tag:blogger.com,1999:blog-3596550435682943926.post3102043265270845389..comments2024-03-21T17:50:42.377-07:00Comments on Hop's Blog: Lunar Ice Vs NEO iceHop Davidhttp://www.blogger.com/profile/12923433894475072056noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-3596550435682943926.post-50605861311925040102015-07-22T15:18:52.096-07:002015-07-22T15:18:52.096-07:00Anonymous, I've downloaded the Kerbal software...Anonymous, I've downloaded the Kerbal software but haven't got into the game yet.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-33713441988711771572015-07-22T14:23:58.141-07:002015-07-22T14:23:58.141-07:00Have you tried your idea for a low delta-vee lunar...Have you tried your idea for a low delta-vee lunar polar orbit out on Kerbal Space yet?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-19948225062981040012014-09-19T11:47:01.449-07:002014-09-19T11:47:01.449-07:00It turns out the answer is that the moon is not ob...It turns out the answer is that the moon is not oblate enough to allow for a highly inclined Sun-synchronous orbit.<br /><br />Sun-synchronous retrograde orbits from 180º up to 130º are possible. For anything above this inclination, the moon's equatorial mass would not provide enough torque on the orbital plane in order for the plane to track the sun.<br /><br />http://space.stackexchange.com/questions/5370/is-a-lunar-sun-synchronous-orbit-possible-at-the-frozen-inclination-of-86%C2%B0/5377#5377<br /><br />It would still be possible to have a satellite in a low lunar polar orbit provided that the satellite is in the frozen orbit of 86°. Your blog post seems to indicate that you can reach a lunar polar orbit using a polar-intersecting hyperbolic coplanar so that the delta-v requirements are lessened. If I understand correctly reaching polar low lunar from LEO would require 4.2 km/s of delta V whereas reaching a more equatorial low lunar orbit (27º) from LEO would require about 4.1 km/s of delta V. Is this accurate?Nydochttps://www.blogger.com/profile/12270771168908012231noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-85673859538929117052014-09-17T08:14:42.175-07:002014-09-17T08:14:42.175-07:00Peter,
I don't know much about orbital preces...Peter,<br /><br />I don't know much about orbital precession. Sorry I can't help much. You might try asking your question at the <a href="http://space.stackexchange.com" rel="nofollow">Space Stack Exchange</a>Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-3031080027334326322014-09-17T06:35:37.800-07:002014-09-17T06:35:37.800-07:00Hello,
I and another member of the orbiter forums...Hello,<br /><br />I and another member of the orbiter forums <a href="javascript:void(0);" rel="nofollow">Sun-synchronous orbit</a> at in one of the lunar frozen orbits at 86°. Such an orbit would pass near the lunar poles and would allow a continuous supply of solar energy. I believe this orbit might be possible with zero eccentricity at a bit above 200 km altitude. This is provided that the satellite can remain in the frozen orbit and would still precess due to the oblateness of the moon. However I am unsure if there might be some reason that this kind of orbital precession might not work as well since I am not very familiar with frozen orbits. Do you know if this type of orbit would be possible?<br /><br />Here is the article regarding frozen orbits:<br /><br /><a href="http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/" rel="nofollow">http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/</a><br /><br /><br />Kind regards,<br />PeterNydochttps://www.blogger.com/profile/12270771168908012231noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-36093044189462796432013-08-27T18:55:05.209-07:002013-08-27T18:55:05.209-07:00And how much experience has the human race had wit...<i>And how much experience has the human race had with mining in a vacuum at 1/6 g?</i><br /><br />None. Of course mining the moon would also be a trial and error process. And of course establishing a lunar mine would take multiple trips. I never said otherwise.<br /><br />Here's the difference: Lunar luanch windows open every two weeks. Trip time is less than a week. 3 trips to a lunar mine could be done in months or even less.<br /><br />3 trips to comet JohnD would take more than 150 years due to rarity of luanch windows.<br /><br /><i>Your conclusion doesn't follow from your antecedent. </i><br /><br />What? You think the engineers will foresee every eventuality? That they can launch in a single payload equipment that will deal with every unforeseen problem?<br /><br />You suffer from the extreme optimism typical of space cadets.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-62734708653191977342013-08-27T16:55:34.548-07:002013-08-27T16:55:34.548-07:00Also microgravity mining in a vacuum is something ...<i>Also microgravity mining in a vacuum is something the human race has zero experience doing. Acquiring the needed experience will be a trial and error process.</i><br /><br />And how much experience has the human race had with mining in a vacuum at 1/6 g?<br /><br /><i> Thus multiple trips would be needed to establish infra-structure.</i><br /><br />Your conclusion doesn't follow from your antecedent. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-10197334507969637682013-08-13T18:14:37.234-07:002013-08-13T18:14:37.234-07:00If the local vertical is in the plane of the eclip...<i>If the local vertical is in the plane of the ecliptic (which it would be for the launch you describe), </i><br /><br />No, I specified earth's equatorial plane. The ecliptic plane and equatorial plane are completely different things.<br /><br /><i>Let's look at this another way: the great circle described by following the lunar 90º longitude lines roughly outlines the face of the Moon that we see.</i><br /><br />Yes.<br /><br /><i>We can treat that as if it were a flat plate (the equivalent of saying that it forms a plane). If that plate were horizontal wrt to Earth, then we wouldn't be able to see the Moon's face.</i><br /><br />The ceiling above me lies in a horizontal plane. It is perpendicular to the local vertical. There just happens to be a circle on my ceiling thrown by my lamp. When I look straight up, I can see this circle perfectly.<br /><br />I also have a mobile hanging from my ceiling that has circular plates hanging vertically. When I stand beneath these plates all I can see is a line.<br /><br /><i>You treat the perilune as apogee (which it is)</i><br /><br />Which it isn't and I don't treat perilune as apogee. Please review the drawings. The ship enters the moon's sphere of influence at apogee. At this point the ship is 60,000 km from the moon's center. The hyperbola's perilune is 633 km above the moon's surface which is 2371 km from the moon's center. The transfer orbit from this height to LLO has perilune 20 km from the moon's surface. <br /><br />I mention two different perilunes for two different orbits. Both of them are clearly quite different from apogee.<br /><br />I don't like talking to an Anonymous. Please name yourself.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-15101789968544963692013-08-13T16:33:01.999-07:002013-08-13T16:33:01.999-07:00No, the local vertical with regard to earth is a l...<i>No, the local vertical with regard to earth is a line passing through moon's center as well as earth's center.</i><br /><br />If the local vertical is in the plane of the ecliptic (which it would be for the launch you describe), then the lunar 90º longitude line <i>has</i> to be perpendicular to the Earth, just as the Earth's 90º longitude line is perpendicular to the equator.<br /><br />Let's look at this another way: the great circle described by following the lunar 90º longitude lines roughly outlines the face of the Moon that we see. We can treat that as if it were a flat plate (the equivalent of saying that it forms a plane). If that plate were horizontal wrt to Earth, then we wouldn't be able to see the Moon's face. We can only see it and see it as a circle if the plate is perpendicular to the Earth/local vertical.<br /><br />There is a way to get around spending the 1.6 km/s but it takes extra time. You treat the perilune as apogee (which it is) and make your plane change wrt Earth. You then complete an orbit which is now polar and then circularize and land on the <i>next</i> apogee/perilune. This saves you the 1.6 km/s but costs you time and an extra two trips through the Van Allen belts.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-71203696985910590882013-08-12T21:05:51.793-07:002013-08-12T21:05:51.793-07:00No, the local vertical with regard to earth is a l...No, the local vertical with regard to earth is a line passing through moon's center as well as earth's center.<br /><br />The 90 degree west longitude line as well as the 90 degree east longitude line form a great circle. This circle lies in a plane passing through the moon's center and this plane is perpendicular to the the local vertical I describe above. It is horizontal wrt earth.<br /><br />If you park in a LLO (Low Lunar Orbit) with 29º inclination, then it would indeed take 1.6 km/s to do a nearly 60 degree plane change to get to a polar orbit.<br /><br />But why spend an extra 1.6 km/s? As I show, it's possible to enter the moon's sphere of influence in a polar orbit. Then a big LLO plane change isn't needed.<br /><br />I believe parking in a low inclination LLO and then doing a big plane change is the route people are thinking of when they say it takes a lot more delta V to reach the poles.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-63175182618726712942013-08-12T18:20:45.609-07:002013-08-12T18:20:45.609-07:00Your illustrations are great; I wish I could draw ...Your illustrations are great; I wish I could draw as well as you do.<br /><br /><i>The lunar 90º longitude line lies in a horizontal plane with regard to the earth.</i><br /><br />Shouldn't the lunar 90º longitude line be perpendicular to the Earth? The longitude 90º runs along the eastern edge of the Moon from the South pole to the North pole.<br /><br />If we follow <a href="http://www.braeunig.us/space/index.htm" rel="nofollow">Braeunig's much simpler derivation</a>, the amount of delta vee required to change an orbit by an angle θ is twice the orbital velocity times the sine of one-halfθ {dV=2Vi sin(θ/2)}. So to change the plane of an orbit around the Moon from 29° to 90°, we'd need 2*1.6*sin([90-29]/2)=1.6 km/s. And that goes up to 2 km/s if we use a 12° orbit.Anonymousnoreply@blogger.com