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The problem is what Konstantin Tsiolkovsky discovered 100 years ago: as speed increases, the mass required (in fuel) increases exponentially. This relation, specifically, is $$\Delta v=v_e\ln\left(\frac{m_i}{m_f}\right)$$ where $v_e$ is the exhaust velocity, $m_i$ the initial mass and $m_f$ the final mass. The above can be rearranged to get $$m_f=m_ie^{-\... 84 The trouble with orbital mechanics is that it rapidly gets exceedingly complicated and hard to make intuitive sense of. However I think there is a reasonably straightforward way to show how little effect GR has on an Earth-Moon transfer orbit. But this takes a little preparation so bear with me while I give a short introduction. I hope everyone who reads ... 70 The whole point to the throat is to increase the exhaust velocity. But not just increase it a little bit -- a rocket nozzle is designed so that the nozzle chokes. This is another way of saying that the flow accelerates so much that it reaches sonic conditions at the throat. This choking is important. Because it means the flow is sonic at the throat, no ... 70 There seems to be a fundamental misunderstanding as to how movement in space works. In space there is no air friction, that is, once you are moving toward your destination, you don't need a continuous source of power to keep going. Landing on a comet doesn't buy you anything, since in order to land you must first match the comet's orbit, at which point the ... 69 It sounds like you are imagining that what satellites do is go up through the atmosphere, break though into outer space, and hang there. That is not right. If you simply go straight up to outer space (say 300 km above Earth's surface), gravity will pull you right back down, even if you've left the atmosphere, and you'll crash back into the Earth. Gravity is ... 69 Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ... 61 Start by considering what is seen by the people watching you from the Earth. Nothing can travel faster than the speed of light, c, so the quickest you could get to Kepler 186f would be if you were travelling at c in which case it would take 490 years. In practice it would take longer than this because you have to accelerate from rest when you leave the ... 57 For the average disposable lighter, when you press the fuel lever a pressurised liquefied gas is released which will create a very small thrust. The combustion, however, will not generate thrust because, unlike a rocket engine, it is not occurring within a chamber. 48 In space you don't just "go somewhere". You have to match orbits, while not wasting too much fuel. If you're in a low circular orbit, and you want to get to a high circular orbit, it takes two tangential burns, one to elongate your orbit into an ellipse, and another at the high point of the ellipse to make it circular again. This is called a Hohman transfer.... 45 The Jet Propulsion Laboratory has incorporated general relativistic effects in its numerical integration of the planets since the mid to late 1960s. For example, the JPL DE19 ephemeris, released in 1967, incorporated relativistic effects in its modeling of the solar system. This didn't help much. Had they ignored relativistic effects there would have been ... 40 Ok David asked me to bring the rain. Here we go. Indeed it is very feesible and very efficient to use an electromagnetic accelerator to launch something into orbit, but first a look at our alternative: Space Elevator: we don't have the tech Rockets: You spend most of the energy carrying the fuel, and the machinery is complicated, dangerous, and it cannot ... 32 Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating. 31 TL;DR: This answer arrives at roughly the same conclusion as Kyle Kanos', i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding to the mass of the rocket itself. This answer, however, is more rigorous in how the \Delta v budget is treated. Developing a relationship between rocket and ... 31 I highly recommend you download Kerbal Space Program and see for yourself (there's a free demo version)! Typically the goal of a satellite is to orbit, and thus as the other answers address, you must build significant horizontal velocity. Indeed if the Earth didn't have an atmosphere, you could orbit a few km above the surface, so the main goal is building ... 28 Has Musk done his homework? With regard to the basic idea of using nuclear weapons to release CO2 and thereby warm Mars, no, he hasn't. I suspect this was either Bored Elon Musk speaking, or perhaps the Elon Musk who didn't quite deny being a super villain ( 1-900-MHA-HAHA Elon Musk?) in that interview with Colbert. CO2's enthalpy of sublimation is about ... 23 I'll start the ball rolling on this one. My GR knowledge is probably not good enough to make this a truly satisfying answer... The gravitational acceleration for an object moving radially at non-relativistic velocities in the Schwarzschild metric is modified by a factor (1 - r_s/r)(3[1-r_s/r] -2), where r_s = 2GM/c^2 = 0.00885 m for the Earth. If we ... 22 Aircraft rely on lift generated by interacting with the atmosphere and on using atmospheric oxygen to burn with fuel they carry. Orbits aren't stable until you are high enough that there isn't enough atmosphere to interact with, and long before that the oxygen content drops too low to be useful. So, to get to a stable orbit, you will need rockets ... 22 When launching into a low Earth orbit only your velocity relative to the Earth matters, as seen from the not-rotating reference frame of the Earth. Your velocity relative to the sun does not matter, because once you are in the orbit your velocity vector relative to the Earth will oscillate between pointing towards and away from the velocity vector of the ... 21 Newton's third law is pretty near to the mark. All of the phenomena you cite stem from the principle of conservation of momentum in an isolated system, itself ultimately a result (through Noether's theorem) of the fact the physical description of that isolated system is unchanged if we shift the spatial origin of our co-ordinate system. So, if you're in ... 21 A few sanity checks without actually computing anything: First, the error due to neglecting general relativity is so small that it didn't affect prediction of lunar eclipses and wasn't actually noticed anywhere except in Mercury's orbit (at least not until they purpose-built experiments to detect minor discrepancies). I know this doesn't give a completely ... 20 When swinging my comfy hammock, I travel all day even up to 0.99 c, some days even more, depending on what particles are passing me by and measuring my exorbitant speeds with their atomic clocks and photons..! 19 At constant 1 g acceleration half-way through, then constant 1 g deceleration the remaining half, it takes 7 years in rocket time, 38 years in Earth time: http://www.cthreepo.com/lab/math1.shtml Scroll down to Long Relativistic Journeys and enter your data. To the Andromeda Galaxy (2.5 mil ly) it's 29 years in rocket time! :) 17 Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then: radius of Earth = 6.4 \times 10^6~\text{m} mass of Jupiter = 1.9 \times 10^{27}~\text{kg} Escape velocity, v_\text{escape} = \sqrt{\frac{2GM}{r}} This gives a value for v_\text{escape}... 16 It's a great way to get gyroscopic stability. NASA has been using this technique for a long time. For instance, the Pioneer spacecraft used this method. Another example is the Juno spacecraft as well. I hope that answers your question sufficiently. 16 Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ... 14 Nowadays, rockets use a Gimbaled Thrust System. The rocket nozzles are gimbaled (An appliance that allows an object such as a ship's compass, to remain horizontal even as its support tips) so they can vector the thrust to direct the rocket. In a gimbaled thrust system, the exhaust nozzle of the rocket can be swivelled from side to side. As the nozzle is ... 14 Deriving the relativistic equations for constant acceleration would be a formidable problem for most non-physicists. If you want to see how it's done then look at Gravitation by Misner, Thorne and Wheeler, chapter 6. For most of us the best option is just to look at John Baez's excellent article on the relativistic rocket. The relevant equation is:$$ d = \...

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One of my mentors likes to say, "Nothing resembles a new effect quite so much as a mistake." Conservation of momentum is a fundamental principle of mechanics supported by hundreds of years of experimental evidence since the language needed to discuss it was codified by Newton. Certainly it's the case that electromagnetic radiation carries momentum and can ...

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There are several issues with using ramjets on rockets: The startup problem. Ramjets rely on the supersonic velocity of the vehicle to create the compression needed for the combustion chamber to operate. So you'll need something (traditional motor stage, air breathing engine, etc) to get the craft moving to the point the ramjets can actually function. ...

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Here is a visualization: Momentum is mass times velocity, so draw it as the area of a rectangle: If we change the mass and velocity a little, we change the momentum: The total change in the momentum is the sum of green, blue, and purple rectangles. Their sizes are just length times width, so overall we have \$\Delta p = m\Delta v + v\Delta m + \Delta ...

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