tag:blogger.com,1999:blog-3596550435682943926.post5845762779608549999..comments2024-08-22T06:13:55.957-07:00Comments on Hop's Blog: Travel on Airless Worlds Part IIHop Davidhttp://www.blogger.com/profile/12923433894475072056noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-3596550435682943926.post-86504634205234817992022-05-09T17:21:26.706-07:002022-05-09T17:21:26.706-07:00Peeking at your list, Europa and Ganymede seem the...Peeking at your list, Europa and Ganymede seem the least-uniform in density. They are differentiated ice/rock such that the rule-of-thumb here needs be adjusted. Can we get below Europa's ocean; okay, knowing the pressure over Europa's crust, how far can we get down into THAT; etc.<br />Lately I've been looking into Ganymede as a colony-prospect if the colonials burrow down to avoid Jovian radiation. As of 2015-18, NASA claims to know near-exactly the composition of Ganymede down to 150 km. Namely: fairly pure water ice. Below that it's liquid saltwater, then rock and a metal core (dynamo no less).<br />Density of water-ice in vacuum is 916.8 kg/m3 which we'll round up to 920. Gravity 1.428 m/s² at shallow (< 150 km) depth. 1313.76 N per meter down so 13 atm/km.<br />I'm calculating you can dig right down to Ganymede's ocean at which point you're not digging anymore, you're swimming.<br />Same as under Europa.Darayvushttps://www.blogger.com/profile/17973750966981889517noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-6260484795926256682021-10-18T13:39:58.884-07:002021-10-18T13:39:58.884-07:00Once inside a massive object, ido not think it is...Once inside a massive object, ido not think it is safe to assume cartesian space any more.<br />Any distance calculations need to take in account the space time curvature.<br />The volume of the earth, for example, is significantly greater than would be calculated from its diameter.flowhttps://www.blogger.com/profile/01125760969179071647noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-47238610340010803872021-10-18T03:55:12.926-07:002021-10-18T03:55:12.926-07:00Pressure, assuming you're introducing an atmos...Pressure, assuming you're introducing an atmosphere, and assuming it goes up to the surface! Since we're starting from a "airless" world, we can keep working in a vacuum, so there's no problem from atmospheric pressure, and the only limiting factor could be the temperature of the rocks.pjbhttps://www.blogger.com/profile/09065388232282458321noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-88225755828753215672015-03-30T11:27:09.163-07:002015-03-30T11:27:09.163-07:00Anonymous, pressure is measured in pascals. This u...Anonymous, pressure is measured in pascals. This unit is kg/(m*s^2), not kg/m^2.<br /><br />However you're correct that my expression gave kilograms. When checking Dr. Seligman's page I see here term for surface gravity is gR. The R's supposed to be subscripted.<br /><br />I mistakenly interpreted gR^2 as g*R^2.<br /><br />I've corrected my text. Now I am simply using g for surface gravity. I will check my spreadsheet when I have the time.<br /><br />Thanks for the heads up.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-40229158564418064762015-03-30T09:27:48.576-07:002015-03-30T09:27:48.576-07:00There is a problem with your equation; the units d...There is a problem with your equation; the units don't work out. G has units of N m^2/kg^2. Gravity has units of N/kg. R has units of m. And pressure has units of kg/m^2; if the equation is correct, then the final units will be kg/m^2.<br /><br />But 3/(8 π G) * g * R2 has units of kg:<br />1/(N m^2/kg^2)*(N/kg)*(m^2)<br />= (N/kg)*(m^2)/(N m^2/kg^2)<br />= (N m^2 kg^2)/ (N m^2 kg)<br />= kg<br /><br />This is also seen in your central pressures which are wrong. See<br />http://cips.berkeley.edu/events/rocky-planets-class09/ch6_depater_lissauer.pdf.obscure<br /><br />Which gives the central pressure of a planet as (3 G M^2)/(8 pi R^4)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-53448563530608070032014-06-27T20:50:02.372-07:002014-06-27T20:50:02.372-07:00To eliminate friction something like this would be...To eliminate friction something like this would be useful.<br />http://en.wikipedia.org/wiki/Inductrack<br /><br />It would also make sure that Coriolis effect wouldn't make the subway car hit the side of the tunnel.Jim Baergnoreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-26333889818232068322014-06-26T10:48:04.590-07:002014-06-26T10:48:04.590-07:00Doug, of course friction can't be entirely eli...Doug, of course friction can't be entirely eliminated. However if the tunnel is airless and coriolis is a small fraction of a g, I believe friction loss could be small.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-82613058343129292912014-06-26T10:42:00.868-07:002014-06-26T10:42:00.868-07:00What about friction?What about friction?DougSpacehttps://www.blogger.com/profile/03057371106251356495noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-28969697330878956612014-06-26T08:39:17.885-07:002014-06-26T08:39:17.885-07:00Good observation jesrad! Googling Coriolis force.....Good observation jesrad! Googling Coriolis force... <br />http://en.wikipedia.org/wiki/Coriolis_effect#Formula says coriolis acceleration is -2W X V where W is the angular velocity vector.<br /><br />I'll imagine a tunnel going from an equatorial point to equatorial point. That makes it simpler--the angle between W and V is 90º and sin(90º) is 1. So the coriolis acceleration becomes simple -2WV.<br /><br />The sidereal day of Ceres is .378 days. 2 pi/(.378 days) comes to an angular velocity of .0001924 radians per second.<br /><br />Our subway car would be moving fastest at Ceres' center, .3659 km/s if we're letting gravity do all the work.<br /><br />-2WV would come to -.14 meters/sec^2 if I did my arithmetic right. (It's a good idea to check my arithmetic, I occasionally make mistakes).<br /><br />About 1/70 of a g. That's a pleasant surprise, I was thinking coriolis acceleration would be high in the rapidly spinning asteroids. <br /><br />Jupiter's moons Metis and Adrastea have an orbital period of about .3 days. The other moons are slower with the slowest Callisto having an period of about 17 days.<br /><br />For Saturn the moons range from Mimas (about a day) to Titan which has an approximately 16 day orbit.Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-24114659712992030402014-06-26T05:21:32.694-07:002014-06-26T05:21:32.694-07:00What about the Coriolis effect for all the cases b...What about the Coriolis effect for all the cases besides the transpolar hole ? I guess they would require curved tunnels.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-58971421990435544442014-06-25T14:15:25.127-07:002014-06-25T14:15:25.127-07:00Charles, it's possible I made an error. Checki...Charles, it's possible I made an error. Checking my spreadsheet...<br /><br />I have 476000 meters as Ceres' radius. 9.43e20 kg. I get a surface gravity of .278 meters/sec^2. These figures seem fairly close to yours. Hop Davidhttps://www.blogger.com/profile/12923433894475072056noreply@blogger.comtag:blogger.com,1999:blog-3596550435682943926.post-19054145033792385052014-06-25T13:31:29.847-07:002014-06-25T13:31:29.847-07:00Question; for the asteroid example, did you vary t...Question; for the asteroid example, did you vary the g force from Earth normal? The figure of over 1400 atmospheres seems a bit too high.<br />Ceres surface gravity is .28m/s versus Earth's 9.8 m/s. Ceres mean radius is 476.2 km versus Earth's mean radius of 6371 km.Chuckhttps://www.blogger.com/profile/16168032045826240483noreply@blogger.com