Why is using a space elevator cheaper than rocket power?

Question

Why is rocket power so much less efficient? In both cases you want to lift a given weight a certain height. What does climbing a tether give you that you don't have when using rockets? Wouldn't a rocket following the same path as the elevator get the centrifugal benefit as well? I don't get it.

Answer 1

With a rocket you have to carry the fuel with you. You are not just propelling the mass of the payload, but also the mass of the fuel. Installing a space elevator is a one-time event that can then be used to propel payloads indefinitely. You no longer have to carry the fuel to get to orbit.

Answer 2

In addition to not requiring fuel:

1. A rocket has to accelerate to orbital speed. This takes a lot of energy. A space elevator can climb at a low, constant vertical speed (albeit for a very long climb), and gets its orbital speed almost for free, from Earth's rotation (see Tom Spilker's answer for far more detail on this).

2. Because a rocket accelerates to high speeds, some of the energy is lost as drag. Another fraction of the energy is lost as gravity losses.

3. An elevator is easily reusable. Fully reusable rockets don't exist yet.

Answer 3

Here's a simple reason:

Most of the rocket's fuel is used just to push the rest of fuel!

It sounds strange for those unfamiliar with Rocket equation. The reality is, if we want to accelerate by exhausting something behind us - then we have a problem when the speed we need to reach (8 km/s orbital speed) is greater than exhaust speed (3-5 km/s). In this case the amount of propellant used grows exponentially. That's why a rocket's payload is no more than 3-5% of total rocket mass, and about 90% is fuel (+oxidiser).

Chemical rockets are rather inefficient when used to fly to space. But they are the only working stuff we have now.

With hypothetical orbital lift we would not need a fuel to push the rest fuel.

Answer 4

It boils down to efficiencies of energy conversion and the cost of the technologies doing the conversions.

If you have a given mass at Earth's surface that you want in geostationary orbit, you have to raise it to the geostationary radius (or altitude, if you prefer to think in those terms), and you have to accelerate it to geostationary orbit velocity. Both of those take energy, a well-determined amount per kilogram of mass you're lofting, ~5.3 X 10$^7$ Joules per kilogram for raising to GEO radius, and ~4.7 X 10$^7$ Joules per kilogram for the orbital kinetic energy. (This is a switch from LEO. At LEO, kinetic energy is much greater than the loft energy. At GEO, loft energy is greater than kinetic energy) That is the fundamental role of both a rocket and an elevator: supply the energy to raise the mass to its desired position in Earth's gravity well, and supply the energy to get it moving with orbital velocity. The rest is "implementation details".

But as they say, "The devil is in the details."

Given the total chemical energy in a rocket's propellants, rocket engines transfer a fraction of that energy to the vehicle, and a very small fraction of that energy to the payload. The precise fraction depends on a number of factors, such as the ultimate ∆V compared to the propellants' exhaust velocity (specific impulse times Earth's gravitational acceleration), ∆V for each stage (if staged) with respect to exhaust velocity, pressure contrast between the nozzle exit and ambient—lots of things. But a typical good performance is around 10% of total available energy transferred to the entire vehicle (payload included), not to just the payload. The fraction transferred to the payload is much smaller still. The energy transferred to the vehicle itself (not the payload) is essentialy wasted. It goes to accelerating all the hardware needed to do the energy conversion (i.e., engines, tanks, pumps, feed lines, avionics, etc.) and, at any given time during the boost, all the remaining propellant. Until recently all that mass became either high-energy junk falling to Earth's surface, high-energy junk in orbit, or high-entropy gases in Earth's atmosphere. Somewhere, somebody paid—a lot!—for all that energy that's now being dissipated as heat, and all that rocket hardware that converted the chemical energy into other forms. Re-usable rocket stages are changing the balance there, but even that has costs associated, such as having to carry extra propellant to perform a landing.

Using an elevator instead, a "car" traveling up the elevator can use electricity to supply the energy to get to geostationary radius, and that energy can come from ground-based or in situ (such as solar) sources.

To get the energy needed for orbital velocity it steals energy from the elevator cable.

That doesn't come free!! More on that later.

Current ground-based electricity sources can convert chemical energy from coal, natural gas, etc. with better than 30% efficiency. None of the fuel, oxidizer (which is almost free: we get it from ambient air) or hardware to do the energy conversion need be lofted. None of the hardware involved, which for a given payload is much lighter than the rocket hardware necessary, becomes high-energy junk! So the loft part of the energy, more than half the total, comes at a much higher efficiency than is available from a rocket.

Many people assume you can get the orbital kinetic energy from the same source. Were that the case, the kinetic energy would be much more efficiently supplied by an electrical system for the elevator than by a rocket.

But the hardware accelerating the payload horizontally (perpendicular to the local vertical) is not the drive motor and drive train on the elevator car. It's the massive elevator cable itself. The higher you go on the cable, the faster it's rotating along with Earth. At the Earth's surface the payload is moving about 450 m/s. At GEO it must be moving ~3100 m/s, as the cable is. As you go up the cable, the cable's local horizontal velocity is proportional to the radius from Earth's center. So as the car moves up the cable, the cable pushes it gently in the orbital velocity direction, gradually adding horizontal kinetic energy to the car. But then the car is pushing on the cable, too, and that has consequences.

If you put a relatively small rocket engine (or some other means of applying a horizontal force) on the car you can make the net horizontal force on the cable zero. This would cancel any of the effects I'm about to discuss. But using a rocket engine you'd need to carry enough propellant for a ∆V of ~2.7 km/s, (non-trivial!) and now you're using rocket propulsion again, for nearly half of the energy needed. For now, assume no such system on the car.

The consequences: the kinetic energy the car is receiving is taken from the cable. The car slows it down—just a little at a time, but over a long time, the time it takes the car to get to GEO. This imparts a non-radial sway to the cable. The sway doesn't behave as if the cable were a rigid rod. Local displacements caused by the vertically moving car propagate as waves to the rest of the cable, vaguely akin to plucking a guitar or piano string. Eventually the cable will wind up slightly inclined westward from its anchor point, and not straight: it will be "wiggling" a little. Some of the cable's kinetic energy and some of the energy of the moving car have been converted from energy of translational motion to energy of vibration.

The non-vertical orientation is not stable. The cable, counterweight and everything connected to it will try to regain that vertical orientation. To do that, everything must accelerate eastward. Accelerating it requires energy. Where will that energy come from?

Earth's rotational energy!

If the cable is tilted slightly to the west, then the tensional force vector on the cable points mostly downward but slightly to the east. If the force on the cable has an eastward component, then the equal and opposite force on the attach point on Earth has a westward component, opposite Earth's rotation. Earth is speeding up the cable, and the cable is slowing Earth's rotation by a miniscule amount.

The cable won't descend (sag) significantly because the counterweight out there somewhere past GEO is putting enough tension on the cable that it will never fully sag, unless some idiot tries to put on a car and payload whose weight is greater than the tension force on the cable. (Aside: but as horizontal movements, i.e. differential movements with respect to the ideal radial-only position, couple into relatively small vertical forces through Coriolis forces, there will be local changes in tensile stress that allow small vertical movements. Also, small displacements from the vertical position will result in small displacements downward in Earth's gravity well; for either westward or eastward horizontal displacements, those downward movements result in small forces in an eastward direction, resulting in small corrective forces on westward displacements, small perturbing forces on eastward displacements)

The cable will be subjected to these horizontal accelerating forces all the way back to its vertical (and wiggling!) position. Which means it won't stop there. Like a simple harmonic oscillator, it continues past that equilibrium point and tilts eastward, eventually stopping and entering the reverse of the westward-tilted recovery process. It's like an upside-down pendulum! In the absence of dissipatory processes (friction, flexural heating, etc.), this oscillation would continue indefinitely. There are dissipatory processes at work in a real elevator system, so eventually the cable would return to a static vertical position.

Yeah. In years, or decades, or even more, depending on the cable material. If you send cars up and down frequently and willy-nilly, not paying attention to timing or ascent/descent profiles, the vibrations and sway excited by these movements could add up to the point they overstress the cable. Needless to say, overstressing the cable is distinctly suboptimal.

How do you stop the sway in a much shorter time span?

You have to apply external forces to the cable!

Those external forces can come at least partly from the elevator car as it goes back down. If the car is carrying the same mass as it was when it went up, there will be a vertical velocity profile that will cancel the sway, and even the vibration. That doesn't mean that this theoretical velocity profile is practical. It might occasionally involve speeds greater than the car technology could handle. It might involve frequent slow-downs, even backing up, and re-accelerations, possibly making the downward trip longer than is desirable. If the optimal profile can't be implemented, then the round trip will leave the cable either swaying, or vibrating, or (most likely) both.

If the descending car's mass is different from that of the upward trip, then no doubt: there will be remnant sway and wiggle from the trip.

Properly timing and profiling another car's upward journey could also damp some of the sway and vibration.

How do you cancel remaining sway and vibration?

Again, you have to apply external forces to the cable.

As the cable approaches the equilibrium (vertical) position, you have to slow down its horizontal speed with respect to that position, so you have to apply a force in the direction opposite its motion. This works for either sway or vibration. But you have to be very careful about the combination of where and when you apply the forces. If you apply "bang-bang" forces (bang-bang control means the control force is either off or 100% on, nothing in between), say at the GEO position, you'll launch waves that travel both up and down the cable from that point, so even though you might be damping lower-frequency vibrations, you're exciting higher-frequency vibrations even more.

You could indeed apply the forces at the GEO point, but not bang-bang. They have to be applied with a profile that damps the sum of whatever motions due to vibration, traveling waves, and sway are happening at a particular time.

You could apply the forces with any system that produces translational forces in a vacuum. You could interact with Earth's magnetic fields or electric fields. That approach will take electric energy. Because you don't get to choose the direction of the magnetic field, your options for the direction of the applied force with a magnetic system are limited. Because translational force from a magnetic field requires a gradient in the field strength, and that gradient is quite small in Earth's magnetosphere, you'd need lots of electric power. Also, during magnetic storms, the magnetic field direction and strength can vary wildly, making it hard to use. There are some similar problems with the local electric field (but not the gradient-required problem), and its direction and magnitude are more variable than the magnetic field's. Either approach would take lots of electric power and somewhere, someone will have to pay for that power.

Or you could use the device most commonly used to apply translational forces in a vacuum: the rocket engine. It would have to be throttleable (no bang-bang!) or be an array of many chambers whose combined duty cycles produce a rough approximation of a continuously-variable thrust-time curve. And because Earth's gravity field is not perfectly cylindrically symmetrical, east-west oscillations would eventually couple into north-south oscillations, so you'd need engines or engine clusters pointed at the four horizontal cardinal points, not just east-west. With this approach, the cars making their rounds would have to carry with them, as part of their payload, propellants for the rocket engines. This eats into the revenue payload capacity of a car, and somewhere, someone will have to pay for the propellants and for replacing engines as they reach their operating lifetimes.

Net result: the orbital kinetic energy of the elevator car and payload does NOT come for free!

One last cost-related aspect: the cost of the technologies for generating electrical energy, distributing it to where it is needed, and converting it into kinetic energy, are significantly less than the cost of rocket technologies. Because mass is such a critical issue for rockets, much money is spent on shaving relatively small amounts of mass from components. That means the manufacture of those components is running with smaller design margins than for Earth-based systems. Running with smaller margins means more precise manufacturing methods (which are generally more expensive than less precise methods), attention to quality control with the attendant increase in inspections, documentation, etc., and the more frequent rejection of a finished part or component. All this makes a Joule of energy supplied by a rocket more expensive than one supplied by an electric motor and the generating station feeding it.

The net result is that indeed, once you have a space elevator in place (and, hoo boy, that's not a trivial task!!), the cost to orbit per kilogram for the elevator system should be considerably less than the cost for rockets. But when you consider the cable dynamics and what you have to do to control it, you find that difference probably isn't quite as large as you first thought.

The "messing with Earth's rotation" aspect makes sense when you view Earth, the elevator, the car, and the payload together as an isolated, spinning system. All that together has a certain amount of angular momentum that won't change unless acted upon by an external force. The magnitude of angular momentum is the product of rotation rate (angular velocity) and moment of inertia. (Actually angular momentum is a vector, the product of an angular velocity vector and an inertia matrix, but we don't need to get into that here!). When you run a mass from Earth's surface out to the GEO station, you increase the moment of inertia of the system by a relatively tiny fraction. Since angular momentum is constant, the angular velocity must decrease by that same tiny fraction. When the car comes back down, assuming it's carrying the same mass it did on the way up, Earth's rotation rate goes back to normal.

Hmm. Thinking about sway and vibration got me thinking about another topic, one I haven't considered before: How would lunar tides affect a space elevator? There just aren't enough hours in a day!

Answer 5

Ultimately an elevator is going to be more efficient, because it doesn't have to deal with gravity losses.

Let me pose a question to you. What does it take for a rocket to hover in place like Blue Origin's New Shepard?

If you've watched any of their launches you know they don't shut off the engine completely, but keep them running the whole time while hovering and even if you haven't watched one of their launches that is the intuitive answer. If you turn off the engine completely you start falling back to earth. This same thing is happening the entire time the rocket is climbing its way up to orbit and we call that energy used up gravity losses.

Because of gravity losses a rocket can't take days or weeks to get to orbit as you'd burn far too much fuel "hovering" the entire time. This is why rockets are big and powerful as they need to get to orbit fairly quickly. In minutes, not hours or days. This means they have to use engines that can meet those characteristics. Powerful and light weight. At the rate rockets consume fuel beamed energy really isn't practical.

We have much more efficient rocket engines that produce a lot more delta-v (change in velocity) for the same amount of fuel and could work on beamed power or actually use solar power, but the amount of thrust they produce is far too low and they often have a thrust-to-weight ratio of less than 1 even just for the engine which means they couldn't even support themselves to hover, let alone lift a large rocket. These types of engines are used by satellites in orbit for station keeping or moving around to different orbits.

A space elevator, if we could build one, opens up many options for removing the weight of the fuel from the payload or even simply take a slow, but more efficient approach. On a tether you can just stop and hang there and not use any power. A simple electrical motor pulling you up the cable and solar panels or beamed power would work. You don't even need the solar or beamed power to provide enough constant power for the climb as you could include some batteries to allow you to charge for a while, climb a bit while draining batteries and then stop to recharge them. Heck even if you had an engine on the crawler with some tanks of gas to power the climber I'd still expect it to be more efficient than a rocket.

Answer 6

As @Evan Steinbrenner points out in his answer, a stopped space elevator needs use no energy to resist gravity. A hovering rocket must burn an enormous amount of energy just to resist gravity. The climber only requires enough energy to move up the cable, essentially slightly over 1g acceleration, and can provide that at it's leisure.

A rocket must expend enough energy not only to counteract earth's 1g pull, but also enough to accelerate to an orbital velocity of at least 7,400 metres per second. 7,400 metres per second is an enormously fast speed, anything less and a rocket is not going to orbit. To get to 7,400 metres per second requires accelerating much of your fuel mass to high velocities as well. The brutal tyranny of the rocket equation means the vast majority of it's fuel is required just to accelerate the fuel needed to accelerate the payload.

Lastly, there are multiple ways to power the climber without requiring it to carry it's own fuel. If the cable can't be charged, beamed power is also a thing. A climber can be powered by microwave radiation from the ground, or solar power. Even if it uses it's own fuel and power plant, it doesn't have to accelerate that fuel and engine to high velocities, freeing it from the "tyranny of the rocket equation".

The current Falcon 9 v5 has a lift-off weight of 549,000KG to put a payload of 22,000 Kg into low earth orbit, fuel weight of roughly 400,000 Kg. The very early 1C Merlin used on the Falcon 1 only generated 40,000 Kg of thrust, burning 140 Kg of fuel per second to do that.

Lets imagine we could build a 100 Km high fueling tower, with a 100 Km long hose hanging from it's top (on an auto-retractor that carries the entire weight of the hose). Now your 22,000 Kg payload requires only a single archaic Merlin 1c to launch with nearly 2G acceleration. It will hit escape velocity in about 7 and a half minutes, and using only 62,000 kg of fuel. Not having to carry your own fuel reduced the amount of energy you needed by 85%!

Now imagine you don't have to accelerate the payload to 7,400 m/sec. How much less energy do you think it will require to lift it to the top of your 100km tower?

Answer 7

To add to the answers above, you can also retrieve energy by sending payloads back down the elevator.

Answer 8

Quite a lot of the answers mention that rockets need to carry the weight of the fuel along with the usual payload, and that requires more energy to lift up. This is correct, but there is also another important thing to consider: rocket engines are thermodynamic engines, and are limited in efficiency by the Second Law of Thermodynamics. Even the most ideal heat engine cannot exceed an efficiency of:

$$\eta = 1 - \frac{T_2}{T_1}$$

where $T_2$ is the temperature of the combustion chamber inside the rocket (high) and $T_1$ is the atmospheric temperature (low). An actual rocket's engine will be far less efficient than this theoretical value due to non-ideal conditions and several other losses.

In contrast, a space elevator would be electromagnetically powered. A technology similar to railguns or linear motors would be used to accelerate objects along the elevator. Here, there are no heat-to-work conversions involved (and hence no need to battle entropy), and such systems are highly efficient.

Another very interesting advantage of electromagnetic systems is: energy can be returned to the source on deceleration. Suppose on approaching the required altitude, a conventional rocket would burn fuel and eject it in the opposite direction (forward) to decelerate the rocket, which further wastes fuel and energy. A space elevator can use regenerative dynamic braking to return the kinetic energy of the load back to the power source.

Answer 9

Another factor that's being overlooked:

Rockets are extremely high energy machines. Many compromises must be made in order to get the energy density needed to make a rocket reach orbit at all. Those compromises generally come at the cost of efficiency. (Off the top of my head--LH2/LOX rockets run their engines quite rich because they actually get more thrust from having a lot of unburned fuel in the exhaust. This is because thrust comes from exhaust velocity, not energy and unburned H2 is a lot lighter than the H2O exhaust and thus moves faster at the same energy level.)

There's also the safety issue. Once again, rockets are very high energy devices. There simply isn't the mass for really good safety systems. Even the best birds sometimes go boom, for manned launches we try to pull the crew away from the boom but that's not always possible. When is the last time you heard of an elevator going boom? About the worst that happens is they get stuck--and that's a headache, not a loss-of-crew incident.

Answer 10

Let's look at the kinetic and gravitational potential energy of a satellite sitting on the launchpad, versus in geostationary orbit. It should be intuitively obvious it has more energy in orbit, so if we can calculate the total change in energy we can establish at absolute minimum bound on the energy required to get to orbit, regardless of what method we use.

Then we can examine how we might supply that energy with a space elevator, versus a rocket.

Minimum Energy Requirement

The total relevant energy of the satellite $E$ is the sum of its kinetic energy $K$ and gravitational potential energy $U$.

$$ E = K + U $$

Kinetic energy should be familiar:

$$ K = {1 \over 2} m_s v_s^2 \tag 1 $$

Gravitational potential energy is negative because it takes energy input to get farther away from Earth.

$$ U = - {\mu m_s \over r} \tag 2 $$

Where:

• $m_s$ is the mass of the satellite
• $v_s$ is the velocity of the satellite
• $\mu$ is the standard gravitational parameter: $3.986 \times 10^{14}\:\mathrm{m^3 \cdot s^{–2}}$ for Earth
• $r$ is the distance from the center of Earth

Sitting on the launchpad at the equator, the satellite has a velocity of 463.2 m/s from the rotation of Earth. At geostationary orbit that velocity is 3070 km/s. So from equation 1, the required change in kinetic energy is:

$$ \Delta E = {1 \over 2} m_s\: 3070^2 - {1 \over 2} m_s\: 463.2^2 = (4.605 \times 10^6)\:m_s $$

Now gravitational potential energy. The radius of Earth is 6,371 km, and geostationary orbit 35,786 km. Using equation 2, the change in gravitational potential is:

$$ \Delta U = - {(3.986 \times 10^{14})\:m_s \over 35786000} + {(3.986 \times 10^{14})\:m_s \over 6371000} = (5.14 \times 10^7)\:m_s $$

So for each kilogram we want to move from the launchpad at the equator to geostationary orbit, we must somehow come up with about 4.6 megajoules in kinetic energy, and 51.4 megajoules in gravitational energy.

Rocket

To get to geostationary orbit we could use a Hohmann transfer.

First the rocket is pointed East at the horizon, and the engines instantly accelerate the rocket. This sends the rocket on an elliptical trajectory with an apogee that intersects the target geostationary orbit.

Pointing East, and not straight up, is most efficient since the rocket is already moving that way due to Earth's rotation. We'll worry about crashing and atmospheric drag some other time.

At apogee the gravitational potential is correct, but the rocket is moving too slow. So a second burn accelerates the rocket again to the target velocity, changing the elliptical orbit to a circular one, and we're done.

I calculated the required change in velocity ($\Delta v$) for the first burn to be 9838 m/s, and the second burn 1503 m/s. That's a total $\Delta v$ of 11,340 m/s.

Now the problem: rockets require reaction mass. To get to the 2nd burn we need to launch not only the satellite, but enough reaction mass (fuel) to perform the second burn. This means the first burn requires a lot more fuel, since it's launching not just the satellite but also all the fuel for the second burn.

This leads to the Tsiolkovsky rocket equation:

$$ \Delta v = v_e \ln{m_0 \over m_f} \tag 3 $$

where:

• $\Delta v$ is the total change of velocity the vehicle can produce
• $m_0$ is the initial mass of the entire vehicle, including payload and fuel
• $m_f$ is the final mass after burning all the fuel (payload + trash, but obviously we try to minimize the trash).
• $v_e$ is the effective exhaust velocity. It's limited by the rocket design.

Of rockets that can produce enough thrust to overcome gravity, current state of the art achieves a $v_e$ on the order of 4400 m/s. So working backwards with our minimum $\Delta v$ of 11,340 m/s we can figure out how many kilograms of fuel are required to launch a kilogram of satellite to geostationary orbit with modern rocket technology:

$$ 11340 = 4400 \ln{m_0 \over m_f} $$

$$ {m_0 \over m_f} = e^{11340 / 4400} = 13.16 $$

Meaning, absolute best case, to launch a 1 kg satellite to geostationary orbit requires 12.16 kg of hydrogen and oxygen. Not bad!

Of course, this is assuming everything that's not payload is fuel. Because of that logarithm in equation 3, it turns out that oversight matters a lot. We have to include the mass of the engines, the tanks, the staging hardware, and so on. Even though all this "trash" isn't payload, we have to accelerate it anyway.

We also did not account for atmospheric drag or gravity drag. In practice it takes about 13600 m/s of $\Delta v$, not the 11340 m/s calculated in the best case. Again the rocket equation makes this matter a lot since fuel requirements go up exponentially.

For a real example, the Falcon 9 can launch a maximum of 8300 kg of payload into geostationary transfer orbit. The entire vehicle is 549,054 kg at liftoff, with 507,500 of that fuel. That's 61 kg of fuel per kg of payload. And this just gets to geostationary transfer orbit -- the payload must use its own propulsion for the final burn, about 1500 m/s of delta-v to circularize the orbit.

Space elevator

The big win with the space elevator is you don't have to take the fuel with you. You just ride the elevator up to geostationary altitude, and let go.

You don't need to bring along fuel because you can send energy electrically up the elevator.

You don't need to go very fast. This means atmospheric drag is not a problem. It also makes engineering the vehicle simpler since it doesn't need to withstand severe aerodynamic forces. It also means you don't need propulsion methods that can produce tremendous thrust, which opens the possibility to more efficient propulsion methods.

Let's consider the efficiencies of various components in the system:

• Electric motor for propulsion: 90%
• Electrical power discribution: 90%. Better with fancy stuff like superconductors.
• Electrical power generation: around 40% for good old fashioned burning hydrocarbons. Hydroelectric plants can reach 95%.

So with modern technology, an overall efficiency of 50% would be easy.

We calculated above it takes 51.4 MJ per kg of payload to geostationary orbit in gravitational potential energy. After inefficiency we're looking at something on the order of 100 MJ per kg of payload.

The kinetic energy can be taken from the cable, rotating Earth, or counterweight. Keeping the elevator stable is complex, and Tom Spilker's answer covers it in some detail. It does still require some energy input, but it's an order of magnitude less than the energy for lifting the payload so we can hand-wave that away as "engineering challenges".

With these numbers, the same 8300 kg payload of the Falcon 9, but launched by space elevator, would require 830 GJ of energy. That's about about a quarter tank of fuel in an Airbus A330.

Answer 11

In addition to other answers.

The rocket fuel is very hot when it leaves the rocket which is an energy loss making the system less effective at pushing.

The rocket compresses the air, it's like jumping in jelly a lot of the energy goes into moving the jelly.

It is safer and requires fewer checks and weather concerns.

It can use cheaper fuel such as coal and nuclear instead of expensive rocket fuels.

Answer 12

There is also a "power" issue. The energy needed to get X to the top is constant but the time needed is flexible. Given the right gearing a small motor will get X to the top, it will just take a long, long time. And as has been said, you don't have to send the motor on the trip. Interestingly you don't even have to send most of the ship. Imagine a vertical ski lift for the math. Stick the astronauts in their suits and ship them off. The chairs falling will balance the chairs being lifted. At the space station, they take off their suits put them in a falling chair seat and re-use them at the bottom. On average, the motor only has to overcome friction.

Answer 13

Other answers mostly concentrate on the energetics of reaching altitude. While true enough, I suggest that these concerns are greatly overrated.

I'd say that everyone is overlooking the enormously-reduced vehicle cost. Orbital rockets are not only cutting-edge, they are bleeding edge. The need to simultaneously show extremely high power output, low weight and high reliablity. Furthermore, they are made in such numbers that each is essentially handmade.

But consider an elevator crawler. Let's do a SWAG (Scientific Wild-Ass Guess) and spec it at 50 tons with a 50 ton payload, with an ascent rate of 10 to 20 mph.

This is not nearly as optimistic as it may seem. Only a relatively small crew space needs to be pressurized: most of the non-perishable cargo can essentially be stacked in CONEXs tacked to the sides of the vehicle. G-forces are extremely low, and presumably vibration will be nowhere near rocket launch levels. Not only is there no need for streamlining, the cargo area doesn't even need to be enclosed. Assuming a start velocity of 10 mph, at the end of 1 hour the altitude will be 50,000 feet, with essentially no atmosphere to produce wind. Accelerating at that point to 20 mph will give a total trip time to ISS orbit of about 13 - 14 hours. A sunshade for the cargo might well be a good idea, but that could be deployed a couple hours after launch with almost no weight.

Power requirements for 100 tons vertically at 10 mph is about 10 MW, or 15k horsepower. This would be distributed into small tractor engines, with a few extra for redundancy. At a modest 3 hp/lb density, motor weight would be about 5,000 lb, or about 1% of the projected vehicle weight. 2% for 20 mph. Wikipedia suggests that electric locomotive engines can run to 1.6 MW, so clearly the technology base is not obviously lacking. Granted, cooling would be a challenge, as it always is in space, as well as challenges raised by operating in vacuum. These are left as an exercise for the reader.

Equally important, such crawlers would be enormously simpler and cheaper than a rocket. No trying-to-tear-itself-apart rocket engine. No finicky cryogenic fuel/oxidizer systems waiting to develop a leak and explode. No attitude control system. No reentry shield. Simple, efficient geometry (due to no streamlining issues) making construction easy. There's no way to do the tradeoffs without a candidate design, but high strength/weight materials are probably unnecessary. I'd suggest it would look something like a very small RORO cargo ship stood on end.

Furthermore, the ground support infrastructure is far cheaper. No fuel storage/transfer system, no radar tracking system. No launch control. There will presumably be other, elevator-unique costs.

Assuming 20 mph speeds, and a day of turnaround at orbit, LEO mission times are about 2 days. There is no obvious reason to consider ground maintenance "refurbishment" as is the case with rockets, more like maintenance on long-haul trucks. Let's say a couple of days. Then you get a mission every 4 days.

As an example of a contrasting system, take the Falcon 9. Payload of 15 tons and a turnaround of about 2 weeks, and the vehicle itself is far more expensive, both to build and to operate.

Answer 14

A space elevator doesn't need kinetic energy to hold its altitude, a rocket must constantly consume fuel just to not start accelerating back towards the earth.

Aerodynamic forces grow as velocity increases, a rocket needs to minimize the time it applies a force against gravity which implies increasing it's vertical velocity, however also increasing velocity reduces efficiency due to aerodynamic drag.

A space elevator is not subject to the losses caused by gravity, and as such can much better minimize aerodynamic losses. It seems to me these are the two main mechanisms by which a space elevator hypothetical can reduce costs.

Answer 15

Since achieving orbit (including the geostationary orbit which we usually think of as the exit point from a space elevator), is all about attaining a target velocity, the following focuses on velocity rather than potential energy change.

Consider a rocket at the first moments of launch. Let's say the vehicle's mass at liftoff is 1,000,000 kg and its engines have roughly comparable performance to the Saturn V F-1, so the exhaust velocity is about 2.5 km/sec (2,500 m/s).

To keep the arithmetic simple, we look at the brief span of time from liftoff until the vehicle attains a vertical velocity of 2.5 m/s, and we'll ignore gravity losses.

To achieve that momentum change, about 1,000 kg of propellant would have to have been expelled (momentum = mass x velocity).

However, energy is one-half mass times the square of the velocity change. So the kinetic energy of that 1,000 kg of exhausted propellant ends up about 1,000 times the kinetic energy acquired by the vehicle, and that's ignoring gravity losses.

Again, keeping the numbers simple, if we say this 2.5 m/s initial velocity change took place in one second, we have an initial launch acceleration of about 0.25 G (Apollo Saturn was something like 0.16), then that 1,000 kg / sec propellant provides only 1/5 of the actual thrust needed (.25 G actual acceleration + 1 G against gravity), so we really need 5,000 kg/s - bringing us to 5,000 times as much energy in propellant as we've put into the vehicle's kinetic energy. In that first second, we've lifted that 1,000,000 kg of rocket about 3 m, so we've added a small amount of potential energy too.

This works out to about 34.4 million joules of combined kinetic and potential energy imparted to the vehicle for about 31.25 billion joules of energy in expelled propellant - still about 1,000 fold difference. Of course, after that first second, the rocket will be 5,000 kg lighter, so it will accelerate a bit faster and therefore acquire more kinetic and potential energy for the same impulse. As the flight progresses, the trend will continue, so the energy efficiency will improve, but it only goes so far.

For a space elevator, the energy expended in the propulsion system could be much more closely matched to the potential and kinetic energy of the vehicle.

The point is that the energy expended in propellant is always greater than the energy acquired by the rocket, and in the early phases of launch it's particularly horrible, despite the fact that rocket engines are remarkably efficient at converting chemical energy into kinetic energy of the exhaust. The inefficiency comes from the relationship between energy and momentum for the vehicle and its exhausted propellant.

High ISP (high exhaust velocity) is generally considered a good thing in rocketry because it means you have to carry less propellant mass to attain a given delta-V, but it comes at a price in terms of the cost of energy needed. Put another way, a very fuel-efficient rocket is actually somewhat energy inefficient, especially when compared with something like an electric motor. That's where space elevators can be more energy efficient.

As for the final question on trajectory... orbit is all about velocity. Given that you have gravity constantly tugging you down to the ground, the most efficient way to get a rocket to orbit is to do it quickly - the less time you spend fighting gravity, the less energy (propellant mass) you need to do it. So, if you want to get to geostationary, you accelerate as quickly as you can (within practical limits) to get into at least a low Earth orbit. At this point, you no longer have to fight gravity, and you can take your time adding velocity until it puts you onto a transfer trajectory (apogee at geostationary altitude). As your trajectory takes you higher, you lose velocity, so that when you arrive at your target altitude, you need to add some more velocity, which circularizes the orbit and you're there. A rocket ascending along a radial line through a fixed position on the ground would be fighting gravity the whole way up, with more or less the sort of efficiency I described for its first second of launch.

Answer 16

Space elevators benefit from the fact that the earth is essentially a rotating "immovable object"*.

To accelerate your craft forwards you must accelerate something else backwards with the same Momentum. Momentum is proportional to velocity but kenetic energy is proportional to velocity squared.

Much of the energy used by a space rocket goes into accelerating the propellant backwards rather than accelerating the rocket forwards. Most of what is left goes to lifting the propellant. Relatively little goes into the final orbital energy (kinetic and potential) of the final stage.

Pushing against an immovable object is far more efficient. Nearly all of the energy goes into lifting the climber and it's payload.

Better still the earth is rotating. So as you climb a vertical tower attatched to the earth as well as gaining height efficiently you also gain horizontal speed essentially for free. Eventually at a height we call "geostationary orbit" you could let go of the tower and stay at the same altitude. Go high enough and letting go of the tower would mean being thrown off into space.

Only problem is how to bulid the tower. We can't simply build a tower from the bottom, the materials that would be needed are simply implausible. A cable held in place by centrifugal force is more feasible but still a massive engeneering challange requiring materials that are on the edge of feasibility.

* That is nothing we puny humans do will have a significant impact on it's velocity or rotation.

Answer 17

Other answers have cited the need to carry its own fuel, but another factor is the need for reaction mass. The fuel for a rocket is used both as fuel and as reaction mass. Something going up a space elevator could use the elevator as reaction mass, making the effective "exhaust" speed much lower and thus needing less energy.

In addition, once we put energy into putting a vehicle in orbit, the only way we can get it down again is by bleeding that energy off. With an elevator, anything we want to bring back to earth can be used as a counterweight to lift a new payload into orbit.