Will we ever have the power to fly a starship?
Consider a basic starship design, with a payload of several tens of thousands of tonnes, thus large enough to carry a human community of some hundreds or thousands of people, and propelled by a high-thrust, high-energy rocket engine.
First, we need to estimate what sort of speed we want to cruise at. From both the technical and the social points of view, a speed in the region of a tenth of light speed seems to be both attainable and desirable (note 5).
Assuming that we can find and use sufficiently energetic starship fuel, a rough calculation puts the energy consumption of our starship at about 3.6 x 1023 Joules, or 1.0 x 1014 MWh (note 6).
This is equivalent to about one fifteenth of an Earthpower-year.
By 20th-century human standards, a fifteenth of an Earthpower-year is an outrageous amount of energy. Fuelling one starship would require all other human activity to cease for about one thousand years.
But will this always be so? The Sun generates 1000 times this amount of energy every second. It has done so for billions of years and will continue to do so for billions of years to come.
Suppose that it is possible to store solar energy for subsequent release in a rocket engine to produce thrust -- an obvious suggestion is to store it in the form of antimatter, which is permitted by the laws of physics as at present understood, but is probably about as far in advance of our present-day engineering capacities as building a nuclear-powered rocket would have been for the pioneers of steam (Thomas Newcomen introduced the first viable steam engine at Dudley Castle coal mine in 1712; the Phoebus 2A nuclear rocket, with a power output of 5000 MW, was tested successfully in 1968).
Imagine that a solar power collector could be built with a radius of 638 km, producing solar energy in some storable form such as antimatter. For an advanced space-based civilisation this should be no trouble at all. Let it be placed in orbit near the Sun, at one tenth of the distance of the Earth. It will then collect energy at a rate of one Earthpower, in other words one starship-load per month. This is beginning to look like a profitable industry of the future (say, 300 years hence).
The essential point of this discussion is that an interplanetary-level civilisation will have access to the full power output of its home star. When one appreciates that our present-day civilisation consumes less than one ten-thousandth of an Earthpower and that the total solar power available is more than a billion Earthpower -- and this does not begin to take into account the possibility of clean nuclear fusion using helium-3 mined on the giant planets -- one begins to appreciate the potential for space-based economic growth over the centuries and millennia to come.
One last point. We do not of course know whether a space-based interplanetary civilisation will ever become a reality. Maybe there is some fatal flaw in human group psychology or biology which will forever restrict us to this one planet. Nobody can predict the future. But what we do know is that the only way to find out is to get out there and try to make it work. And we also know that no fatal flaws have emerged so far; quite the opposite, as early predictions that astronauts would become disoriented or sick were resoundingly quashed.
People have lived comfortably in orbiting space-stations for periods of over a year; they have flown successfully to the Moon and back and worked efficiently on its surface. The dangers are understood; the rewards potentially astronomical.
To describe our present-day society -- with all its economic problems, its drug-taking, its famines, its poverty, its wars and its terrorist outrages -- as a society on the threshold of infinite possibilities is to be no more than a hard-headed realist.
Return to contents list.
--------------------------------------------------------------------------------
Useful units of energy and power
1 calorie = 4.185 J
1 erg = 10-7 J
1 kW = 8766 kWh/year (and similarly for W and MW)
1 kWh = 3.6 MJ
The energy content of TNT is about 4 MJ per kilogram (various sources consulted).
(J = Joules, of energy; W = Watts, of power, i.e. energy per unit time; kWh = kilowatt-hours; kg = kilograms, MJ = Megajoules.)
--------------------------------------------------------------------------------
Energy content of a 20-kiloton atomic bomb (the Hiroshima bomb, except that Duncan Steel, Rogue Asteroids and Doomsday Comets (Wiley, 1995), gives that as 13 kilotons) = 80 x 106 MJ.
Energy content of the Tunguska impactor (1908) = between 10 and 20 megatons (Steel, p. 44); 15 megatons = 60 x 109 MJ.
Energy content of most powerful nuclear device exploded by Man = 67 megatons (Steel, p. 59) = 268 x 109 MJ.
Energy content of impact of Fragment G of Comet Shoemaker-Levy 9 on Jupiter (Scientific American, Oct. 1994, p. 8) is thought to have been at least 6 million megatons = 24 x 1015 MJ.
Energy expenditure of starship which accelerates a 30,000 tonne payload onto an interstellar cruise at a speed of 13% that of light and decelerates it at its destination = 360 x 1015 MJ.
Energy content of the Chicxulub dinosaur-killer = at least 100 million megatons, and probably very much greater (Steel, pp. 56-57), = about 100 million megatons (CCNet, 18 Dec. 2000, item 2), = a force of impact equivalent to an earthquake about 10,000 times stronger than the one that leveled San Francisco in 1906 (CCNet, 18 Dec. 2000, item 3); 100 million megatons = 400 x 1015 MJ.
--------------------------------------------------------------------------------
One Sunpower = the total power radiated by the Sun at the present epoch = 3.85 x 1020 MW.
Equivalently: one Sunpower = 3.375 x 1024 MWh/year.
The Solar Irradiance = the solar power passing through unit area at the orbital distance of the Earth = 1.368 kWm-2.
--------------------------------------------------------------------------------
The definition of an Earthpower is not so straightforward. Earth receives energy from two main sources: radiant sunlight and internal radioactivity. About a third of the sunlight falling on Earth is immediately reflected back into space (i.e. Earth's albedo is about 0.33, though varies somewhat depending on the degree of cloud and snow cover); the rest is absorbed into the atmosphere, oceans or land and is eventually re-radiated as heat after driving the climate and Earth's biosphere. Internal radioactive decay heats Earth's interior and drives geological processes such as volcanoes and continental drift, before again eventually escaping to space as waste heat.
I propose the following definitions:
One Earthpower = the amount of sunlight falling on Earth = the Solar Irradiance times the area of the disk of Earth as seen from the Sun = 1.748 x 1011 MW = 1.533 x 1015 MWh/year (as per the calculation given at the top of this page).
The purpose of this unit is to act as a convenient and easily visualised unit of large amounts of power, particularly in the context of the large-scale harvesting of solar power by space-based collectors for industrial use.
One Earthpower-year = one Earthpower sustained for one year = 1.533 x 1015 MWh = 5.52 x 1024 J.
--------------------------------------------------------------------------------
One Absorbed Earthpower = the amount of sunlight actually absorbed into Earth's atmosphere and surface = approximately two-thirds of an Earthpower, or around 1.0 x 1015 MWh/year. (With the Absorbed Earthpower we might make comparisons between the energy input by the Sun and that input by human industrial activities on Earth, currently around 1011 MWh/year, i.e. one ten-thousandth of an absorbed Earthpower.)
Geothermal power is relatively small. Don Anderson, Theory of the Earth (Blackwell, 1989), gives it as about 1013 calories/second. If he means gram-calories, then this is 4 x 107 MW. John S. Lewis, Mining the Sky (Helix Books, Addison-Wesley, 1996), p.224, gives it as 1028 ergs/year, or 1015 MJ/year. Since 1 W = 1 Js-1, 1 W = 31.5576 MJ/year, and his figure for the geothermal heat flow is about 3 x 107 MW, in reasonable agreement.
Return to contents list.
--------------------------------------------------------------------------------
Estimate of energy available in the helium-3 reserves of the giant planets
A mature space industrial civilisation will have access to considerable reserves of fuel suitable for nuclear fusion reactors. The reaction of deuterium with helium-3 is especially suitable for power generation, whether in an electrical power plant or for rocket propulsion, since the main branches of the reaction do not produce free neutrons and thus relatively little unwanted radioactivity is created. The only snag is that, whereas deuterium is widely available, helium-3 is virtually non-existent on Earth.
Robert Parkinson MBE gives an estimate of at least ten thousand trillion tonnes (1016 tonnes) of helium-3 in the atmosphere of Jupiter (Daedalus Report, Journal of the British Interplanetary Society, 1978, p.84). His estimate assumes that 17% of the Jovian atmosphere is helium and that the abundance of helium-3 relative to helium-4 is a cautious one part in 100,000, though he expects the actual abundance to be nearer one part in 10,000.
John S. Lewis points out that helium-3 is just as accessible in the atmospheres of Uranus and Neptune (Mining the Sky, Helix Books, Addison-Wesley, 1996, ch.13). The longer flight times between these outer planets and the inner solar system would be irrelevant once a regular supply line had been set up, while their lower escape velocities would greatly facilitate lifting the helium-3 collected off the planet and returning it to where it was required. The escape velocities of the four giants are:
Jupiter: 60 km/s;
Saturn: 36 km/s;
Uranus and Neptune: both 20 km/s;
Compare with Earth: 11.2 km/s.
This means that the rocket energy necessary to launch a tonne of helium-3 off either Uranus or Neptune is only about one-ninth that required to launch it off Jupiter, and therefore that Uranus (the nearer of the two) is likely to be the first planet to supply helium-3. Nevertheless, the rocket power required is still considerable, and will demand a nuclear engine. Hydrogen for the return journey (the reaction mass for the nuclear rocket) will not be a problem, since it is the commonest constituent of the atmospheric gas and will be a byproduct of the helium-3 extraction process.
How long does it take to reach the giant planets? For regular flights by automatic spacecraft we cannot rely on the gravitational slingshot effect used by the Pioneer and Voyager probes. Jupiter is only aligned correctly with Saturn once every 19.9 years, with Uranus once every 13.8 years, and with Neptune once every 12.8 years. So our space tankers will probably fly the low-energy Hohmann transfer ellipse between Earth and the outer planet of choice. A complete mission would involve a flight out, a period of time spent in the giant planet's atmosphere extracting helium-3 or loading it from an extractor plant already in place, and a return to Earth (assuming that the helium-3 was to be delivered to Earth). The time for the flights out and back is easy to calculate: it is simply the period of an ellipse whose perihelion is 1 astronomical unit and whose aphelion is the distance of an outer planet (all of which move in almost exactly circular orbits). Those periods are as follows:
Jupiter: 5.5 years;
Saturn: 12.1 years;
Uranus: 32.1 years;
Neptune: 61.3 years.
Given that we may already be within a decade or two of building a practical fusion reactor, these lengthy round-trip times, especially to Uranus and Neptune, plus the time required on site to load up with helium-3, suggest that we could well start right now with a pilot project to bring back the first samples.
How much of the stuff is there? John Lewis considers the atmosphere of Uranus down to a depth where the pressure is 12 atmospheres, and estimates the helium-3 resource to be 16 trillion tonnes (16 x 1012 tonnes).
How much energy will this generate in a fusion reactor? The reaction is 2D + 3He –> 4He + 1p; mass fraction converted to energy is 0.0039 (Daedalus, p.47); burnup efficiency may be around 10% (Daedalus, p.57-60). One tonne of helium-3 in combination with 2/3 tonne of deuterium then yields 5/3 (tonnes) x 1000 (kg per tonne) x 0.0039 (fraction converted to energy) x 0.1 (efficiency = fraction of fusion fuel actually consumed) x 9 x 1016 (speed of light squared) = 5.85 x 1016 J, or 1.85 GWyear.
If current annual global industrial energy consumption is 10,000 GWyear, this is equivalent to the use of about 5400 tonnes of helium-3 in fusion reactors (John Lewis, p.211, makes it 450 tonnes per year, but starts from a power figure of 8500 GW and probably assumes that the fusion fuels can be reacted at 100% efficiency, which is unlikely).
Let us assume that John Lewis's figure for the helium-3 resource of Uranus is accurate (i.e. that atmospheric mixing is only efficient down to a depth where the pressure is 12 atmospheres), and that the resources of the other giants are in proportion with their masses. Then the energy reserves in the atmospheric helium-3 of the giant planets, expressed in years at the present-day rate of use, are as follows:
Jupiter: mass 318 Earth masses, estimated helium-3 resource 350 trillion tonnes, equivalent to 65 billion years;
Saturn: mass 95 Earth masses, estimated helium-3 resource 104 trillion tonnes, equivalent to 19 billion years;
Uranus: mass 14.6 Earth masses, estimated helium-3 resource 16 trillion tonnes, equivalent to 3 billion years;
Neptune, mass 17.2 Earth masses, estimated helium-3 resource 19 trillion tonnes, equivalent to 3.5 billion years.
On the subject of mining helium-3 on the Moon, see John Lewis, p.137-41, for reasons why this is unlikely to become practical.
From Space Age: the Energy Page
Basically, the concentration of helium-3 absorbed into the lunar regolith is expected to be about one part in 100 million, whereas its concentration in the atmospheres of the giant planets seems to be about one part in 100,000, or 1000 times greater. Lunar regolith, being solid, is very much harder to work than atmospheric gases. A lunar helium-3 mine would consume large amounts of electrical power, and also require constant maintenance. The energy advantage of extracting helium-3 is little greater than that of devoting the same effort to collecting solar power and using it directly, with the difference that the helium-3 is a non-renewable resource whereas solar power will continue for billions of years to come. The helium-3 sources of choice are therefore, in Lewis's view, the atmospheres of the outer giants, starting with Uranus.
Return to contents list.
