I’m laughing about the idea already.
If you change thrust to energy I would agree with you. If you expended X amount of energy putting an object into orbit, you will have to dissipate X amount of energy to return it to earth safely.
You can dissipate this energy several ways. You can get a 1950’s rocket and burn fuel all the way down. Or you can use a small rocket to slow you enough that the upper atmosphere creates drag which further slows you until you get low enough and slow enough that you can either use parachutes or wings to dissipate the final amounts of energy to assure a safe, survivable landing.
The return vectors weren’t linear. It’s not as if they reached orbit, then pointed at the moon, said “go”, and suddenly the orbital energy vanished. After they reached low earth orbit, they added a bit of extra speed to get into an even higher-energy orbit shaped like a figure-8 between the earth and the moon. They returned on a similar path, thus still had a great deal of speed to dump, and no way other than atmospheric braking to dump it.
As far as the simple physics, consider a particle in a vacuum. If it takes X gallons of fuel to accelerate it from 0 to 17,000 MPH, it takes X gallons of fuel to decelerate it from 17,000 MPH to zero. There is nothing magic about outer space that changes the fundamentals.
That is not how rockets work. You also have to accelerate the fuel and reaction mass. In rockets we are used to the reaction mass is also the fuel. So for a round trip from zero to 17,000MPH back to zero more fuel is use on the zero to 17,000 MPH leg than from the 17,000MPH leg back to zero.
This questions changes a little bit if you have access to different (but realistic) technology and don’t care if your passengers remain solid or intact. I’m too tired to even attempt the math here, but the faster you get into orbit the less energy you have to expend while getting there in a purely theoretical sense (correct me if I’m wrong here). Consider a rocket that provides 1.0g of thrust (while at takeoff weight) – it will hover until it gets light enough to fly away and basically keep expending fuel and get there very slowly. Shouldn’t a rocket that is capable of providing 100g of thrust (at takeoff weight) but burns the fuel 100x as fast, be much more efficient at getting into orbit (disregarding structural integrity) or am I missing something?
I did want to point out one thing:
Someone earlier was asking about why there is no burning atmosphere when the shuttle goes up? The answer is that there IS a heating of the atmosphere to the point that plasma can be formed and is visible. A recent shuttle launch video shows it quite clearly. It’s just not as pronounced because the direction of movement and orientation of the shuttle minimize the friction of the atmosphere (for obvious reasons).
Here are a few old threads I have in my subscription log that are at least tangentially related to the topic:
[thread=402535]Why don’t rockets burn up on the way out?[/thread]
[thread=378910]Could the Space Shuttle reach lunar orbit? Mars? Pluto?[/thread]
[thread=399784]Explain Escape Velocity[/thread]
[thread=328586]Could the Space Shuttle go to the Moon?[/thread]
[thread=329349]Overheating on re-entry[/thread]
No point waiting, really. Jurph, David Simmons, robby, and others have already covered essentially all the basics. Reentry speeds are high because they’re essentially orbital speeds, and it is impractical to carry enough fuel to bring a spacecraft to a halt before entering the atmosphere, as this would require not only slowing down but thrusting against gravity to maintain altitude. In orbital ballistics, “you slow down to speed up, and you speed up to slow down.” Using the atmosphere to do the work of slowing your craft down–essentially, transfering the orbital momentum of your craft to the rotational momentum of the Earth–is free…as long as your heat shield doesn’t break apart or burn up.
As others have already noted, Newton’s Laws of Motion apply even more exactly in the vacuum of orbital and interplanetary space, down to several decimal places (when you then have to account for the distortions on spacetime by the presense of mass-energy per General Relativity), so the amount of energy required to accelerate or deccelerate a spacecraft aren’t “fractional”. In fact, it’s really more instructive to look at the kinetic and graviational potential energies of objects in orbit rather than their speeds; an object in the orbit of the Moon will have an enormous energy compared to one in Low Earth Orbit.
And getting to the Moon, once you’ve gotten to orbit is, as has been previously noted, comparitively simple; you essentially adjust your orbit via a long burn while coming around from the opposite side of the planet so that you’re in a really eccentric ellipse which just happens to intersect with the Moon’s sphere of influence, and then it does the rest of the work at bringing you in. It’s true that you do need to thrust in order to slow down and remained “captured” (in orbit) of the Moon, and usually another to circularize the orbit or get in position for deploying the LEM, but because the velocity with respect to the Moon is much lower (and the mass of the LEM is much smaller than the Apollo CSM) it doesn’t require all that much propellant to inject into lunar orbit and land on the Moon without aerobraking. Even then, the mass of the LEM is mostly propellant, and even at that they leave the landing platform and engines on the Moon while the LEM upper stage ascends back to rendevous with CSM. All of this complexity (as opposed to a single, unitary vehicle that would perform a direct ascent. As the Wikipedia article notes, this approach was abandoned both by the US and the USSR as being physically impossible with chemical propellants, which are limited to a (weight) specific impulse of few hundred seconds.
Since Jurph brought up the Tsiolkovsky rocket equation (and I have nothing further to add in salient content), I’ll make note of the work of Konstantin Tsiolkovsky, who is the first person to have conceived and developed of using multistage rockets for orbital injection (though Hermann Oberth and Robert Goddard came up with the same concept independent of him and each other). Another Russian, the beleagured Yuri Vasilievich Kondratyuk, was the first to conceive of using modular spacecraft and orbitial rendevous, disposable staged landing craft (rather than the direct ascent approach mentioend above) and most notably the “Kondratyuk’s loop”, the low energy freefall orbit that allowed the Apollo craft to get from Earth orbit to Luna with its pitiful engines (and what the Apollo XIII mission used to abort and return when the #2 oxygen tank exploded on their outbound leg. This was a precursor to the planetary swingby used by many NASA/JPL interplanetary missions, the theory for which Kondratyuk also developed.
I can’t think of any more useful trivia to jam in so I’ll just cut it off here.
Stranger
The Space Shuttle orbiter has a landing mass of about 100,000 kg (that’s the spaceship itself, plus payload, astronauts, life support, etc.). The fully-fueled vehicle, including the external tank and SRBs, is about 2 million kg. That means that the total launch mass is about 20 times the mass of the thing we want to get into orbit and get back. If we didn’t use atmospheric braking, we would need 2 million kg total in orbit, to get that 100,000 kg back. To get that 2 million kg into orbit, we would need a total of 20 times that on the launch pad.
Actually, it’s a fair bit more than what’s necessary to escape Earth’s grasp. Escape speed at any given height is equal to sqrt(2) times the circular orbit speed at the same height. So if you’re already in a circular orbit, you need to change your speed by .414 times your current speed, to escape. But you need to change your speed by 1 times your current speed to stop in your orbit and come straight down. So it would take over twice as much change of speed to stop in place as to escape.
True enough. I was oversimplifying a bit by considering things only from the point of view of the payload (like a Mercury capsule, for example). As far as the payload is concerned, it takes just about as much fuel and thrust to slow it down as took to speed it up.
Of course, if you were actually going to use a rocket for the deceleration phase, you would also have to get all of that fuel to orbit, which takes a tremendous amount of fuel.
Chronos has summed it up pretty clearly in his last post.
That’s sometimes called “gravity loss” since the principle is most clearly visible in the case of a rocket wasting it’s fuel by hovering against gravity; however it also holds in free fall. The most efficient possible use of reaction mass would be if you could instantly expel enough mass to reach the velocity change you needed. Even if you could do this however there is still a fundamental limit to how efficient a rocket can be based on the velocity of the reaction mass (which is measured in the technical term called specific impulse). And in practice doubling the thrust of a rocket engine means for a given technology doubling the weight of the engine. Not to mention making the rocket strong enough to hold up under high acceleration, aerodynamic stress for a rocket lifting through atmosphere, etc.