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Re: Fly to the moon question

What they used to do for the Apollo missions was boost to LEO, then 
when everything was lined up for lunar transit, do a final burn called 
"translunar injection" (TLI) that gave the spacecraft the boost to 
escape velocity.  The translunar trajectory would then carry the craft 
on a "free return" transit that would coast all the way to pericynthion 
and back to reentry with nothing but very small midcourse correction 

TLI was an *enormous* delta-V, and required almost all of the fuel and 
thrust of an entire S-IVB booster stage (the S-IVB, LM, thrust 
structure, and CSM together was larger than almost all of the previous 
Gemini and Apollo mission's entire launch weight) to get from LEO to 
escape velocity.  The "lunar orbit insertion" (LOI) and "transearth 
injection" (TEI) burns near pericynthion were much smaller because of 
the much smaller orbital velocities involved due to the moon's smaller 
mass, and were well within the capability of the SPS engine on the SM, 
as well as leaving behind the 20,000 pound LM for the TEI burn.  Apollo 
13 and the later missions used a "hybrid free return" trajectory which 
did require a burn near PC to get back to a zero-energy true free 
return (which is why 13 needed a PC+2 burn after the tank explosion and 
abort decision) but allowed a somewhat wider launch window and range of 
landing sites.

But .. it's all done at the ends of the trajectory.  The spacecraft did 
slow down to about 5000 (?) fps at the equigravisphere, and speed up 
somewhat toward pericynthion, but the only burns done in midcourse were 
very small magnitude, just enough to fine tune the trajectory.  The 
changes in speed were planned for in the basic calculations of the 
trajectory, and enough delta-V was put in at the beginning to overcome 
the deceleration over the inital part of the transit .. that's it.  
Requires some vector calculus to quantify, but the basic concept is 
fairly simple.

Now .. for some real fun .. the translunar path is a free return back 
to earth, give or take a bit, but the *cislunar* path *gains* energy 
from the moon's motion in exactly the same way the translunar path 
loses energy.  If you launch a payload on a cislunar trajectory that 
follows a hyperbolic path very close to the lunar surface, you'll gain 
quite a bit of delta-V coming out of it .. exactly how much is a little 
beyond what I can calculate with any confidence, but I suspect it's 
enough to save a bit of fuel on interplanetary launches .. heh .. as 
long as you're comfortable with looking like you're going to hit the 
moon until the very last moment ..

On Thursday, Nov 20, 2003, at 18:28 US/Central, William Leijenaar wrote:

> Hi AMSATs,
> I'm now in San Salvador on my world-travel.
> During my travel I'm thinking out some ideas hihi.
> I have been trying to find out what is needed to get
> something in a moon orbit. Its already very long time
> ago I did these physic calculations, and I'm sure many
> amateurs are better in it than me hihi.
> On internet I find that the escape velocity of the
> earth is abt 11km/s , and that of the moon some 2km/s.
> (I have not the exact numbers here with me now, so it
> can not be 100% correct)
> My question:
> When I see it right, an object first has to gain speed
> to exit the earth gravity, and later when aproaching
> the moon it has to be slowed down again with more than
> 8km/s.
> This means you need double engine power, first to exit
> the earth, and later to enter the moon orbit ???
> Kind regards,
> 73, de PE1RAH
> William in San Salvador
> ===
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Heard from a flight instructor:
"The only dumb question is the one you DID NOT ask, resulting in my 
going out and having to identify your bits and pieces in the midst of 
torn and twisted metal."

           --... ...-- -.. . -. ..... ...- -...
                   Bruce Bostwick N5VB

"Many that live deserve death. And some that die deserve
life. Can you give it to them? Then do not be too eager to deal out
death in judgement. Even the very wise cannot see all ends." -- Gandalf

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