SpaceX Falcon 9 v1.1 - SES-8 - DISCUSSION THREAD

[russianhalo117]

SES posted a few pictures of the satellite arriving at the Cape: https://www.facebook.com/media/set/?set=a.677725285592988.1073741840.257596920939162&type=1

Are we seeing in the photos both halves of the payload fairing, one marked with the "SES^" logo and one marked with an American flag?

yes, though there are currently two sets of payload fairings present for the next two CCAFS SLC-40 launches

[smoliarm]

How many restarts will SES-8 do?

just one

What's the smiley for?  Don't spacecraft typically perform multiple burns to get from GTO to GSO?  More so if they need to lower apogee?  It it true that SES-8 will only burn twice (one burn, one restart  ?

The question contains too bold statement, that invites a smile.
Compare to "How many restarts SES-8 flight profile requires"?, the answer is "just one", without smile.

Lots of people will feel uneasy until the engine actually restarts. The only previous attempt failed.

Edit: better English.

The answer to "how many restarts will SES-8 do" will probably be tens of thousands, over ~15 years.

Oh, this is too funny (and so typical)!  You guys are using the same name -- SES-8 -- for two different things (the satellite (correct) and the booster (Incorrect™)), and talking past each other.  Oh, well ... as Dark Helmet once said, "Keep firing!" 

Can I use one more smiley?
Thank you   

Now, there was question directed to me:

...
What's the smiley for?
...

Raphael Aloysius Lafferty, "Seven-Day Terror"
Read the last sentence of the story - this is the answer

[sdsds]

I'm not certain how to do the math [...] is the required delta-v to get to 0 degrees the orbital velocity times twice the sine of half the angle?

My prior analysis mis-guessed the perigee altitude, and thus underestimated the apogee velocity. For 295 x 80,000 km (per the SpaceX mission book), apogee velocity is 814 m/s. The mission book also provides a more exact inclination: 20.75 degrees. So if my method is correct the inclination change requires 814 * 0.3602 = 293.2 m/s.

FWIW I calculate the Hohmann transfer burns to be 926.34 m/s (prograde at apogee) followed by 489.76 m/s (retrograde at perigee) for a sub-total of 1,416.1 m/s and a grand total of 1709.3 m/s from the delivery orbit to geo-stationary orbit.

[johnmoe]

I'm not certain how to do the math [...] is the required delta-v to get to 0 degrees the orbital velocity times twice the sine of half the angle?

My prior analysis mis-guessed the perigee altitude, and thus underestimated the apogee velocity. For 295 x 80,000 km (per the SpaceX mission book), apogee velocity is 814 m/s. The mission book also provides a more exact inclination: 20.75 degrees. So if my method is correct the inclination change requires 814 * 0.3602 = 293.2 m/s.

FWIW I calculate the Hohmann transfer burns to be 926.34 m/s (prograde at apogee) followed by 489.76 m/s (retrograde at perigee) for a sub-total of 1,416.1 m/s and a grand total of 1709.3 m/s from the delivery orbit to geo-stationary orbit.

Elon suggests somewhat less:

@elonmusk Uh… why that high for a geostationary satellite launch? o_O

Elon Musk ‏@elonmusk 22 Nov
@DarkSapiens Easier to make the plane change to equatorial orbit. Done in order to reduce satellite side delta V from 1800 m/s to 1500 m/s.

[sdsds]

Elon suggests somewhat less:

@elonmusk Uh… why that high for a geostationary satellite launch? o_O

Elon Musk ‏@elonmusk 22 Nov
@DarkSapiens Easier to make the plane change to equatorial orbit. Done in order to reduce satellite side delta V from 1800 m/s to 1500 m/s.

Ah, thanks for this hint!

Only now do I see they must be combining the prograde burn at apogee with the inclination change burn. Those don't sum as absolute values, but like the sides of a triangle. I used http://www.mathwarehouse.com/triangle-calculator/online.php to generate the result: a single apogee burn of 1020.7 m/s. (See attached.) Adding that to the retrograde perigee burn: 1020.7 + 489.76 = 1510.4, so at least Elon and I are now in the same ball park!

Thanks again for your help!

[Arthree]

Elon suggests somewhat less:

@elonmusk Uh… why that high for a geostationary satellite launch? o_O

Elon Musk ‏@elonmusk 22 Nov
@DarkSapiens Easier to make the plane change to equatorial orbit. Done in order to reduce satellite side delta V from 1800 m/s to 1500 m/s.

Ah, thanks for this hint!

Only now do I see they must be combining the prograde burn at apogee with the inclination change burn. Those don't sum as absolute values, but like the sides of a triangle. I used http://www.mathwarehouse.com/triangle-calculator/online.php to generate the result: a single apogee burn of 1020.7 m/s. (See attached.) Adding that to the retrograde perigee burn: 1020.7 + 489.76 = 1510.4, so at least Elon and I are now in the same ball park!

Thanks again for your help!

Shouldn't the angle be 90-10.375 = 79.625 deg, and the result be 919.91 m/s?

[Kabloona]

And for those of us who have forgotten orbital mechanics, why isn't the angle 90 degrees? I thought the plane change delta V was always applied perpendicular to the orbital plane.

[sdsds]

And for those of us who have forgotten orbital mechanics, why isn't the angle 90 degrees? I thought the plane change delta V was always applied perpendicular to the orbital plane.

I'll try to describe how I'm imagining a pure inclination change burn. Tell me if I go amiss.

Consider the plane perpendicular to the radius vector at the apogee location. That's not the orbital plane, in fact it's at 90 degrees to the orbital plane, but like the orbital plane it too contains the velocity vector. In fact both the initial velocity vector and the final velocity vector are in that plane.

If you believe that, then consider that since the spacecraft will be delivered such that apogee occurs right above the equator, the angle between the initial velocity vector and the final (equatorial) velocity vector is the orbital inclination.

So then to see why the trigonometric approach works: consider that the two velocity vectors form the two equal-length sides of an isosceles  triangle. The engine burn is going to apply delta-v matching the third side of that triangle, so the length of that side represents the required delta-v.

Shouldn't the angle be 90-10.375 = 79.625 deg, and the result be 919.91 m/s?

My kingdom for a good drawing program!

The first attachment attempts to show the pure inclination change. In the second I've attempted to represent the combined burn. The vector lengths are approximated. The angle you're asking about is, I think, the one shown in blue. Isn't it clearly larger than 90 degrees?

[AJA]

 Even though the space-craft is making these propulsive transfers, I don't think orbital mechanics comes under the purview of 'spacecraft discussions'. So porting this over here from the Orbital SES-8 thread.

Can someone explain to me how you can raise the perigee AND lower the apogee (or vice versa) with the same burn? Or even multiple burns in the same direction (i.e. all of them being prograde or retrograde)?

Let's suppose you're in a circular orbit. Do a single burn thrusting away from the center of the Earth, as if trying to raise the orbit the intuitive but wrong naive way. Immediately after the burn the altitude will clearly increase so apogee must have increased. To first order this burn only changes the direction of velocity, not its magnitude. The vis-viva equation therefore says that the semi-major axis (i.e. average of perigee and apogee) has not changed. We deduce that perigee must have decreased.

Thanks. I figured it'd involve thrusting in an 'unconventional' direction. (Still, I gotta say that the idea of doing a burn, and spending prop - only to dump ALL the chemical energy of the prop into orbiting the exhaust - is revolting But hey -  if it works, it works)

sdsds's post immediately preceding this one, shows the isosceles triangle of the pre and post burn velocity vectors, with the third side being the delta-v. This kind of eccentricity change (or inclination change), while maintaining the same space-craft energy still requires a retrograde component of delta-v. You'll also notice, that (due to equilateral triangle geometry) as the angle change in the velocity vector approaches 60 degrees, the delta-v required approaches the orbital velocity at the point of burn - i.e. the orbital velocity where you want to change direction!

Since the new GEO orbit and the old GTO orbit intersect at this point (where you're changing only the direction of S/C velocity), we're able to calculate the angle with some simple co-ordinate geometry. I went ahead and graphed the GEO orbit, and the elliptical super-sync GTO orbit using the graphical calculator at www.desmos.com, and worked out the slopes of the tangents to the curves at the point of intersection. Turns out, that the angle of intersection of these orbits (assuming they share the same orbital plane) is, in fact, 61 degrees.

I would've assume the delta-v required would be beyond an upper-stage, let alone a satellite.


1. So what's going on here?

Here're two more questions.

2. In summary, would it be accurate to say that the super-sync GTO orbits are useful only if there's an inclination change required? (Make the burn at apogee, where the triangle of velocities is low, and hence delta-v is low)

3. What's the launch/burn sequence? Here's what I'm guessing.

LV Phase:


Launch->1st stage burnout->2nd stage burn to reach a preliminary orbit with 28 degree (launch site inclination)-> coast along until you get to a height of 295 km -> restart 2nd stage at this point to change the inclination to 20.75 degrees as well as raise the perigee of the preliminary orbit, making it 80,000 km - and thus making it the apogee of the new GTO super-sync orbit -> Release payload into a 250 km x 80,000 km x 20.75 degree orbit (Call this Orbit A)

Satellite Phase:


Satellite then does a pure inclination change burn (at the 80,000 km apogee, to knock the inclination down to 0 degrees - assuming this is the final orbital inclination) - to reach an elliptical 295 km x 80,000 km x 0 degree orbit (Orbit B) followed by a circularising burn, performed at the intersection of the final GEO orbit, and Orbit B (the two of which are co-planar).

You could potentially do it with one burn, if the final GEO orbit desired, and Orbit A intersect at two points, but since they're not co-planar, this isn't a given. [You could even do the inclination change at the point in Orbit A that's closest to the final orbit, but then you'd still have to make another burn, and changing inclination gets costlier, as you move further away from apogee.]

 
(An advance thank you to whoever reads this long post and clarifies the doubts )






EDIT: Equation 11 in the second image should actually be s₁ = -(X/Y)(b^2/a^2) = slope of the Ellipse at X,Y
However, the anglecorresponding to the slope of the tangent to the Ellipse at (X,Y) only changes by about 3 degrees, and so too, the angle between pre-and post burn velocity vectors (at the point of intersection) - which is now ~58 degrees, as opposed to 61 degrees. Still a large delta-v required.

Also, I worked the numbers out with an perigee of 250 km for the super-sync GTO, but wrote, and (inconsistently at that) 295 km in the post.

[llanitedave]

And for those of us who have forgotten orbital mechanics, why isn't the angle 90 degrees? I thought the plane change delta V was always applied perpendicular to the orbital plane.

Could it be because we're dealing with spherical trigonometry?

[mr. mark]

This is a view from 22,000 miles out. I believe that on this mission second stage should achieve 25,000 miles out. If the second stage actually goes out that far it should take a great picture. (If there is a camera on it)

[Garrett]

This is a view from 22,000 miles out. I believe that on this mission second stage should achieve 25,000 miles out. If the second stage actually goes out that far it should take a great picture. (If there is a camera on it)

Just looked at the mission patch and noticed the similarites with your Earth and Moon image. SpaceX seem to be showing how excited they are to be sending Falcon so far from Earth and a quarter of the way to the Moon! 

[MP99]

This is a view from 22,000 miles out. I believe that on this mission second stage should achieve 25,000 miles out. If the second stage actually goes out that far it should take a great picture. (If there is a camera on it)

50,000 miles:-

https://twitter.com/NASASpaceflight/status/404398072534269953

cheers, Martin

[deltaV]

Since the new GEO orbit and the old GTO orbit intersect at this point (where you're changing only the direction of S/C velocity), we're able to calculate the angle with some simple co-ordinate geometry. I went ahead and graphed the GEO orbit, and the elliptical super-sync GTO orbit using the graphical calculator at www.desmos.com, and worked out the slopes of the tangents to the curves at the point of intersection. Turns out, that the angle of intersection of these orbits (assuming they share the same orbital plane) is, in fact, 61 degrees.

I would've assume the delta-v required would be beyond an upper-stage, let alone a satellite.


1. So what's going on here?

The expensive direction changing burn you're concerned about will probably be replaced with two burns: one at apogee to raise perigee, and one at the new perigee to circularize the orbit.

Here're two more questions.

2. In summary, would it be accurate to say that the super-sync GTO orbits are useful only if there's an inclination change required? (Make the burn at apogee, where the triangle of velocities is low, and hence delta-v is low)

I'll leave that question for someone who actually knows what they're talking about to answer.

3. What's the launch/burn sequence? Here's what I'm guessing.

LV Phase:
Launch->1st stage burnout->2nd stage burn to reach a preliminary orbit with 28 degree (launch site inclination)-> coast along until you get to a height of 295 km

More importantly coast until you're near the equator.

-> restart 2nd stage at this point to change the inclination to 20.75 degrees as well as raise the perigee of the preliminary orbit, making it 80,000 km - and thus making it the apogee of the new GTO super-sync orbit -> Release payload into a 250 km x 80,000 km x 20.75 degree orbit (Call this Orbit A)

That matches my guess.

Satellite Phase:

Satellite then does a pure inclination change burn (at the 80,000 km apogee, to knock the inclination down to 0 degrees - assuming this is the final orbital inclination) - to reach an elliptical 295 km x 80,000 km x 0 degree orbit (Orbit B) followed by a circularising burn, performed at the intersection of the final GEO orbit, and Orbit B (the two of which are co-planar).

You could potentially do it with one burn, if the final GEO orbit desired, and Orbit A intersect at two points, but since they're not co-planar, this isn't a given. [You could even do the inclination change at the point in Orbit A that's closest to the final orbit, but then you'd still have to make another burn, and changing inclination gets costlier, as you move further away from apogee.]

(An advance thank you to whoever reads this long post and clarifies the doubts )

I don't believe that's right. Here's my guess for the spacecraft phase: first the spacecraft will burn at apogee, simultaneously lowering the inclination somewhat and raising perigee to GEO altitude. Then coast until perigee. At perigee burn again, simultaneously lowering the inclination to zero and lowering the apogee to GEO altitude.

[Avron]

I don't believe that's right. Here's my guess for the spacecraft phase: first the spacecraft will burn at apogee, simultaneously lowering the inclination somewhat and raising perigee to GEO altitude. Then coast until perigee. At perigee burn again, simultaneously lowering the inclination to zero and lowering the apogee to GEO altitude.

does it not cost more in delta v to do the inclination change at perigee than apogee, where the velocity is at its lowest?

[Lars_J]

I don't believe that's right. Here's my guess for the spacecraft phase: first the spacecraft will burn at apogee, simultaneously lowering the inclination somewhat and raising perigee to GEO altitude. Then coast until perigee. At perigee burn again, simultaneously lowering the inclination to zero and lowering the apogee to GEO altitude.

does it not cost more in delta v to do the inclination change at perigee than apogee, where the velocity is at its lowest?

Correct.

[cleonard]

I think that just about every post includes at least one "ideal" condition.  The reality will be a but more messy.  For example the apogee will not be exactly over the equator and the impulses are not instantaneous.  There will be many burns to get into the exact final orbit, but I agree that the first two will do the majority of the changes.  None of them will start or end exactly at apogee or perigee.  They will not even be exactly over the equator.

Assuming all goes well SpaceX will deliver the payload into an orbit.  That orbit will be characterized and the needed burns computed.  These burns will not be the textbook single dimensional examples many of us know.  They will be in a 3D vector that minimizes the total fuel consumption to get where they need to go.

[dsobin]

I don't believe that's right. Here's my guess for the spacecraft phase: first the spacecraft will burn at apogee, simultaneously lowering the inclination somewhat and raising perigee to GEO altitude. Then coast until perigee. At perigee burn again, simultaneously lowering the inclination to zero and lowering the apogee to GEO altitude.

It seems to me, based on the knowledge gained from reading many expert posts (thank you!) on this forum, that one would never want to do a substantial portion of the plane-change burn at perigee, where the cost in delta-V is  higher than at apogee. To make good use of the very high apogee, it seems that you would want to complete zeroing out the inclination at apogee. For added efficiency, you would combine that with the perigee-raising burn, completing both at apogee. Then, at next perigee, burn retrograde and lower the apogee to GEO.

I can see other options to reflect spacecraft and/or orbital constraints, but I think that the plane change burn (or burns) would only occur at apogee. To do otherwise would waste the extra delta-V cost invested in getting to the super-sync orbit.

[Joffan]

A bi-elliptic transfer can lower the delta-v required even without any inclination change.  Because the apogee of the transfer half-orbits is higher up the gravity well, inclination change can use much lower delta-v also at that point.

[Avron]

Looks llke I was posting in the wrong thread ,, maybe this is the right place..

Re: SpaceX Falcon 9 v1.1 - SES-8 - November 25 - UPDATE THREAD
« Reply #172 on: Today at 06:15 PM »

Quote from: tigerade on Today at 05:41 PM
Jonathan Amos ‏@BBCAmos

Falcon-9 telecon tonight with SES CTO Martin Halliwell: Entry of @SpaceX to commercial launch market will "shake industry to its roots".



Since my last msg was "lost", let me re-phrase my reply... if this launch and the next go well, thats the #2 sat service provide aligning with Spacex and three in a row for F9 v.1.1 means the airforce et. al can use Spacex services.. now that must make ULA/LMT worried.. we are looking for proof that history is been made.. watch LC40
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