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#840
by
Lars-J
on 05 Jan, 2018 17:09
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At any rate, why would the departure burn be at the terminator? At that point, the vehicle will be flying directly away from (or towards) the Sun.
Optimal Earth departure into a heliocentric Hohmann transfer towards an outer planet requires that the LEO velocity vector be aligned as closely as possible with the direction of the Hohmann orbit at periapsis. This Hohmann orbit is tangent with the the Earth's orbit, meaning the LEO orbit direction needs to be as close as possible to the Earth's heliocentric orbit direction which only happens at local midnight.
The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnght. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
That's a great illustration. So it seems like a late afternoon launch would be close to perfect. (It also allows them to launch during daylight hours, which is probably an internal requirement for collecting the maximum visual data on the launch)
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#841
by
envy887
on 05 Jan, 2018 17:10
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At any rate, why would the departure burn be at the terminator? At that point, the vehicle will be flying directly away from (or towards) the Sun.
Optimal Earth departure into a heliocentric Hohmann transfer towards an outer planet requires that the LEO velocity vector be aligned as closely as possible with the direction of the Hohmann orbit at periapsis. This Hohmann orbit is tangent with the the Earth's orbit, meaning the LEO orbit direction needs to be as close as possible to the Earth's heliocentric orbit direction which only happens at local midnight.
The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnght. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
Thanks, I wasn't accounting for that. That places the optimal injection burn time, but what about the launch time?
For a launch due east out of Canaveral to be as close to the ecliptic plane as possible at 1800 local time, the Earth would need to be near the vernal equinox (when the difference is about 4 degrees). But this launch will be much nearer the winter solstice, so SpaceX will either
1) launch N orbits early (when Canaveral is in plane) and coast to the injection burn - not sure what N values work here
2) launch direct to injection but inject out of the ecliptic or
3) launch ~1/4 orbit early (around 12:00 local) to near the equator and do a plane change to match the ecliptic, rather like a GTO launch
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#842
by
Lars-J
on 05 Jan, 2018 17:13
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envy887, the Roadster is not going to Mars. It is going to be more of a Mars-like transfer orbit. (In fact, the less risk of impacting Mars, the better - Musk wants it to be up there for a billion years)
So there is no need to be so exact.
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#843
by
octavo
on 05 Jan, 2018 17:40
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envy887, the Roadster is not going to Mars. It is going to be more of a Mars-like transfer orbit. (In fact, the less risk of impacting Mars, the better - Musk wants it to be up there for a billion years)
So there is no need to be so exact.
Unless you're also trying to demonstrate pinpoint long range trajectories maybe?
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#844
by
RoboGoofers
on 05 Jan, 2018 18:00
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At any rate, why would the departure burn be at the terminator? At that point, the vehicle will be flying directly away from (or towards) the Sun.
Optimal Earth departure into a heliocentric Hohmann transfer towards an outer planet requires that the LEO velocity vector be aligned as closely as possible with the direction of the Hohmann orbit at periapsis. This Hohmann orbit is tangent with the the Earth's orbit, meaning the LEO orbit direction needs to be as close as possible to the Earth's heliocentric orbit direction which only happens at local midnight.
The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnght. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
That's a great illustration. So it seems like a late afternoon launch would be close to perfect. (It also allows them to launch during daylight hours, which is probably an internal requirement for collecting the maximum visual data on the launch)
It gets scoffed at a lot as just a video game, but Kerbal Space Program is an excellent teacher of basic orbital mechanics. Elementary school kids that play this "game" can teach this lesson to their apollo-era grandparents.
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#845
by
hektor
on 05 Jan, 2018 18:14
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envy887, the Roadster is not going to Mars. It is going to be more of a Mars-like transfer orbit. (In fact, the less risk of impacting Mars, the better - Musk wants it to be up there for a billion years)
So there is no need to be so exact.
Unless you're also trying to demonstrate pinpoint long range trajectories maybe?
I have the impression that the later FH launches, the closer it is possible to bring it to Mars.
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#846
by
envy887
on 05 Jan, 2018 18:43
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envy887, the Roadster is not going to Mars. It is going to be more of a Mars-like transfer orbit. (In fact, the less risk of impacting Mars, the better - Musk wants it to be up there for a billion years)
So there is no need to be so exact.
Unless you're also trying to demonstrate pinpoint long range trajectories maybe?
No, since they can define a arbitrary target orbit and then measure how close to it they got.
The only reason to not do direct launch to injection is if they want to demonstrate LEO long coast, or want to get closer to Mars' orbital plane without the performance hit of a plane change. I don't think either of those are necessary: they demonstrated long LEO coast on NROL-76, they have plenty of performance for a plane change, and there's no particular reason to get in plane with Mars since that increases the chance of actually hitting Mars, which they don't want to do.
So direct launch to injection is likely, IMO. which puts launch time around 1800 local time in Florida, +/- an hour or so.
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#847
by
kevinof
on 05 Jan, 2018 18:51
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Will there still be daylight at 18:00 local time? I really can't see them launching in the dark - my thinking is they will want cameras recording everything especially the separation events so sending this thing up in the dark would scupper that.
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#848
by
ZachS09
on 05 Jan, 2018 18:55
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Will there still be daylight at 18:00 local time? I really can't see them launching in the dark - my thinking is they will want cameras recording everything especially the separation events so sending this thing up in the dark would scupper that.
Around the end of January 2018, sunset is approximately 6:00 PM Eastern Time (23:00 UTC). So, there's technically still daylight at that time.
Find the coordinates to LC-39A and enter them in the location box on this link.
https://www.esrl.noaa.gov/gmd/grad/solcalc/
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#849
by
clongton
on 05 Jan, 2018 19:07
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The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
<snip> </snip>
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnight. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
That's a great illustration, thanks.
I would add that without an additional phase burn that the trajectory would still not be in the solar plane of ecliptic, but rather will enter a heliocentric trajectory on the earth orbital inclination, and will thus not encounter Mars at all, demonstrating only that the Falcon Heavy has the ability to send a payload out as far as Mars.
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#850
by
LouScheffer
on 05 Jan, 2018 19:10
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Will there still be daylight at 18:00 local time? I really can't see them launching in the dark - my thinking is they will want cameras recording everything especially the separation events so sending this thing up in the dark would scupper that.
I agree - I think a daylight launch, a short coast, then a terminator-ish Mars injection. This also gives the possibility of a big launch window.
So I'd guess a window of 1000 to 1400 local, a coast (10--30 minutes) that depends on the time of launch, the an injection at about 1800 local time for the payload.
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#851
by
rakaydos
on 05 Jan, 2018 19:11
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Is the roadster mounted at the angle it is to keep the 2nd stage thrust vector through the CoG?
I suspect it is to aid Horizontal Integration. The tesla may be able to hang vertically from it's jack points when the fairing is sideways, but why risk it?
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#852
by
envy887
on 05 Jan, 2018 19:11
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The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
<snip> </snip>
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnight. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
That's a great illustration, thanks.
I would add that without an additional phase burn that the trajectory would still not be in the solar plane of ecliptic, but rather will enter a heliocentric trajectory on the earth orbital inclination, and will thus not encounter Mars at all, demonstrating only that the Falcon Heavy has the ability to send a payload out as far as Mars.
They can also demonstrate orbit injection accuracy relative to a predefined target orbit. There's no need to define the target orbit relative to Mars.
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#853
by
Space Ghost 1962
on 05 Jan, 2018 19:15
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Also, this is a test launch.
You want good illumination, good weather / visibility so that you can see any deviation.
Also, you want to track in the eastern range (especially to F9US engine start), so not so much downrange with the core as you otherwise might for such a light payload.
It's not about the payload on this flight, its about the flight data.
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#854
by
rsdavis9
on 05 Jan, 2018 19:43
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I think people (or me) are missing the 28deg of Cape Canaveral. At the winter solstice Cape Canaveral is just about exactly on the ecliptic at midnight. (28-23.5=4.5). If you boost straight east you get the maximum assist from the earth's rotation in the desired direction. So the following changes this position:
1. not at solstice.
2. not instantaneous delta-v application
3. not infinite velocity so gravity curves the trajectory
So I guess somewhere around 1800-2400 lies the maximum ecliptic delta-v vector addition.
So given any season to launch in the ecliptic. What season gives the most delta-v in the earths orbit vector?
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#855
by
mme
on 05 Jan, 2018 19:53
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Also, this is a test launch.
You want good illumination, good weather / visibility so that you can see any deviation.
Also, you want to track in the eastern range (especially to F9US engine start), so not so much downrange with the core as you otherwise might for such a light payload.
It's not about the payload on this flight, its about the flight data.
This. A million times this.
This mission is not targeting Mars, it's not about maximizing delta-v and it's not about the upper-stage doing something new or different. It's about proving the FH works as expected and collecting the maximum amount of data to verify that and/or be able to determine what went wrong.
Also the last thing they want is to actually impact Mars.
My money is on a mission that has a daylight launch with a long launch window without the US having a longer than already proven linger time.
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#856
by
john smith 19
on 05 Jan, 2018 21:05
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I think that graphic is the view from the Sun at the Autumnal Equinox, with the "shiny spot" being a representation of the Sun's specular reflection.
The image is generic and always true.
The viewpoint changes with the seasons.
It was the view back along the Eart’s orbit two weeks ago at the the Solstice.
Today it is swung around ~15 degrees.
I never really understood why the Tropics of Cancer and Capricorn were special.
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#857
by
1
on 05 Jan, 2018 21:13
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At any rate, why would the departure burn be at the terminator? At that point, the vehicle will be flying directly away from (or towards) the Sun.
Optimal Earth departure into a heliocentric Hohmann transfer towards an outer planet requires that the LEO velocity vector be aligned as closely as possible with the direction of the Hohmann orbit at periapsis. This Hohmann orbit is tangent with the the Earth's orbit, meaning the LEO orbit direction needs to be as close as possible to the Earth's heliocentric orbit direction which only happens at local midnight.
The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnght. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
But that's the thing. As others mentioned, we don't actually have a target to reach since this is a vehicle test. We're just going for a generic heliocentric orbit where aphelion reaches out towards Mars. I don't think we want to go 'left'. We want to go 'down'. And if Space Ghost's thoughts about keeping the launch over Florida range assets as long as possible are plausible, we might not even want to launch
east but rather do something like launch at 6 in the morning going more or less straight up. Energetically efficient? Probably not, but who cares as long as the vehicle works properly?
Edit:
Also, to some others, yes, that first image I dug up was just a generic image; not intended to show where the Earth is today. My post subsequent to that first one explained my assumptions on where the Earth is now.
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#858
by
Comga
on 05 Jan, 2018 22:08
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At any rate, why would the departure burn be at the terminator? At that point, the vehicle will be flying directly away from (or towards) the Sun.
Optimal Earth departure into a heliocentric Hohmann transfer towards an outer planet requires that the LEO velocity vector be aligned as closely as possible with the direction of the Hohmann orbit at periapsis. This Hohmann orbit is tangent with the the Earth's orbit, meaning the LEO orbit direction needs to be as close as possible to the Earth's heliocentric orbit direction which only happens at local midnight.
The problem is that the Earth's gravity turns the direction of the vector after the burn. If you add infinite velocity, then the Earth's gravity has no impact, and it's best to burn at midnight. But in practice, you are just above Earth escape, and you get a bend of almost 90 degrees. Imagine a really high apogee orbit as shown below, where the bend is exactly 90 degrees. The dot shows where you need to burn to go from the circular parking orbit to the high apogee orbit. This is the same place the direct injection needs to terminate.
Now as you add velocity above Earth escape, the optimum point rotates around, until with infinite added velocity it's at midnght. But a Mars apogee orbit is very much closer to earth escape than infinity, so the burn position is close to that shown. On the other hand, as another poster pointed out, the launch takes a finite amount of time, which rotates the launch point further clockwise. Overall you should launch much closer to 1800 than midnight.
Thanks, I wasn't accounting for that. That places the optimal injection burn time, but what about the launch time?
For a launch due east out of Canaveral to be as close to the ecliptic plane as possible at 1800 local time, the Earth would need to be near the vernal equinox (when the difference is about 4 degrees). But this launch will be much nearer the winter solstice, so SpaceX will either
1) launch N orbits early (when Canaveral is in plane) and coast to the injection burn - not sure what N values work here
2) launch direct to injection but inject out of the ecliptic or
3) launch ~1/4 orbit early (around 12:00 local) to near the equator and do a plane change to match the ecliptic, rather like a GTO launch
We'll have what's behind door number three.
(except no plane change is needed. That's only relative to the Earth, which on the scale of solar orbits is small.)
If you want the orbit to cross the equator and terminator you launch due east at solar noon.
After a quarter orbit the spacecraft is at the that point, the equator and terminator.
That makes for the most efficient launch "forward" with respect to the Earth's orbit around the Sun.
Note that with a light payload, SpaceX doesn't have to be maximally efficient, so they can do this slightly differently.
Also, as pointed out by Space Ghost and myself, there is the issue of the inclination of the orbit and the location of the perihelion relative to the orbit of Mars. To avoid contact, or gravitational perturbation which will eventually be chaotic (unpredictable with available precision of launch injection) SpaceX may further modify the launch timing and aiming.
edit: Launching "N orbits early" requires a second stage operational lifetime that has not been demonstrated to date.
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#859
by
the_other_Doug
on 06 Jan, 2018 00:19
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My guess would be about 6:00 PM (18:00) local time. My reasoning is this: injection is most efficient when exactly the opposite of the intended target. To get maximum throw, they need the target to be tangent to the Earth's orbit. This puts the opposite point at about 18:00 local time. This is only true for direct injection - with a parking orbit, the earth departure burn needs to happen about 18:00 local time (in the probe's time zone) but the launch can be anytime.
Shouldn't that be midnight local time? For a Hohmann transfer, apoapsis and periapsis are on opposite sides of the sun, and the tangent to Earth is on the far side from the Sun.
You want to leave Earth at midnight local time, but the way to do that is to burn at 18:00 local time, then the Earth bends the trajectory around 1/4 turn. To see this, imagine a super long and skinny orbit around the Earth with a 40,000,000 km apogee. If the long axis of this orbit is tangent to the Earth's orbit, then the perigee is at the trailing edge of Earth, or about 18:00 local time. Since firing at perigee is the most efficient time, that's where the injection should be.
Hopefully, this point hasn't been made yet. Dangers of responding two pages up from the end of the thread...
I think SpaceX will launch during daylight. This is first and foremost a demonstration of the launcher. They will want the best overall visibility of the rocket for the tracking cameras (in case of an RUD and also to gather the most info they can from the flight), and you just don't get that with night launches. With a night launch, the light of the exhaust completely overwhelms detailed imagery of what the rocket itself is doing. Not a good idea on a first flight.
All they want to demonstrate is that they can push a payload out to the "neighborhood" of Mars. It doesn't have to be exactly in Mars' orbital inclination. Remember, no one at this point needs to demonstrate that FH can target a payload actually to the planet Mars, as it seems it will never be called upon to do so. Just getting the payload out as far away from the Sun as Mars in a heliocentric orbit is all SpaceX really has to worry about.
For that, you can launch pretty much any time of day you like, as far as inclination to the ecliptic is concerned. And I imagine you could launch pretty much any time past local noon to get a direct injection into a solar orbit with an aphelion of one and a half AU.
So, as I say, since there is no reason to actually target the planet Mars with this launch, I can't at all imagine they will launch the FH first check-flight at night, when they can't see the dynamics of the elements of the rocket, except insofar as they impact the plumes.