Author Topic: Orbits Q&A  (Read 177096 times)

Offline Proponent

  • Senior Member
  • *****
  • Posts: 7277
  • Liked: 2782
  • Likes Given: 1462
Re: Orbits Q&A
« Reply #320 on: 03/05/2017 09:30 am »
Pioneer P-30

You mean the failed 1960 Atlas-Able lunar probe?  You're gonna have to explain that one....  Was its lunar-orbit-insertion engine ignited before it fell back to Earth?
« Last Edit: 03/05/2017 09:30 am by Proponent »

Offline Jim

  • Night Gator
  • Senior Member
  • *****
  • Posts: 37440
  • Cape Canaveral Spaceport
  • Liked: 21451
  • Likes Given: 428
Re: Orbits Q&A
« Reply #321 on: 03/05/2017 12:30 pm »
Pioneer P-30

You mean the failed 1960 Atlas-Able lunar probe?  You're gonna have to explain that one....  Was its lunar-orbit-insertion engine ignited before it fell back to Earth?

Actually, it doesn't matter.  Got a better one. 
Transit 1B  Thor Able-Star

Offline Nicolas PILLET

  • Member
  • Senior Member
  • *****
  • Posts: 2454
  • Gien, France
    • Kosmonavtika
  • Liked: 670
  • Likes Given: 134
Re: Orbits Q&A
« Reply #322 on: 03/05/2017 07:33 pm »
Transit 1B  Thor Able-Star

Yes, indeed, Ablestar made the first engine restart.

But it is not really a "maneuvering satellite"...
Nicolas PILLET
Kosmonavtika : The French site on Russian Space

Offline as58

  • Full Member
  • ****
  • Posts: 835
  • Liked: 300
  • Likes Given: 186
Re: Orbits Q&A
« Reply #323 on: 03/22/2017 12:38 pm »
I noticed this question about an equation appearing in Thomas Pynchon's Gravity's Rainbow on mathoverflow. Does anyone recognise the equation as something related to rocketry?

Offline Dalhousie

  • Senior Member
  • *****
  • Posts: 2766
  • Liked: 780
  • Likes Given: 1131
Re: Orbits Q&A
« Reply #324 on: 12/09/2017 01:18 am »
Earth to Mars launch widows occur every 26 months (780 days).

What about the Earth return launch window? Is that also every 780 days, or another value? 


Apologies in advance for any lack of civility - it's unintended

Offline Ultrafamicom

  • Member
  • Posts: 73
  • Liked: 22
  • Likes Given: 27
Re: Orbits Q&A
« Reply #325 on: 02/15/2018 04:31 am »
May anyone do a calculation on how much can FH launch  to a Jupiter direct transfer orbit with Boosters ASDS landing and central core expended ?  Much thanks :P

(My estimation is some 5-6 tons, based on the TMI performance on official website and Musk's twitter https://twitter.com/elonmusk/status/963094533830426624 But when using the 3500kg to Pluto capacity figure,it gives me an astonishing 9-10t, which is obviously counter-intuitive)
« Last Edit: 02/15/2018 04:40 am by Ultrafamicom »

Offline Robotbeat

  • Senior Member
  • *****
  • Posts: 39270
  • Minnesota
  • Liked: 25240
  • Likes Given: 12115
Re: Orbits Q&A
« Reply #326 on: 02/15/2018 05:05 am »
Direct transfer to Jupiter is really high energy. Without stretching the upper stage and perhaps using a kick stage, FH would get very little payload even if fully expendable.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

To the maximum extent practicable, the Federal Government shall plan missions to accommodate the space transportation services capabilities of United States commercial providers. US law http://goo.gl/YZYNt0

Offline Ultrafamicom

  • Member
  • Posts: 73
  • Liked: 22
  • Likes Given: 27
Re: Orbits Q&A
« Reply #327 on: 02/15/2018 05:21 am »
Direct transfer to Jupiter is really high energy. Without stretching the upper stage and perhaps using a kick stage, FH would get very little payload even if fully expendable.

https://en.wikipedia.org/wiki/Hohmann_transfer_orbit

according to the table in this page, Pluto direct transfer requires even far higher energy(2.1km/s delta-v more), but Musk has confirmed FH can do a direct insertion. Even if the direct-to-Pluto payload is as little as 500kg, given the 348s Isp and 4.5-5t dry mass of S2, it can still be converted into about 5 tons of payload to Jupiter.

Offline Demidrol

  • Member
  • Posts: 21
  • Ukr
  • Liked: 26
  • Likes Given: 485
Re: Orbits Q&A
« Reply #328 on: 02/15/2018 09:17 am »
Direct transfer to Jupiter is really high energy. Without stretching the upper stage and perhaps using a kick stage, FH would get very little payload even if fully expendable.
https://en.wikipedia.org/wiki/Hohmann_transfer_orbit
according to the table in this page, Pluto direct transfer requires even far higher energy
A Hohmann transfer orbit is not a direct transfer orbit - https://www.quora.com/What-is-the-hohmann-transfer

Offline Ultrafamicom

  • Member
  • Posts: 73
  • Liked: 22
  • Likes Given: 27
Re: Orbits Q&A
« Reply #329 on: 02/15/2018 09:35 am »
Direct transfer to Jupiter is really high energy. Without stretching the upper stage and perhaps using a kick stage, FH would get very little payload even if fully expendable.
https://en.wikipedia.org/wiki/Hohmann_transfer_orbit
according to the table in this page, Pluto direct transfer requires even far higher energy
A Hohmann transfer orbit is not a direct transfer orbit - https://www.quora.com/What-is-the-hohmann-transfer
Sorry, I intended to mean heading for Jupiter without gravity assist by "direct"...

So any idea on FH's capacity to Hohmann transfer orbit to Jupiter?

Offline monkey55

  • Member
  • Posts: 3
  • UK
  • Liked: 0
  • Likes Given: 0
Re: Orbits Q&A
« Reply #330 on: 06/30/2018 12:57 pm »
In the circular restricted three body problem (CR3BP) of Sun-Earth system, how can I get the initial conditions for a distant retrograde orbit around Earth? The state vector is X=[x0 0 0 0 vy0 0]. What values do I choose for x0 and vy0, and how do i choose them?

Thanking you in advance
« Last Edit: 06/30/2018 01:05 pm by monkey55 »

Offline Proponent

  • Senior Member
  • *****
  • Posts: 7277
  • Liked: 2782
  • Likes Given: 1462
Re: Orbits Q&A
« Reply #331 on: 07/03/2018 04:36 am »
There is no unique DRO.  The value x0 is up to you,  Unless you choose x0 to coincide with L1 or L2 (I'm guessing you're choosing the x-axis to correspond to the Earth-sun line), the orbit will not have a simple, repeating shape like a circle or an ellipse.  Orbits that are somewhat inside of the L-points but not too close to the moon probably won't be too drastically different from the circles or ellipses that would result if Earth were the only body present, though their orientations and shapes will vary a bit over time.
« Last Edit: 07/03/2018 04:38 am by Proponent »

Offline monkey55

  • Member
  • Posts: 3
  • UK
  • Liked: 0
  • Likes Given: 0
Re: Orbits Q&A
« Reply #332 on: 07/04/2018 07:26 pm »
thank you Proponent. And what about the vy0 component?

Offline Proponent

  • Senior Member
  • *****
  • Posts: 7277
  • Liked: 2782
  • Likes Given: 1462
Re: Orbits Q&A
« Reply #333 on: 07/07/2018 03:06 am »
There is no unique vy0 either.  Different values of vy0 will give you different orbits, none of which will be a simple closed shape.  If I recall correctly, the discussions of lunar DROs a few years ago when they were first proposed for the Asteroid Redirect and Rendezvous Mission contain some links to on-line calculators that may allow you to calculate orbits of DROs.

Offline monkey55

  • Member
  • Posts: 3
  • UK
  • Liked: 0
  • Likes Given: 0
Re: Orbits Q&A
« Reply #334 on: 07/07/2018 04:30 pm »
i will search for these papers. Do you have any in mind?

thank you again!

Offline yarg

  • Member
  • Posts: 1
  • Ireland
  • Liked: 0
  • Likes Given: 0
Re: Orbits Q&A
« Reply #335 on: 07/24/2018 09:05 am »
Hi,

I was wondering how GTO vs GTO- vs GTO+ is defined. From what I have read (and I am no subject matter expert) it would seem that the biggest differentiator is the delta v requirement to GEO from the given transfer orbit. I have seen that -1,800 m/s dv is typical for GTO, -2,200 m/s dv seems to be common for GTO- and -1,500 m/s dv is common for GTO+. Am I right in assuming that a) this is the best way to differentiate between GTO / GTO- / GTO+, and b) if the numbers above are approximately correct.

Thanks,

Yarg

Online mmeijeri

  • Senior Member
  • *****
  • Posts: 7772
  • Martijn Meijering
  • NL
  • Liked: 397
  • Likes Given: 822
Re: Orbits Q&A
« Reply #336 on: 09/23/2018 11:02 am »
One advantage this architecture shares with EML-1 or EML-2 rendezvous is no huge Earth departure stage is needed.

Yep, and that's a big deal. In addition, Lagrange point staging allows you to save energy if you have large, reusable transfer ships that travel between the edges of the gravity wells of Earth and Mars and separate smaller ones that take you from low orbits at the bottom of those gravity wells to their edges.

Quote
Also compared to EML-1 (respectively EML-2) rendezvous you save ~1.5 km/s (resp. ~0.7 km/s) of delta-vee entering and exiting the Lagrange points.

Bear in mind that there exist slower (~3 months) but cheaper (~3.2 km/s) quasi-ballistic transfers to EML1/2 that don't need a perigee-raising maneuver at the end, because that's taken care of by the small perturbation of almost Keplerian orbits created by the attraction of the sun. This should take care of the bulk of the mass, which would be propellant. Additionally, on the outgoing leg, you can lower your perigee considerably more cheaply by doing a powered flyby of the moon.

Quote
The second advantage is you don’t have to schedule everything around the moon being in the right place. In particular this should help reach high-declination departure asymptotes, i.e. missions that go in a direction that’s far from being in the plane of the moon's orbit.

Does this objection apply to high-energy orbits like EML-1/2? The orbital velocity of circular orbits near that of the moon is ~1 km/s. Even a 90 degree plane change from there doesn't sound prohibitive. Not cheap obviously, but also not prohibitive. I recall reading that the cost of this too can be reduced with a powered flyby of the moon. It may have been in the latest revision of NASA's DRM 5.0 for Mars, which has some good comments on Lagrange point staging. And if not, you could always do a cheap apogee raising maneuver to roughly the distance of SEL1/2, where orbital velocity is ~600 m/s and perigee lowering should also be a bit cheaper.

And SEL1/2 brings me to another topic, one of my favourite links to post here. Are you familiar with the following study by Wes Huntress:

The Next Steps in Exploring Deep Space

It is an incredibly well-thought-out incremental approach, that makes great use of the efficiency and flexibility afforded by Lagrange point staging. Even after so many years it easily remains the most impressive and comprehensive plan I've seen.
« Last Edit: 09/23/2018 11:03 am by mmeijeri »
Pro-tip: you don't have to be a jerk if someone doesn't agree with your theories

Online mmeijeri

  • Senior Member
  • *****
  • Posts: 7772
  • Martijn Meijering
  • NL
  • Liked: 397
  • Likes Given: 822
Re: Orbits Q&A
« Reply #337 on: 09/23/2018 12:10 pm »
Quote
Trans-Mars injection delta V is typically several hundred meters per second, depending on the interplanetary C3. Departure from L2 is relatively insensitive to the escape declination because the outbound lunar flyby can change the inclination from near the plane of the Moon into the departure plane. However, the orbital motion of the Moon limits the low delta V escape opportunity to about 3 days, as shown in Figure 2-13. The TMI maneuver should be performed near perigee to keep the delta V minimized, providing a single nearly instantaneous opportunity to depart on the orbit following lunar flyby.
(emphasis added)

Human Exploration of Mars Design Reference Architecture 5.0 Addendum #2 p. 35
« Last Edit: 09/23/2018 12:22 pm by mmeijeri »
Pro-tip: you don't have to be a jerk if someone doesn't agree with your theories

Offline speedevil

  • Senior Member
  • *****
  • Posts: 4406
  • Fife
  • Liked: 2762
  • Likes Given: 3369
Re: Orbits Q&A
« Reply #338 on: 09/23/2018 09:34 pm »
Quote
The second advantage is you don’t have to schedule everything around the moon being in the right place. In particular this should help reach high-declination departure asymptotes, i.e. missions that go in a direction that’s far from being in the plane of the moon's orbit.

Does this objection apply to high-energy orbits like EML-1/2? The orbital velocity of circular orbits near that of the moon is ~1 km/s. Even a 90 degree plane change from there doesn't sound prohibitive.
Also, can't you get a fair degree of plane change 'free' with a flyby of earth?

Offline deruch

  • Senior Member
  • *****
  • Posts: 2422
  • California
  • Liked: 2006
  • Likes Given: 5634
Re: Orbits Q&A
« Reply #339 on: 10/03/2018 08:34 am »
Hi,

I was wondering how GTO vs GTO- vs GTO+ is defined. From what I have read (and I am no subject matter expert) it would seem that the biggest differentiator is the delta v requirement to GEO from the given transfer orbit. I have seen that -1,800 m/s dv is typical for GTO, -2,200 m/s dv seems to be common for GTO- and -1,500 m/s dv is common for GTO+. Am I right in assuming that a) this is the best way to differentiate between GTO / GTO- / GTO+, and b) if the numbers above are approximately correct.

Thanks,

Yarg

I've not seen "GTO-" or "GTO+" notations used before.  Usually when discussing the orbits, people will just use GTO.  Really, things like GTO-1500 is understandable, but bad, shorthand.  What is actually meant is that it is "a GTO which is GEO-1500" or, explicitly, a GTO which needs an additional 1500m/s of delta-v to reach GEO.  The difference between GTO-1500 and GTO-1800 is launch site.  For GTO launches from Florida (CCAFS/KSC), the standard GTO with apogee at GEO altitude and not much inclination change during launch will end up being about 1800m/s from GEO.  The same is true for similar launches out of Kourou, French Guiana (e.g. Ariane 5 launches), except that they will be ~1500m/s short.  Exact shortfalls will depend on the payload mass, rocket capabilities and performance, and mission requirements. 

Anything that someone talks about as "GTO+" is maybe some unusual notation for a supersynchronous transfer orbit, which is a type of GTO where the apogee of the transfer orbit is above the GEO altitude.  And maybe if someone is talking about a "GTO-" orbit without indicating a delta-v shortfall, they might be talking about a sub-synchronous transfer orbit where the apogee is below GEO.  But those are just guesses, as like I wrote I've never seen those notations used before.
« Last Edit: 10/03/2018 08:34 am by deruch »
Shouldn't reality posts be in "Advanced concepts"?  --Nomadd

Tags: inclination 
 

Advertisement NovaTech
Advertisement Northrop Grumman
Advertisement
Advertisement Margaritaville Beach Resort South Padre Island
Advertisement Brady Kenniston
Advertisement NextSpaceflight
Advertisement Nathan Barker Photography
0