Article View: sci.space.shuttle
Article #3833Re: Re : To the moon.....
From: Chris Marriott
Date: Sun, 24 Jan 1993 20:31
Date: Sun, 24 Jan 1993 20:31
65 lines
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3334 bytes
In article <PAULI.RAMO.93Jan24190410@vipunen.hut.fi> Pauli.Ramo@hut.fi writes: >In article <74006@cup.portal.com> BrianT@cup.portal.com (Brian Stuart Thorn) > >The *thrust* is not a problem, after getting to LEO. If you have less >thrust, it only takes a little longer to get to the Moon. In fact, you >could make both TLI and LOI with only one OMS (or even RCS) engine, if >you had enough fuel to it and it lasted a long continuous firing. The >problem is the amount of fuel needed. > >There is a simple equation to calculate the amount of velocity change >for a given amount of fuel and mass of vehicle: > > dv = ln (m/m0) * I > >where >dv = velocity change >m = initial mass (with full tanks) >m0 = mass without fuel (empty tanks & payload) >I = the impulse of the engine used > >With chemical rockets, I is some thousands of m/s. For example, the impulse >of SSME is about 4400 m/s. With SRB's it is lower. > >For a complete trip to moon and back, you must sum the velocity changes >needed in it's various parts together and use that to calculate your >fuel requirements. Unfortunately, I don't have the exact requirements >for TLI, LOI, TEI and EOI on hand (maybe someone can help me here), but >even LEO requires a total velocity change in the order of 9000 m/s. > >Thus, to only get to LEO with a single stage booster equipped with SSME's, >you had to have the mass ratio of about 7.73:1. That means that of the >initial mass of the vehicle, 87% had to be used as fuel just to get to >LEO, and the remaining 13% would include all the tanks and engines for >the fuel. > >If someone has the velocity change requirements for the rest of our trip >to moon, we can continue this little number game. > > Pauli >-- >Disclaimer fault - lawyers dumped > Your equation is too simplistic - it only really applies either in the absence of gravity, or if the velocity change occurs instantaneously (ie, an "impulse"). If the fuel is burned over a finite time, you have to take into account the losses due to gravity and (if in atmosphere) friction. For example, if you have a rocket on a launch pad, and that rocket produces <1g of acceleration, it can burn all its fuel, but its velocity will stay at zero! Best wishes, Chris -- -------------------------------------------------------------------------- | Chris Marriott | chris@chrism.demon.co.uk | | Warrington, UK | BIX: cmarriott | | | CIX: cmarriott | | Save the whales. Collect the whole set. | CompuServe: 100113,1140 | --------------------------------------------------------------------------
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Newsgroups: sci.space.shuttle From: chris@chrism.demon.co.uk (Chris Marriott) Path: arc1-ca.pugleaf.net!sparky!uunet!europa.asd.contel.com!paladin.american.edu!darwin.sura.net!spool.mu.edu!agate!doc.ic.ac.uk!pipex!demon!chrism.demon.co.uk!chris Subject: Re: Re : To the moon..... Distribution: world References: <PAULI.RAMO.93Jan24190410@vipunen.hut.fi> Organization: None Reply-To: chris@chrism.demon.co.uk X-Mailer: Simple NEWS 1.90 (ka9q DIS 1.19) Lines: 65 Date: Sun, 24 Jan 1993 20:31:40 +0000 Message-ID: <727907500snz@chrism.demon.co.uk> Sender: usenet@demon.co.uk