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- Path: sparky!uunet!elroy.jpl.nasa.gov!ames!titan.ksc.nasa.gov!hollis
- From: hollis@titan.ksc.nasa.gov
- Newsgroups: sci.space.shuttle
- Subject: Re : To the moon... (And you thought that was the last of it...)
- Message-ID: <1993Jan21.163339.5025@titan.ksc.nasa.gov>
- Date: 21 Jan 93 16:33:39 EST
- Organization: NASA, Kennedy Space Center
- Lines: 109
-
- Greetings and Salutations:
-
- From: davem@ee.ubc.ca (Dave Michelson)
- >I've been corresponding with Ken about this point and I think he was confused
- >by the way "injection velocity" is described in Fundamentals of Astrodynamics.
- >
- >I, for one, agree that TLI from LEO only requires a velocity increment some
- >something less than 7500 mph or just over 2 miles/sec. Ken was under the
- >impression that TLI required an *additional* (rather than a net) velocity of
- >of 25,000 mph...
-
- It is very kind of you to explain my confusion on the matter (I usually just
- let the audience wonder ;->) I understand now where the complete confusion came
- from, and I have gotten better data. Reference McDonnell Douglas Astronautics
- Company - West - Propulsion engineering manual - Development engineering -
- Manual M8069.ACD. For a 100 NM orbit, we require 25,581 ft/second velocity.
- For lunar missions (depending on how much time you wish to be in transit to the
- moon) 90 hour transfer = 35,850 to 35,910 ft/sec with 8,310 ft/sec additional
- velocity to land. 50 hour transfer orbit is 36,070 to 36,160 ft./sec. with
- 9,410 ft/sec. additional velocity to land. Note that the Apollo missions never
- reached true Earth escape velocity (36,178 ft./sec.) but would have been in a
- highly elliptical orbit if something had gone badly wrong.
-
- Now for the confusion factor. Here is the original post:
-
- From: ritley@uimrl7.mrl.uiuc.edu
- >Would it be possible for the Space Shuttle to journey to the Moon?
- >
- >Obviously it could not easily land there, but could it at least make
- >the trip?
- >
- >If so, would it be possible to fit the Shuttle with equipment which
- >would permit it to land there --- possibly some sort of rocket
- >engine located in the cargo bay? Or possibly a second landing
- >vehicle could be taken up (perhaps on a second flight) and
- >``shuttled" to the Moon in the cargo bay?
-
- My reply was, essentially, that it would take at least twice (if not more)
- ENERGY to get to the moon than the shuttle expends for launch into low earth
- orbit. I got wrapped around the axle when energy was equated with velocity in
- some of the other posts. This is like discussing apples & pineapples. They
- are somewhat related, but not directly. You must expend energy to get more
- velocity, but that energy expenditure can be an efficient expenditure or not so
- efficient. Once again my McDonnell reference gives a graph of "Rocket vehicle
- payload fraction versus velocity relationship". When you get into the
- approximately 35,000 ft/sec range, the ratio gets pretty nasty. The first hint
- is that the payload fraction side of the scale is logarithmic. Basically you
- burn an exponential amount of fuel for linear velocity increase. Even with an
- Isp of 460, you get only about 0.0025% payload into 34,000 ft/sec orbit,
- assuming the "good" step structure factor of 0.10, of course.
-
- To put it differently, Reference "Rocket Propulsion Elements", Sixth edition,
- George P. Sutton. Assuming you have enough thrust to put 100% payload in orbit
- (300 nautical miles), to soft land that on the moon, you would only be able to
- get about 10% to 20% of the payload there. To land and return, you can only
- get about 1% to 4%.
-
- By the way, as a matter of general interest, even if you run the OMS system dry
- you only get about an additional 1,500 ft/sec extra boost with no payload.
- I.e. The OMS will get you about 10% of the way to the moon...
-
- Now, I can see how far off my SWAG (Scientific Wild Ass Guess) was. I am going
- to go for pure numbers.
-
- Lets, just for grins, go for a empty Orbiter.
-
- 171,205 = Orbiter weight (dry, with engines)
- 1,655,600 = External Tank with propellants
- 1,300,000 = SRB (Each)
- 1,300,000 = SRB #2
- ----------
- 4,426,805 Lbs total
-
- 171,205 = Orbiter weight (dry, no OMS propellants, with engines)
-
- Payload fraction = Payload weight/launch weight. This gives us a payload
- fraction versus 25,000 ft/sec fraction of 0.038 to low Earth orbit.
-
- The SSMEs have a Isp of 453.5 at altitude. The SRBs have a Isp of 265.5. One
- SRB has 3 million Lbs thrust, but is optimized for atmosphere. One SSME has
- 470,000 Lbs thrust in vacuum at 100% thrust and is optimized for upper
- atmosphere / vacuum.
-
- To get a payload fraction of 0.0025% into 34,000 ft/sec orbit (as mentioned
- above) it looks like you would have to have about:
-
- 171,205 / 0.0025 = 68,482,000 Lbs total. That is assuming a Isp of somewhere
- around 460, and a GREAT step structure factor. This is also neglecting the
- weight of the propellant tank(s) that are required.
-
- Gee. Looks like I was off by a factor of 5 (conservatively). You would need
- about 11 times the amount of propellants, and have it all lifted by SSMEs, no
- SRBs. That still doesn't get you to the moon.
-
- Now if you wish to discuss refueling options in low earth orbit, to get an
- additional 10,000 ft/second for moon injection, you make life a little easier.
- But that would be cheating. You STILL have to send the propellant up, and
- count that as total energy.
-
- The absolute best case would be where you had an infinite number of propulsion
- filling stations all long the flight path to the moon, but that sort of
- thinking starts getting ridiculous.
-
- --
- -----------------------------------------------
- Ken Hollis INTERNET: HOLLIS@TITAN.KSC.NASA.GOV
- SPAN/HEPnet: KSCP00::HOLLIS
- Due to additional budget constraints, the light at the end of the tunnel
- will be turned off until further notice...
-