0
$\begingroup$

The screen shot and block quote below are from Oribal ATK's Mission Update: Minotaur IV ORS-5 Launch. ORS-5 or "SensorSat" is a technology demonstration for a class of satellites that will sit in LEO and keep an eye on the GEO belt for any "debris" which I suspect means pranksats or worse. ORS-5 will sit in a 600 x 600 km equatorial LEO.

The graphic suggests that after stage 5 (Orion 38) burnout and payload separation, the velocity will be 7.05 km/s and the altitude 600 km and that checks out; the sidereal rotation speed of the Earth-Fixed frame at this altitude is another 0.51 km/s. That means the speed of the orbit in inertial coordinates is 7.05 + 0.51 = 7.56 km/s.

Orbital ATK's Minotaur IV space launch vehicle will launch the ORS-5 mission for the U.S. Air Force as a part of the Orbital/Suborbital Program-3 (OSP-3) contract. ORS-5, also known as SensorSat, is designed to scan for other satellites and debris to aid the U.S. military’s tracking of objects in geosynchronous orbit. For the ORS-5 launch, the Minotaur IV will use two Orion 38 upper stage motors. The final Orion 38 motor burn will reduce the angle of the ORS-5 satellite’s orbit, redirecting the spacecraft to equatorial orbit.

If I understand correctly, "The final Orion 38 motor burn" does not suggest that the solid fuel motor can burn twice, but that both the 4th and 5th stages are Orion 38 motors, and "the final burn" really means the 5th stage, or the burn of "the final Orion 38 motor".

I'm curious what "angle" means in last sentence, and guessing that it means inclination. For such a light satellite (something like 110 - 140 kg) would the standard thrust of the Orion 38 happen to provide just the right amount of delta-v to change the inclination to equatorial? Since the table shows the velocity unchanged during the 5th stage burn, it seems all of the thrust should be perpendicular to the direction of motion and purely for inclination change.

Question: Could someone help to estimate the change in inclination angle that an Orion 38 can provide this satellite, starting with the specs in English units :-( and see if it's even in the right ballpark for this launch?

Approximate mass of the payload:

Approximate mass of the Orion 38:

enter image description here

below: Image of Orion 38 motor from Orbital ATK's Propulsion Products Catalog.

enter image description here

below: Orion 38 motor specifications from Orbital ATK's Propulsion Products Catalog.

enter image description here

below: ORS-5 during assembly from Orbital ATK tweet.

enter image description here

$\endgroup$
4
$\begingroup$

For inclination change in circular orbit, $$\Delta v = {2v\, \sin \left(\frac{\Delta{i}}{2} \right)}$$

Delta v from a burn done all in one direction is given by our old friend the Tsiolkovsky equation: $$\Delta v = v_\text{e} \ln \frac {m_0} {m_f}$$

Here, $m_0$ is the mass at the start of the burn, combining the total motor mass of the Orion 38 (873 kg) and the mass of the payload (I took the 113 kg figure as the most accurate) for 986 kg. $m_1$ is the mass at the end, 94 kg from the motor + 113 kg payload for 207 kg.

$v_e$ is the effective exhaust velocity, which is 9.81 m/s2 times the effective $I_{sp}$ given in seconds: 2815 m/s.

So the motor can deliver to this payload $$2815 \ln \frac {986} {207}$$ or a very respectable 4391 m/s.

If all that ∆v is applied perpendicular to the circular orbit, we get a pure inclination change maneuver; our circular orbit velocity at 600km is 7558 m/s, so:

$$ 4391 = {2} \times {7558} \sin \left(\frac{\Delta{i}}{2} \right)$$

Rearranging:

$$ 0.290 = \sin \left(\frac{\Delta{i}}{2} \right)$$ $$\Delta{i} = 2 \arcsin 0.290$$

or 33.77º.

If you throw in a little extra unaccounted mass, e.g. a payload adapter between the Orion 38 and the satellite, the rocket equation figure might come out closer to 4000 m/s; this still gets you about a 30º inclination change.

As either of these figures is greater than Cape Canaveral's 28º latitude, it's sufficient to push the satellite into equatorial inclination. I assume the rocket was initially launched to just the right inclination for the Orion stage to cancel, and did the correction burn while crossing the equator early in its first orbit. Launching to an initially higher inclination effectively sacrifices a tiny bit of the "free" velocity you get from Earth's rotation in exchange for not having to do anything excessively clever with the Orion 38.

I note also that while the velocity figure is unchanged across the 5th stage burn, the altitude goes from 578 km at 4th stage burnout to 600 km at 5th stage burnout; if the final orbit of the ORS-5 is in fact 600km circular, then the Orion 38 stage must be doing some circularization as well as plane change.

$\endgroup$
  • $\begingroup$ Great! It seems that the Orion 38 and ORS-5 were made for each other, they seem to match very nicely. I'm guessing the orbit is not so critical, so if the final $\Delta i$ were of by a fraction of a degree it might be OK. Any idea if the ORS-5 would need a small thruster to de-orbit itself at end-of-life (thus more mass), or at 600 km is eventual decay pretty much a given? $\endgroup$ – uhoh Aug 26 '17 at 13:34
  • 1
    $\begingroup$ The EOPortal article implies it has no propulsion, but the calculator at lizard-tail.com/isana/lab/orbital_decay suggests that orbital decay time for a sat that size at 600km is on the order of 200 years (though this could vary drastically with solar activity and other variables). $\endgroup$ – Russell Borogove Aug 26 '17 at 14:54
  • $\begingroup$ @RussellBorogove isn't there some kind of established guideline about providing for decay this these days? Presumably it's going to need some gyros and computer to do high quality spying on small objects in GEO and send them all the pics to Earth, so it may have some nice solar panels which "improve" drag. Or perhaps something else pops out when it's EOL. Anyway, thanks for the great answer! $\endgroup$ – uhoh Aug 26 '17 at 17:39
  • 1
    $\begingroup$ @uhoh - I didn't catch the inertial velocity correction on the first pass, and have now updated my figures. The plane change is a touch more expensive, but still very much achievable. $\endgroup$ – Russell Borogove Aug 27 '17 at 14:24
  • 1
    $\begingroup$ @uhoh - Those are the secondary payloads? That implies a dog-leg ascent, and something else missing from my analysis -- either the circularization costs more than I think, or the payload adapter is heavier, or ballasted to get exactly the right ∆v from the stage, or something. I don't work in more than two dimensions at a time, so I won't try and refine further. $\endgroup$ – Russell Borogove Aug 27 '17 at 15:44

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.