Talk:Explorers on the Moon/Archive 1

Reference about...

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It's not like I'm an expert on this theme (not even far), but in this line: "However, this rocket provides continuous thrust since the enormous energy produced by the nuclear fission explosions somehow can be released gradually."

Is there any reference to afirm that there's no way (with the current technology) to release gradually the energy of a nuclear fission explosion? --Ferran (talk) 23:14, 19 February 2008 (UTC)Reply

Explanation of my edit

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I removed the following from the scientific accuracy / inaccuracy section:

Second, if the spaceship truly was accelerating at such a rate, the flight time to the moon would be a matter of hours, not days as depicted in the English version of the book. In the original French version, however, flight director Baxter calls the landing site shortly after the rocket has taken off from the Moon to inform them that it will arrive 'in four hours.' The French dialogue also establishes that the rocket is moving at 45 kilometres per second some twenty minutes before the halfway point, when its maximum speed -- assuming a constant 9.8m/s acceleration -- would be about 57km/s. The rocket would thus be amazingly capable compared to current technology; travelling at ~5 million kilometres a day it would cross the gulf between Earth and Mars in a few weeks.

(1) I don't know where you got the idea that the flight time (in the English version) is days. Professor Calculus's first log book entry (page 32) is dated 3 June, 2345 GMT, i.e. less than 48 hours after they launched and this is clearly many hours after they arrived on the Moon (since by then they have already unloaded the cargo (breaking off for a rest/sleep at one point), walked on the Moon for the first time, and started installing the observatory and assembling the tank).

(2) No justification is given for the assumption that the acceleration is 9.8m/s/s. Yes, it "creates a sort of artifical gravity", but no-one says how much gravity, and in particular, no-one says that it's exactly the same as the Earth's gravity. It could easily be, say, three-quarters as much, which would make the 45 km/s figure perfectly realistic. 91.107.160.101 (talk) 19:15, 18 May 2008 (UTC)Reply

WikiProject Comics B-Class Assesment required

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This article needs the B-Class checklist filled in to remain a B-Class article for the Comics WikiProject. If the checklist is not filled in by 7th August this article will be re-assessed as C-Class. The checklist should be filled out referencing the guidance given at Wikipedia:Version 1.0 Editorial Team/Assessment/B-Class criteria. For further details please contact the Comics WikiProject. Comics-awb (talk) 16:29, 31 July 2008 (UTC)Reply

C-Class rated for Comics Project

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As this B-Class article has yet to receive a review, it has been rated as C-Class. If you disagree and would like to request an assesment, please visit Wikipedia:WikiProject_Comics/Assessment#Requesting_an_assessment and list the article. Hiding T 14:03, 23 February 2009 (UTC)Reply

Inaccuracies

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In you article you state as an inaccuracy, "Finally, the mission is allowed to continue unabated even when it is discovered that extra passengers are aboard, resulting in increased oxygen consumption. In reality, space missions are terminated as soon as possible if it is believed that oxygen supplies may be inadequate." On page 32, 1st panel Prof. Calculus states, "Gentlemen, our plan was to stay on the moon for a whole lunar day - that's the equivalent of fourteen terrestrial days. But our oxygen supplies were intended for four people and one dog, and not for six people, which is our present number. So we shall have to restrict our stay to six days." —Preceding unsigned comment added by Mjwciw (talkcontribs) 11:14, 8 November 2010 (UTC)Reply

Mare Nectaris ou Hipparchus (lunar crater) ?

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The vessel landed on the Mare Nectaris in an english version, it is in the French that the vessel landed in the Hipparchus (lunar crater). — Preceding unsigned comment added by 82.120.140.87 (talk) 16:50, 29 April 2012 (UTC)Reply