

Orbital elements: 2020-049A = NORAD 45935 = Tianwen-1 spacecraft (geocentric escape)
Perigee 2020 Jul 23.21849682 +/- 1.48e-5 TT = 5:14:38.125 (JD 2459053.71849682)
Epoch 2020 Jul 24.0 TT = JDT 2459054.5 Find_Orb
q 6571.98920246 +/- 0.827 (J2000 equator)
H 26.4 G 0.15 Peri. 249.91904731 +/- 0.0031
Node 341.68008391 +/- 0.0045
e 1.1972718520 +/- 4.52e-5 Incl. 29.42030776 +/- 0.00012
110 of 113 observations 2020 July 23-30; mean residual 0".472

Could someone help me to translate this into Orbiter ? It seems to be an escape trajectory (not a parking orbit), with 249.91 km as Pe and and Inclination (Equatorial, Ecliptic ?) of 29.42°. Could someone make a scenario file from this data ? Many thanks, that would help me do determine a plausible ascent profile.

 

[BrianJ]

Hi,
I can't give you a direct conversion of those elements, but they appear to be from here:

https://www.projectpluto.com/pluto/mpecs/tianwea.htm#r008

and JPL Horizons says:

References for trajectory segment data:
Jul23-Aug01: https://www.projectpluto.com/pluto/mpecs/tianwea.htm#o001

I can give you the first available data point in the JPL Horizons data.
Orbiter scenario date and state vectors:

Date MJD  59053.25
STATUS Orbiting Earth
RPOS  19481445.9302127  2405635.38145612 -7767511.22959222
RVEL  6864.87454252779  1167.29399600163  1102.61737023795

or if you prefer to use elements in your scenario:

Date MJD  59053.25
STATUS Orbiting Earth
ELEMENTS -33379750.69659633 1.197410648548793 10.06001523952199 297.9803293025521 233.4936814246468 249.59452282854065 59053.25

(but check the elements look right - I haven't done that conversion for a long time!)

P.S. The given inclination of 29.42 is equatorial - you want ecliptic for Orbiter.

Cheers,
Brian

EDIT: I forgot to put in the "STATUS Orbiting Earth" line ;-)

 

[N_Molson]

Many thanks !

Given the launch date (MJD 59053.195139) it seems awfully like they had no time for a LEO Parking Orbit. Which also explains why there are no elements for that on JPL or elsewhere. Seems very much like we have a direct TMI ! Won't be easy to simulate.

Edit : What seems a little weird is that given those elements, the probe is approaching Earth, not ejecting...

Edit2 : here's a file with Orbiter's stock Carina in place of Tianwen (same mass, 5000 kg)

Edit3 : from Wikipedia :

Launch date : 23 July 2020, 04:41:15 UTC

Which gives, using the "Date" tool in Orbiter/utils : MJD 59053.195313

Edit 4 :

Using pen-and-paper, or, maybe more conveniently, an online calculator (https://instacalc.com/42678), for the given inclination we get :

North : 67.62°
South : 112.38°

From Wenchang, the only viable option is South, because North flies directly over China's very densely populated coast. Also I know they use atolls and boats in China's Sea, right South/Southwest form Wenchang, as tracking stations.

 

[BrianJ]

Many thanks !

Given the launch date (MJD 59053.195139) it seems awfully like they had no time for a LEO Parking Orbit. Which also explains why there are no elements for that on JPL or elsewhere. Seems very much like we have a direct TMI ! Won't be easy to simulate.

Edit : What seems a little weird is that given those elements, the probe is approaching Earth, not ejecting...

Edit2 : here's a file with Orbiter's stock Carina in place of Tianwen (same mass, 5000 kg)

Edit3 : from Wikipedia :



Which gives, using the "Date" tool in Orbiter/utils : MJD 59053.195313

Edit 4 :

Using pen-and-paper, or, maybe more conveniently, an online calculator (https://instacalc.com/42678), for the given inclination we get :

North : 67.62°
South : 112.38°

From Wenchang, the only viable option is South, because North flies directly over China's very densely populated coast. Also I know they use atolls and boats in China's Sea, right South/Southwest form Wenchang, as tracking stations.

It's not valid to propagate the elements back in time further than the end of the final burn of the launcher, which would be close to (or soon after) perigee at ~ 05:16 UTC. (assuming the final burn is relatively short).

Looking at the map, perigee occurs about 1/4 orbit from the launch site, so that's about a 20min parking orbit at ~218km.
Which leaves ~15mins for the launcher to get into parking orbit (from launch at 04:41 UTC)
Enjo's "Azimuth Calculator" tells me launch azimuth should be about 113.6deg from latitude 19.6N (Wenchang aprox.) for equatorial inc. 29.42.

So very roughly I'd say launch to 113.6deg azimuth at 04:41 UTC to 218km parking orbit and aim to make the final burn around 05:15 UTC over the south pacific. That's my initial guess (without knowing much about the launcher).
(Of course, I'd use TransX or IMFD once in parking orbit to get a more accurate solution, provided the parking orbit plane is not too way off).

Cheers,
Brian

 

[N_Molson]

Thank you, I must say a Parking Orbit is the most convenient solution for me, as it leaves the player the responsability to use IFMD/TransX to plan the TMI burn ? I was a bit scared to have to compute a clean interplanetary transfer by myself ?

I entered the 113° value in my first stage ascent program and we get real close from the 29.42 value, and the second stage will easily correct the remaining error. I'm already very consistent with the 218km. So far so good ! ?