Yourself

No, the gravitational field of an object that size would be exceptionally weak and easily overpowered by electrical forces. Not only that, but the sphere of influence of a sphere that size likely wouldn't be much bigger than the sphere itself, so you couldn't have stable orbits around it anyway, even if you just had it orbiting earth.

Judging from the questions asked in this topic, I don't think he's actually trying to model the temperature of planets as much as he's trying to model temperature as a function of distance from the sun.

How exactly is it that the integers (or natural numbers) are more representative of reality than Z2? For example, the statement 1 + 1 = 0 in Z2 is the same as the statement "the sum of two odd integers is even". In some sense that statement also comes up in binary computer systems, since it resembles the XOR operator.

GPS receivers can do some pretty complicated modelling of the Earth's atmosphere. Specifically there are ionospheric correction terms in the almanac section of the GPS signal. The almanac isn't real time data and the ionospheric correction terms can only be as accurate as our models and measurements of the ionosphere, but this correction is usually better than nothing.

I know $80 isn't cheap, but that's a steal for a textbook. I also recommend this book because it's the one I used in Orbital Mechanics. I also took Orbital Mechanics with the Conway that wrote the book (I also had several other courses with him, he's a very good lecturer) also met Professor Prussing a few times but don't think I had any classes with him. Anyway, I still have my copy so if anyone has any questions about something in it I might be able to answer.

That would be a very basic lift model. In real life an airfoil will stall between about 10-20 degrees angle of attack. 45 degrees is almost certainly well past anything resembling the stall point of even a flat plate. I hope KSP's lift model doesn't follow this, because that would be quite weird.

To be specific, the Lagrange points are not where the gravitational pull from both bodies is equal. At the L1 point (the point between Kerbin and the Mun), Kerbin's gravitational pull is actually slightly stronger. It has to be stronger in order to provide the necessary centripetal force to keep the spacecraft moving in a circle. Lagrange points are simply those points where a low-mass object can be placed such that it will remain in that spot relative to the other two co-orbiting bodies (although some of these points are unstable, so if you're not exactly on top of them you'll gradually move away from them).

The ISS is not the ideal place for a high performance telescope. Largely because the ISS is not easier to control (it\'s very hard to turn around and to make it point somewhere specific, even if you have a movable base on the telescope, you\'re pretty limited in the directions you can look). But then one major issue with the ISS is the vibration. It\'s full of people that are moving around and life support systems circulating air. It would be extremely difficult to get a sharp image out of that. Not to mention that it\'d probably just be cheaper to actually build a standalone space telescope and place it somewhere more conducive to observations.

Apollo didn\'t use a slingshot because there\'s nothing to slingshot around to get you to the moon. Your only real choice is a straight transfer.

There\'s a huge difference in the two directions you can go. First of all, the rotation of Kerbin can save you some fuel for getting into orbit, but that\'s not the biggest issue. The biggest issue with flying at 270 is that you\'re flying the opposite direction that the Mun is orbiting, so when you get out to the Mun you\'ll have a much larger relative speed which means it\'s much harder to slow down (the difference is having a relative velocity of about 368 m/s as opposed to 722 m/s).

Back as an undergrad we were actually show a complete derivation of angular motion starting only from Newton\'s laws (the strong form, assuming that forces between point masses are equal, opposite, and aligned with the vector connecting the points) and an arbitrary set of point masses (more specifically we were walked through an analysis of the system and then we defined terms in the resulting equations to be angular momentum, torques, and moments of inertia). I really wish I could remember this derivation, but I remember it taking quite a bit of time to complete. The concepts of angular momentum and inertia really fall right out of Newton\'s laws

I think this line of reasoning is questionable. What\'s so special about making the magnitude of the drag force and the gravitational forces equal? It doesn\'t follow that this minimizes the amount of time fighting gravity. I\'ve run some tests and this strategy does offer improvement, but I\'ve noticed that I can get an improvement on that (albeit a small one) if I keep the aerodynamic forces about 3% lower than the gravitational forces. I\'m unconvinced that this is optimal. In any case, here are the results: http://imgur.com/a/ZiZNP I think my model is reasonably close to the actual one (which is what the first few images are about, I was just tweaking model parameters to get things close). I haven\'t actually attempted any optimization so far, just testing out models.

From an actual optimal control standpoint this problem is impossible analytically. Numerically is merely difficult. In any case there are some general principles to keep in mind. If not for the atmosphere, the optimal control would be to just full throttle it all the way. Gravity losses are merely a function of time spent in the gravity well. The longer you spend in higher gravity, the more fuel you waste fighting it. The real complicating factor here is the addition of the atmosphere, if there\'s any chance for a singular arc in the solution (a singular arc is one where the control, in this case throttle, is not at one of its limits), it\'s introduced by the atmosphere. The atmospheric and drag models will highly influence the optimal solution. Now, I can give you the math, but it won\'t help much. I can tell you that a singular arc exists when: L2 / m - L3 / Isp = 0 Where L2 and L3 are the Lagrange multiplier costates, m is the total mass of the rocket, and Isp is the specific impulse of the engine. Whoopee. I could give you the differential equations governing the evolution of the costates too, but they\'re too heavily tied up in the exact atmospheric model to be useful. Now, there\'s a few things I can definitely say about the solution: It\'ll start at full throttle and it\'ll end at full throttle. I have a hunch that the optimal control isn\'t bang-bang (meaning you go from full throttle instantly to no throttle), though it could very well be. If the optimal control deviates from full throttle, it\'ll be somewhere in the lower atmosphere. Now, if someone gave me the drag and atmospheric models I could go ahead and let a computer smack the problem around until it finds a solution. Now, I haven\'t searched the forum too much, but I did find this. However, the atmospheric model has been updated since then (not to mention that the given model is for pressure and I\'m really after density), so I\'m not sure how accurate this is and I\'m still left with the question of the drag model.

I know I\'m a bit late to the game on this, but the Prescott or Daytona campus? I\'m actually going to be on the Daytona campus in December between semesters to do an upgrade of ERAUs simulator fleet (at least that\'s the plan right now).

I have a bachelor's in aerospace engineering, but I work as a software engineer for a company that builds flight training devices (high end simulators).