162

Aerodynamic heating depends on how dense the atmosphere is and how fast you are moving through it; dense air and high speed mean more heating. When the rocket is launched, it starts from zero velocity in that portion of the atmosphere which is densest and accelerates into progressively less dense air; so during the launch profile the amount of atmospheric ...


116

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^{-\...


100

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 ...


79

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 ...


78

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 ...


78

It is often said that the escape velocity at the event horizon is the speed of light, but while this is true in a sense it is not very useful. The problem is that the speed is an observer dependent quantity. An observer far from the black hole would say the escape velocity at the event horizon was zero, which is obviously nonsensical and proves only that ...


77

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 ...


71

It sounds like you are imagining that what satellites do is go up through the atmosphere, break through 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 ...


68

When an astronaut bumps against the wall of the spacecraft, the spacecraft does gain whatever momentum the astronaut transfers to the wall. However, the astronaut loses momentum-or gains it in the opposite direction. The net result is that the center of mass of astronaut- plus-spacecraft does not move, and the combined momentum does not change. It is ...


63

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 ...


63

Recently I read up on spacecrafts entering earth using a heat shield. However, when exiting the earth atmosphere, it does not heat up, so it does not need a heat shield. Why is this so? A spacecraft on launch does heat up, just not to the degree that it does on reentry. And it heats up for the same reason--atmospheric drag, which includes adiabatic air ...


56

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.


53

How do astronauts, especially those inside small spacecraft like the Crew Dragon, not “push” the spacecraft when they bounce and push off walls? You're right that when an astronaut collides with the walls of the spacecraft, some of their momentum is transferred to the spacecraft and in turn their momentum either reduces or gets reversed in direction. ...


51

Reentry speeds are fast. Astonishingly fast. The shuttle reentered at 7.8km/s. Now note the units. That's "per second." That's 28,158km/hr. And you have roughly 100 vertical kilometers to do that braking in. Yes, the braking gets to be done at a very shallow angle, which means you have more linear distance to break than the 100km would ...


46

The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast? To understand this it is useful to consider a “toy model” ...


43

TL;DR: This answer arrives at roughly the same conclusion as Kyle Kanos' answer, i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding the mass of the rocket itself. This answer, however, is more rigorous in how the $\Delta v$ budget is treated. The rocket equation: Consider the Tsiolkovsky ...


42

Why is it necessary to burn the hydrogen fuel coming out of the engine for the lift of the rockets ? Hydrogen isn't the only fuel possible, so I presume your question is more general, why is any fuel burned? If it is done to create a greater reaction force on the rocket then why can't we do the same lift with just adjusting the speed of the hydrogen gas ...


40

It is not feasible because it would cost an enormous amount of energy to accelerate the spacecraft. To prove this let's calculate with some concrete numbers. Very optimistically estimated, your spacecraft may have a mass of $m=1000\text{ kg}$ (enough for a few people and a small space capsule around them, but neglecting the mass of the fuel needed). And ...


33

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 ...


33

Consider the energy $E_1$ required to remove $1$kg from Earth's gravity. This is given by: $$ E = \frac{GM}{r} $$ where $r$ is the radius of the Earth, and this works out to be about: $$E_1 = 6.3 \times 10^7 \,\text{J} $$ Now consider the energy $E_2$ required to accelerate that $1$kg to $0.99c$. The total energy is give by the relativistic equation for ...


32

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 ...


30

You've noted that at high velocities, a tiny change in velocity can cause a huge change in kinetic energy. And that means that the thrust due to burning fuel seems to be able to contribute an arbitrarily high amount of energy, possibly exceeding the chemical energy of the fuel itself. The resolution is that all of this logic applies to the fuel too! When ...


29

Re-entry velocity from LEO is $~7,800 \frac m s$, from lunar space it is as high as $~11,000 \frac m s$ [1]. Different books give the terminal velocity of a skydiver as about $56 \frac m s$ or $75 \frac m s$ [2, 3]. The exact value isn't material, but the fact that it is two powers of ten smaller then re-entry velocity, is. The difference between skydiving ...


29

So, the energy used to propel the planet must be much more than the energy used to propel the feather. Yes. But that is because the rocket is a special thing. See below. It follows that the feather will traverse a distance of 1m much faster than the planet [...] Work is not about duration. It does not depend on time or on how fast. Pushing with 1 N and ...


29

That's why "escape velocity > speed of light" is not a good way to describe the event horizon of a black hole. It's convenient for understanding, but it's not precise. You simply cannot leave a black hole once you're within the horizon. That's because all possible trajectories point inwards. See Wiki for an even more precise version of what I ...


28

Releasing compressed gas will produce some thrust. But when the gasses are combusted they expand much more. This produces a much higher exhaust velocity which gives a much greater thrust.


27

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 ...


27

I think you are mis-understanding. It dosn't matter which way you launch; it matters which way you orbit. No matter what direction you launch you still have Earth's eastward rotational energy. The reason you launch vertically is you want to get out of the thickest part of the atmosphere as quickly as possible to reduce your atmospheric drag; the fastest way ...


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