SpaceX Falcon 9 : GPS III SV01 : SLC-40 : Dec. 23, 2018 - DISCUSSION

[envy887]

I don't know how else the team would be able to "get performance back".  Get it back from where?

Get it back from margin reserves.

That's what I'm saying, SpaceX already knows their margin reserves.  USAF and LM are saying they need this flight to determine how much lift they actually need.  They're lucky the F9 has margin to spare.  Since it seems that they don't have hard numbers yet.
Or to Gongora's point, they're not willing to take any chances with this bird, but will take more risks with later flights.

Do they, though? They don't seem to have pushed a Block 5 to its limits yet.

[OccasionalTraveller]

Secondly, SpaceX's goal is to reduce the price of launches through reuse.  Assuming they do reduce the price of recoverable launches eventually, there is a fleet of identical GPS satellites.  Wouldn't it be cheaper for the Air Force to accept the original performance (without margins) and, if one of them fails to reach the original contract orbit, build an extra satellite and launch it on SpaceX's dime?  Even though it could have done so if SpaceX hadn't reserved fuel for recovery?  The Air Force would get a whole bunch of cheaper launches in return, saving millions of dollars in launch costs.

Back in September, Lockheed Martin was awarded a contract to build two more GPS III satellites after the initial run of 10 satellites, with options for a further 20. Bloomberg reported a possible maximum price of $7.2bn for 22 satellites. I make that $327m per satellite. Clearly that is much more than any possible saving from landing boosters - it's more than 3 times the total price SpaceX are being paid for the second launch ($96.5m).

The lead time for new satellites is also likely to be longer than the lead time for SpaceX to manufacture a new booster.

Failures in the launch industry are generally insured against, rather than the launch vehicle operator eating the cost of the launch. (SpaceX have offered a free launch to Spacecom, the owners of AMOS-6, but this is a goodwill gesture rather than a contractual requirement.) However, the US Government does not buy launch insurance; rather, as the issuer of the currency, they simply borrow/tax/print the money to make it happen (generally much lower cost than the premiums for insurance, which run at around 4% for Ariane 5 and slightly higher for SpaceX, at least in 2016). Essentially the government budget is so big that an additional launch, even an additional satellite, is a rounding error.

[gongora]

Back in September, Lockheed Martin was awarded a contract to build two more GPS III satellites after the initial run of 10 satellites, with options for a further 20. Bloomberg reported a possible maximum price of $7.2bn for 22 satellites. I make that $327m per satellite. Clearly that is much more than any possible saving from landing boosters - it's more than 3 times the total price SpaceX are being paid for the second launch ($96.5m).

That contract is for GPS IIIF, which will be an updated design.  It's $1.3B for the first two satellites with an option to bring it up to a total of $7.2B for 20 more satellites.  That's cheaper per unit than the GPS III satellites.

[abaddon]

The US Air Force says this is the first national security launch for SpaceX.  What about NROL-71 and OTV-5?

IIRC those flights were not directly purchased by the US Govt.

[Brovane]

At least the question if this payload would be horizontally or vertically integrated has been answered that it was horizontally integrated. 

[Jim]

At least the question if this payload would be horizontally or vertically integrated has been answered that it was horizontally integrated. 

never was in question.  Was known from day 1

[LouScheffer]

With the press kit, we can now see the sequence of burns:

Launch into an LEO x 4000 km orbit (about 2 hour period).  Takes LEO + 830 m/s.

Wait one hour until apogee, then go 4000 x 20000 km (6.9 hour orbit) .  Takes 1960 m/s.  Release satellite.

Second stage coasts to apogee (3.45 more hours).   Then a retrograde burn at -480 m/s to a 100 x 20000 orbit (5.8 hour period).   Wait 2.9 hours and re-enter.

Total delta-V is about LEO+3270 m/s, as predicted.  Re-entry at launch + 6.5 hours, as stated.  Satellite needs about 970 m/s to circularize.

[pb2000]

With the press kit, we can now see the sequence of burns:

Launch into an LEO x 4000 km orbit (about 2 hour period).  Takes LEO + 830 m/s.

Wait one hour until apogee, then go 4000 x 20000 km (6.9 hour orbit) .  Takes 1960 m/s.  Release satellite.

Second stage coasts to apogee (3.45 more hours).   Then a retrograde burn at -480 m/s to a 100 x 20000 orbit (5.8 hour period).   Wait 2.9 hours and re-enter.

Total delta-V is about LEO+3270 m/s, as predicted.  Re-entry at launch + 6.5 hours, as stated.  Satellite needs about 970 m/s to circularize.

4000x35000 is a pretty standard second stage graveyard orbit for centaur, so if the deorbit burn is what pushed this expendable, it seems a bit wasteful...

[Alexphysics]

With the press kit, we can now see the sequence of burns:

Launch into an LEO x 4000 km orbit (about 2 hour period).  Takes LEO + 830 m/s.

Wait one hour until apogee, then go 4000 x 20000 km (6.9 hour orbit) .  Takes 1960 m/s.  Release satellite.

Second stage coasts to apogee (3.45 more hours).   Then a retrograde burn at -480 m/s to a 100 x 20000 orbit (5.8 hour period).   Wait 2.9 hours and re-enter.

Total delta-V is about LEO+3270 m/s, as predicted.  Re-entry at launch + 6.5 hours, as stated.  Satellite needs about 970 m/s to circularize.

One thing is the timings are a little bit off with that first orbit you said. I started my calculations at a 200x3300km orbit and then the second stage gets to apogee just an hour after shutdown (which I assumed it was at the perigee altitude). Then at apogee a burn to a 3300x20100km which would be at about 18:30 UTC (T+4h19min). Then another burn by the second stage at the new 20100km apogee to deorbit itself after sat deployment. This all gives a reentry between 21:20 UTC (T+7h9min) and 21:30 UTC (T+7h19min) which is on the higher end of the reentry window for the second stage.

[zubenelgenubi]

Is the 2nd stage de-orbit a requirement from the USAF, or is it a choice by SpaceX?  (Or something else?)

[RDMM2081]

If excess performance of the first stage is only reserved for margin, could SpaceX still execute entry and landing burns, sans ASDS, as an “I told you so”?

No. That's not how it works. You always transfer your margins, aka you impart as much velocity to the second stage as possible to cover for an eventual 2nd stage or satellite performance shortfall.
<more>

Of course that all makes perfect sense.  Thank you for taking the time to respond to my silly question with such a well reasoned and informed refutation.

[edkyle99]

With the press kit, we can now see the sequence of burns:

Launch into an LEO x 4000 km orbit (about 2 hour period).  Takes LEO + 830 m/s.

Wait one hour until apogee, then go 4000 x 20000 km (6.9 hour orbit) .  Takes 1960 m/s.  Release satellite.

Second stage coasts to apogee (3.45 more hours).   Then a retrograde burn at -480 m/s to a 100 x 20000 orbit (5.8 hour period).   Wait 2.9 hours and re-enter.

Total delta-V is about LEO+3270 m/s, as predicted.  Re-entry at launch + 6.5 hours, as stated.  Satellite needs about 970 m/s to circularize.

Also note that the 55 deg orbit inclination requires more delta-v just to get to LEO than a 28.5 deg GTO type inclination.  Something like 240 m/s extra, maybe?

 - Ed Kyle

[LouScheffer]

With the press kit, we can now see the sequence of burns:

Launch into an LEO x 4000 km orbit (about 2 hour period).  Takes LEO + 830 m/s.

Wait one hour until apogee, then go 4000 x 20000 km (6.9 hour orbit) .  Takes 1960 m/s.  Release satellite.

Second stage coasts to apogee (3.45 more hours).   Then a retrograde burn at -480 m/s to a 100 x 20000 orbit (5.8 hour period).   Wait 2.9 hours and re-enter.

Total delta-V is about LEO+3270 m/s, as predicted.  Re-entry at launch + 6.5 hours, as stated.  Satellite needs about 970 m/s to circularize.

Also note that the 55 deg orbit inclination requires more delta-v just to get to LEO than a 28.5 deg GTO type inclination.  Something like 240 m/s extra, maybe?

I get 148 m/s more is needed compared to a GTO launch.  Here is my thinking (this is the first time I've tried this particular calculation, so mistakes are possible):

A 55 degree orbit passes over the cape (lat 28.5^o) at an azimuth of 40.74^o, from azimuth = sin^-1(cos(55^o)/cos(28.5^o)).  We need about 7800 m/s for an LEO, at this angle (in an inertial frame).

Now the North-South component of this is 5910 m/s.  The East-West component is 5090 m/s, but the Earth rotation provides 409 m/s of this at latitude 28.5^o.  That leaves 4681 m/s to be provided by the rocket.  So the rocket needs to provide sqrt(5910^2+4681^2) = 7539 m/s.  So the Earth's rotation here saves 7800-7539 = 261 m/s.   Compared to the 409 m/s savings of a due-East GTO launch, that's 148 m/s more that the rocket needs to provide.

As a side effect, we can find the launch azimuth in the Earth-rotating frame as 90^o-tan^-1(5910/4681) = 38.38^o.

Also, the last (de-orbit) burn has no payload mass.  So if the final mass of stage-2 plus residuals is 5000 kg,and the burn is 480 m/s at an ISP of 348, then the stage starts with 5000*exp(480/348/9.8) = 5750 kg.  If the 3900 kg payload was still attached, they would get only  348*9.8*ln((5750+3900)/(5000+3900)) = 276 m/s.  So they are getting about 204 m/s that they can use for de-orbit, but could not use for the payload.

[deruch]

Just a question on classification of this mission:

The US Air Force says this is the first national security launch for SpaceX.  What about NROL-71 and OTV-5?  Is this an admission that, while classified, they weren't national security?  Or is it more proper to say that GPS III SV01 was SpaceX's first competitively won national security mission?

Just trying to keep this straight with what the company's launched since this contract was bid on (2015) and awarded (April 2016).

NROL-76 (not L-71 which is launching on D4H) was technically a commercial launch procured by Ball Aerospace, the sat builder.  Same as the Zuma launch I believe, only that was via Northrop Grumman.  I'm not sure about the OTV-5 launch.  I believe that was actually a direct NSS launch, only it wasn't put out for bid.  IIRC, the AF sole sourced it to SpaceX.  So, this launch will be the first that SpaceX was awarded through the AF's EELV procurement process. 

[Semmel]

Also note that the 55 deg orbit inclination requires more delta-v just to get to LEO than a 28.5 deg GTO type inclination.  Something like 240 m/s extra, maybe?

[...] A 55 degree orbit [...]

Careful, Spaceflight Now reports a 55 deg ground track to the equator:

The satellite, valued at more than a half-billion dollars, is heading to an orbit around 12,550 miles (20,200 kilometers) above Earth, with a ground track angled 55 degrees to the equator.

Is this the same as a 55 deg orbit? I might have my mental gymnastics confused, but this sounds to me that the ground track and not the orbital plane is angled at 55 degrees. When imagining the ground track of a ~12h orbit, it should make a figure - 8 type track that ranges from far north to far south. Since the Earth is rotating, an orbital plane at 55 degrees should not produce a ground track that crosses the equator at 55 degrees.

Or maybe Spaceflight Now just used unlucky wording and they meant a 55deg orbital plane.

[Ben the Space Brit]

From Ben Coopers Facebook page:

A new era in GPS is finally set to begin: The first block III Global Positioning System satellite is poised for liftoff for the US Air Force Tuesday morning at 9:11 a.m. EST aboard this Falcon 9 rocket. This will mark the 20th and final space launch from Cape Canaveral this year, the most since 1998.

https://www.facebook.com/photo.php?fbid=10101611102824218&set=a.912747205518&type=3&theater

It gives me good Titan-III flashbacks to see the USAF symbol on the Falcon 9's PLF.

[LouScheffer]

[...] A 55 degree orbit [...]

Careful, Spaceflight Now reports a 55 deg ground track to the equator:

The satellite, valued at more than a half-billion dollars, is heading to an orbit around 12,550 miles (20,200 kilometers) above Earth, with a ground track angled 55 degrees to the equator.

Is this the same as a 55 deg orbit? I might have my mental gymnastics confused, but this sounds to me that the ground track and not the orbital plane is angled at 55 degrees. When imagining the ground track of a ~12h orbit, it should make a figure - 8 type track that ranges from far north to far south. Since the Earth is rotating, an orbital plane at 55 degrees should not produce a ground track that crosses the equator at 55 degrees.

Or maybe Spaceflight Now just used unlucky wording and they meant a 55deg orbital plane.

Well, the government definition of the GPS orbits says 55 degree inclination.  Inclination is measured from the plane of the equator.

Ground track is not the same thing as inclination.  Imagine, for example, that the earth was rotating once per second.   Then the ground track, even for a polar satellite, would be nearly a direct East-West line.  In real life, for low earth orbits the satellite goes around many times per earth revolution, and the ground track at the equator is pretty close to the inclination (a few degrees).  But for higher orbits it gets worse, as the Earth's rotation is a larger portion of the orbital time, and above synchronous orbits it exceeds it.  The moon, for example, orbits west to east, but its ground track goes the other way (east to west).

In this case, the orbit at 20000 km, 55 degrees has an orbital speed of 3888 m/s.  Of this, 3185 m/s is north-south and 2230 m/s is east-west.  But if you are standing on the equator, you are moving with the satellite at 485 m/s, so you only see 1745 m/s of this movement.  So the ground track is moving at angle atan(3185/1745) or 61.28 degrees.

So yes, SFN was just sloppy and equated inclination with ground track.

[OneSpeed]

A 55 degree orbit passes over the cape (lat 28.5^o) at an azimuth of 40.74^o, from azimuth = sin^-1(cos(55^o)/cos(28.5^o)).  We need about 7800 m/s for an LEO, at this angle (in an inertial frame).

Now the North-South component of this is 5910 m/s.  The East-West component is 5090 m/s, but the Earth rotation provides 409 m/s of this at latitude 28.5^o.  That leaves 4681 m/s to be provided by the rocket.  So the rocket needs to provide sqrt(5910^2+4681^2) = 7539 m/s.  So the Earth's rotation here saves 7800-7539 = 261 m/s.   Compared to the 409 m/s savings of a due-East GTO launch, that's 148 m/s more that the rocket needs to provide.

As a side effect, we can find the launch azimuth in the Earth-rotating frame as 90^o-tan^-1(5910/4681) = 38.38^o.

An azimuth of 40.74^o looks right, but have you swapped your sin and cos of that angle?

I get:
vertical component = sin(40.74) * 7800 = 5090 m/s
horizontal component = cos(40.74) * 7800 = 5910 m/s
Subtract rotation: 5910 - 409 = 5501 m/s

From this, sqrt(5090^2 + 5501^2) = 7494 m/s.
7800 - 7494 = 306 m/s.
409 - 306 = 103 m/s, so only 103 m/s more required than zero azimuth at 28.5^oN.

[CorvusCorax]

For the record, I'm predicting the countdown will be scrubbed somewhere around the T-1:00 mark due to upper level winds being out of bounds.
Question is if they let it go down all the way to -30s or even further or scrub before engine chill and auto-sequence before 1:00.

[LouScheffer]

A 55 degree orbit passes over the cape (lat 28.5^o) at an azimuth of 40.74^o, from azimuth = sin^-1(cos(55^o)/cos(28.5^o)).  We need about 7800 m/s for an LEO, at this angle (in an inertial frame).

Now the North-South component of this is 5910 m/s.  The East-West component is 5090 m/s, but the Earth rotation provides 409 m/s of this at latitude 28.5^o.  That leaves 4681 m/s to be provided by the rocket.  So the rocket needs to provide sqrt(5910^2+4681^2) = 7539 m/s.  So the Earth's rotation here saves 7800-7539 = 261 m/s.   Compared to the 409 m/s savings of a due-East GTO launch, that's 148 m/s more that the rocket needs to provide.

As a side effect, we can find the launch azimuth in the Earth-rotating frame as 90^o-tan^-1(5910/4681) = 38.38^o.

An azimuth of 40.74^o looks right, but have you swapped your sin and cos of that angle?

I get:
vertical component = sin(40.74) * 7800 = 5090 m/s
horizontal component = cos(40.74) * 7800 = 5910 m/s
Subtract rotation: 5910 - 409 = 5501 m/s

From this, sqrt(5090^2 + 5501^2) = 7494 m/s.
7800 - 7494 = 306 m/s.
409 - 306 = 103 m/s, so only 103 m/s more required than zero azimuth at 28.5^oN.

No, since azimuth is measured from north, not the equator, the N-S component is sin(90-azimuth), and the horizontal cos(90-azimuth).  This is equivalent to swapping sin() and cos().

A quick check is that since the azimuth is less than 45 degrees, the N-S component should be bigger than the E-W component.  This is not so in the solution above, but is in my calculations.