tag:blogger.com,1999:blog-987241699123718287.post1123469986838911461..comments2023-10-23T12:57:43.485-07:00Comments on Planet Plots: Export industry for Marsken_anthonyhttp://www.blogger.com/profile/07612961297952294600noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-987241699123718287.post-8519747801565136662015-01-25T15:18:18.697-08:002015-01-25T15:18:18.697-08:00Doug, first, let me try to explain some of the iss...Doug, first, let me try to explain some of the issues. <br /><br />A lunar polar launch, unlike an earthbound polar launch, is largely unaffected by rotational aspects that give low latitude launch sights an edge on earth. However, "largely"does not equal "zero". The earth rotates once a day, giving an equatorial launch site an eastbound speed of 1070 miles per hour. For, say, Kennedy, it's at 28 degrees north, so rotational speed (and thus speed the LV does not have to produce) is reduced to 915mph (assuming a 28 degree eastbound launch inclination). The speed at the north or south pole would of course be zero. <br /><br />So this gets us to a lunar polar launch. The moon rotates once per orbit of the earth (27.3 days), so it's equatorial ground speed is the circumference (6783 miles) divided by the time (655 hours) giving us an equatorial ground speed of 10.35 mph. That's all you lose by launching from the lunar poles, BUT... that's not including what kind of LEO orbit you're looking for. LEO could be low or high inclination (ISS is an example of high inclination). Getting from the moon's equatorial zones to a high inclination LEO requires some additional Delta-V, which is why the Russian Zond sample probes returning from the moon used skip reentry in order to land in Russia (the alternative was a significant delta/V burn at the earth-moon libation point). However, from a lunar polar region, you're basically going into lunar polar orbit first, so a high inclination LEO target would be easier, and a low inclination one harder. <br /><br />To be honest, I don't know how to calculate the various possible burn profiles and orbits exactly, so I can't give a precise number. What I can do is base the figures on a launch from the moon's low latitudes (such as the Apollo landing sites)to a low inclination LEO (approx 2.54 kps, or 1.57 mps). To go from the lunar poles you lose a negligible rotational advantage of 10mph (.0027 MPS). But, if you're aiming at a low inclination LEO, you're going to at need two mid-course burns, and I don't know how to calculate the needed delta/V for that, though it's probably under .3kps<br />The above figures assume airocapture into LEO. <br /><br />What this shows is that if water (and thus hydrogen/O2) is the desired commodity, and it's easily obtainable at the lunar poles, it'd be a lot less delta/V to get it from the moon (ballpark it at 2.8 kps) than from Mars (ballpark it at 4.1 kps)<br /><br />On the other hand, the moon is believed to be carbon-poor, so methane would be problematic, as would food production (and the even lower lunar grav might be an issue there as well). <br /><br />My own personal take on all this; resource acquisition in space is not much different from on Earth; what you get, and from where, varies according to what it is you're after. So, for some things, the Moon is a better case, for others, Mars, and for others, asteroids. <br /><br /><br /> <br /><br /> C Jhttps://www.blogger.com/profile/09526212511734492238noreply@blogger.comtag:blogger.com,1999:blog-987241699123718287.post-41851788911493593072015-01-24T21:37:20.871-08:002015-01-24T21:37:20.871-08:00What is the delta-v for water ice at the lunar pol...What is the delta-v for water ice at the lunar poles to LEO using aerocapture?DougSpacehttps://www.blogger.com/profile/03057371106251356495noreply@blogger.comtag:blogger.com,1999:blog-987241699123718287.post-32536829554222103852015-01-24T14:05:20.395-08:002015-01-24T14:05:20.395-08:00Thanks!
Not brilliant though... I was just pulli...Thanks! <br /><br />Not brilliant though... I was just pulling together a few details I hadn't seen together anywhere else. <br /><br />I'm planning a post on MCT soon, outlining my reasons why I doubt it will be what so many hope; a spacecraft that lands on both Earth and Mars. I'll need to get more heavily into the math for that, but... for a colony, this has major ramifications. If it's entirely space-based (basically a habitat) you'll need a dedicated system for descent/ascent, and fortunately, the reusable upper stage of the BFR will have to have that capability anyway. <br /><br />In fact, I'll do a post on the whole system, including SpaceX's mention of fuel depots being needed. I've got a strong hunch I know what they have in mind for the system architecture, and it's not what the speculation is envisioning. The good news; I think what they have in mind will reduce the per-mission cost by a heck of a lot. <br /><br />Here's a teaser for part of it; I think MCT's design is already on the internet, hidden in plain sight. It's something you're quit familiar with, too. :) <br /> C Jhttps://www.blogger.com/profile/09526212511734492238noreply@blogger.comtag:blogger.com,1999:blog-987241699123718287.post-74914840343651682912015-01-24T10:25:30.694-08:002015-01-24T10:25:30.694-08:00CJ, this is what I need a brilliant guy like you f...CJ, this is what I need a brilliant guy like you for. I play it too safe, assuming no exports (to get on base) and you come along and knock it out of the park. That's two more points for our team. Well done. This info is going to be used on my trip.<br /><br />You and I are going to make it happen while dragging others kicking and whining across the goal. I've got four weeks before I leave to put some pressure on my local attorney to give me some satifying referrals.ken_anthonyhttps://www.blogger.com/profile/07612961297952294600noreply@blogger.com