SpaceX And/Or Rocketry In General

robl
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Another Launch scheduled for

Another Launch scheduled for today March 1 at 10:50 PM EST. No discussion about 1st stage recovery that I could find.

[edit] Here is a link to a live feed of the launch.

Mike Hewson
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Nice one. Fairly long

Nice one. Fairly long secondary burn still going at end of webcast ~ 12 minutes ie. probably lighter payload going to higher orbit. Have to find out what 'super-synchronous transfer orbit' is ....

Cheers, Mike.

( edit ) Period longer than synchronous ie. 'above' geostationary.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

archae86
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A pretty good article on

A pretty good article on today's mission can be found at:

nasaspaceflight.com article on March 1 falcon 9 flight

The article says that the mission requirements (mostly the payload mass and required transfer orbit) were near enough to limits of the launcher to require launching without the hardware needed for the recovery we are all so interested in.

Mike Hewson
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Yes. I should have noticed

Yes. I should have noticed that. No legs, no fins ..... :-)

Interestingly, even with the highest punt available from the Falcon :

Quote:
Following separation, the spacecraft will take around eight months to manoeuvre into their final geostationary orbits – this long orbit-raising phase a consequence of the less powerful but more efficient electric propulsion systems they are using.


So it seems 'super-synchronous transfer orbit' is an orbit which enables one to later go to super-synchronous. This implies high ellipticity : 408 km by 63,928 km is quoted. A geostat is low ellipticity @ ~ 42,000 km. A long chug up hill .....

Cheers, Mike.

( edit ) 'Eutelsat 115 West B' is the clue ie. that is the final geostat spot/slot. NB That geostationary is circular and above equator ( more or less ) but that is a special case of 'geosynchronous' which includes other orbits of period one day.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

Mike Hewson
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Here's some data on Eutelsat

Here's some data on Eutelsat 115 West B and ABS-3A.

Tricky questions of the day :

#1 If the Earth did not rotate around it's own axis, could you have any geostationary orbits ? :-)

#2 How do you turn a highly elliptical orbit into a circular one ?

So it looks like the ~$60 M USD launch cost has been split two ways. Of course the craft are doing some of the lifting, but a fair slab of the job has been done.

Cheers, Mike.

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

Gary Roberts
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RE: Tricky questions of the

Quote:
Tricky questions of the day :


Calling them 'tricky' makes me believe that you'll scare off most people for fear of providing the obvious dumb answers :-). However, methinks it's my day to look dumb ....

Quote:
#1 If the Earth did not rotate around it's own axis, could you have any geostationary orbits ? :-)


Yes. I imagine a non-rotating earth still travelling around the sun in one year. The earth would still have 'days' - they would be exactly one year long. Neglecting for the moment how people would cope with this, to get a geostationary satellite, you would just have to put it in the same orbit around the sun as the earth but leading or trailing the earth by whatever distance you decided was appropriate.

Quote:
#2 How do you turn a highly elliptical orbit into a circular one ?


In general terms you would need to decelerate the craft (perhaps over a series of progressively less elliptical orbits) by providing reverse thrust at just the right time to cancel a fraction of the 'slingshot' effect that is making the orbit elliptical in the first place :-). Please don't ask when the exact 'right time' is. I'm sure there are plenty of computer programs that can work this out quite precisely :-).

Well, Master - did I get close??? :-).

Cheers,
Gary.

archae86
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Pending the Master's grading,

Pending the Master's grading, I'll offer some comments:

Quote:
to get a geostationary satellite, you would just have to put it in the same orbit around the sun as the earth but leading or trailing the earth by whatever distance you decided was appropriate.


But those would not be satellites, useful though they might be.

Quote:
would need to decelerate the craft (perhaps over a series of progressively less elliptical orbits) by providing reverse thrust at just the right time


Backwards--the circular orbit has more energy than does the highly elliptical one of same apogee. Timed burns is right, but need to get the sign right.

archae86
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RE: I'm sure there are

Quote:
I'm sure there are plenty of computer programs that can work this out quite precisel


Apparently one option is the Astrogator module of STK, which apparently can be downloaded for free at some level of functionality. I confess I've not tried it, nor downloaded it.

Gary Roberts
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RE: Pending the Master's

Quote:
Pending the Master's grading, I'll offer some comments:


The Master seems unresponsive - he's probably sitting back and waiting to see how many contradictory answers will be forthcoming :-).

Quote:
Quote:
to get a geostationary satellite, you would just have to put it in the same orbit around the sun as the earth but leading or trailing the earth by whatever distance you decided was appropriate.

But those would not be satellites, useful though they might be.


Yes they would :-). They'd be satellites of the sun which are geostationary wrt earth ;-). Come to think of it, the tricky question asked about 'geostationary orbits' without specifically mentioning orbits of what?? If we were meant to assume orbits of a non-rotating earth, I can't see that something actually orbiting the earth could be geostationary as well.

Quote:
Quote:
would need to decelerate the craft (perhaps over a series of progressively less elliptical orbits) by providing reverse thrust at just the right time

Backwards--the circular orbit has more energy than does the highly elliptical one of same apogee. Timed burns is right, but need to get the sign right.


Ahhh, there's the rub. The question didn't actually specify whether the radius of the circular orbit would be based on apogee or perigee (or something in between). I assumed perigee because I was thinking of a craft on a journey to a distant object that needed to be 'captured' by that object, giving a highly elliptical orbit and then end up in a close circular orbit around that object in order to perform the useful part of its mission.

Cheers,
Gary.

Mike Hewson
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Ummm ... errr .. Master (

Ummm ... errr .. Master ( ???? ) sezz :

#1 Short answer : no.

Long answer : think for a moment of just the Earth and our candidate satellite. No Sun, no rest of universe, nuthin. If the Earth doesn't rotate but we require the satellite to remain fixed overhead wrt some Earth location ( definition of 'geostationary' ) then it can't have any angular motion. Ergo no orbit in the usually understood sense. Can you hold a satellite in 'hover mode' over a specific Earth surface point ? Yes of course, but you'd have to spend energy all day, every day and one day you would run out of fuel and come straight down on a radial line, ending in a full RUD event upon whomsoever was underneath looking up at it. As orbiting is a type of falling where you always miss the planet, call this mode 'temporary geostationary' ... :-)

What I think Gary was alluding to was fixed positions with respect to the Earth's centre when circling the Sun aka The Lagrange Points :

.... these exist as stable relationships in the Earth-Sun system, whether the Earth rotates or not. Do you consider them as orbits around the Earth in the instance where the Earth doesn't rotate ? Read on ....

The Earth will have fixed orientation wrt distant stars. So if I'm standing on the Earth and look up one day seeing an object at L4 say more or less directly above me ( I have extremely good eyes evidently ) then if I stare at it ( without blinking ? ) for three months say, the distant stars will have moved behind the L4 object. The L4 object will no longer be directly above me, it will be directly above another point on the Earth some 90% of longitude away from me. Because we have all orbited a quarter way around the Sun. So the L4 object will appear to go around the Earth - indeed just as the Sun appears to also - with a period of one year.

But it will never hold a constant position over some specific point on the Earth's surface when that surface does not rotate .....

.... but they would if the Earth did rotate with a one year period in the correct direction ! :-)

So that question was a tad evil. But it brings out a few interesting concepts.

#2 A useful way to think of this is to recall one of Johannes Kepler's laws : in a two body system the radius vector ( from one body to the other ) sweeps out equal areas in equal times.

In a precisely circular orbit this is rather self evident eg. a quarter of the period moves a body a quarter of a full circle around thus the area sweep is one quadrant of a circle. The genius bit is to extend this to ellipses. When the bodies are far apart ( near 'apogee' say ) then the angular rate of movement is slower but the radial vector length is longer, and this happens to just compensate for the close-in case ( near 'perigee' say ) where the angular rate is faster but the radial vector length is somewhat shorter.

Now I say 'just compensate' but this is no fluke. Closer analysis, which I will spare you, reveals that this area law of Kepler's is really the Conservation Of Angular Momentum in disguise. If you leave the system alone that won't change. Linear momentum is the instantaneous velocity vector scaled by the mass at the time. Newton used to call this 'the quantity of motion' implying that it increased if there was more mass of something at a given direction and speed of travel.

Any specification of angular momentum first requires that an origin is selected, about which you will refer all rotations to. Plus a 'positive' sense of rotation needs to be defined as well. Then the magnitude of the angular momentum ( at some particular time ) is the product of the length of the radius vector to said object and the length of the component of linear momentum which is perpendicular to that radius vector. That's a mouthful for sure, but you are seeking the vector part of the linear momentum that contributes to circulation around that chosen origin.

Now the very curious bit is what is the direction of the angular momentum vector. I've found it's length as above, but where does it point ? Convention dictates that one places the tail of the angular momentum vector at the origin point and it points ( in a right handed sense ) away from the plane of the rotation.

So for our satellite trying to get to geostationary territory you could conveniently imagine a vector at Earth's centre pointing to the North pole say, with a length that represents the satellite's angular momentum. I've reasonably assumed that the satellite will orbit in the plane of the Earth's equator, and that the Earth's center is our preferred origin to describe rotations.

Thus to get from highly elliptical to circular - while making the orbit 'larger' overall I'll add - then you need to add angular momentum. Well you need to add energy overall to get the satellite mass to come away from nearby the Earth. Therein lies a clue for burn timings.

If one fired the thrusters when travelling along the long sides of the ellipse then that will add to the eccentricity. You'd get a longer ellipse. Yes you've added energy but not in the way we desire. Hence ...

... you fire at around apogee and/or perigee giving acceleration in the rotation sense that you are already going ( call this 'prograde' ) and that will be pretty well at right angles to the radius vector at those portions of the orbit. Keep doing that per orbit and one day you will come to circular and higher up.

Gary is correct in that 'retrograde' will give a circular orbit too. But down in close above Earth. However that's a correct answer as per the wording of my question.

Other bits to think about :

- you have to throw fuel mass away in some direction to change things. So the satellite mass will vary accordingly.

- it's quite crowded up there in Geostat City. You don't want to arrive like a bowling ball. For that matter the initial apogee is above geostationary so the orbits first up will go through there on the way up and down ....

- if you've deduced that the total area enclosed by a full orbit is a proxy for the orbit's energy, then well done ! Just remember that energy here includes gravitational potential energy plus kinetic. So while the satellites up at Geostat City are circulating slower than say, the ISS at low Earth orbit ( period ~ 90 minutes ) they have way more stored potential. And that was achieved by spending rocket fuel to push them 'uphill'.

Here's hoping I've (a) not made a mistake and (b) have been clear ..... :-)

BTW the 'gee' in perigee and apogee here refer to the Earth. Likewise in context : 'peri' = around, about ie. close, while 'apo' = apex, summit ie. far.

Cheers, Mike.

( edit ) FWIW : there's a far more subtle underlying principle here : rotations are absolute ( well, accelerations generally ). You can't transform away rotation without also losing any original inertial aspect. See Ernest Mach say, and/or Newton's bucket, and of course Einstein or Poincare or .....

I have made this letter longer than usual because I lack the time to make it shorter. Blaise Pascal

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