Also on Centauri Dreams here: Getting There Quickly: The Nuclear Option

The Solar Gravitational Lens amplifies signals from distant stars and galaxies immensely, thanks to the slight distortion of space-time caused by the Sun’s mass-energy. Basically the Sun becomes an immense spherical lens, amplifying incoming light by focussing it hundreds of Astronomical Units (AU) away. Depending on the light frequency, the Sun’s surrounding plasma in its Corona can cause interference, so the minimum distance varies. For optical frequencies it can be ~600 AU at a minimum and light is usefully focussed out to ~1,000 AU.

One AU is traveled in 1 Julian Year (365.25 days) at a speed of 4.74 km/s. Thus to travel 100 AU in 1 year needs a speed of 474 km/s, which is much faster than the 16.65 km/s that probes have been launched away from the Earth. If a Solar Sail propulsion system could be deployed close to the Sun and have a Lifting Factor (the ratio of Light-Pressure to Weight of Solar Sail vehicle) greater than 1, then such a mission could be launched easily. However, at present, we don’t have super-reflective gossamer light materials that could usefully lift a payload against solar gravity. Carbon nanotube mesh has been studied in such a context, as has aerographite, but both are yet to be created in large enough areas to carry large payloads. The ratio of the push of sunlight, for a perfect reflector, to the gravity of the Sun means an areal mass density of 1.53 grams per square metre gives a Lifting Factor of 1. A Sail with such an LF will hover when pointing face on at the Sun. If a Solar Sail LF is less than 1, then it can be angled and used to speed up or slow down the Sail relative to its initial orbital vector, but the available trajectories are then slow spirals – not fast enough to reach the Gravity Lens in a useful time.

Absent super-light Solar Sails, what are the options? Modern day rockets can’t reach 474 km/s without some radical improvements. Multi-grid Ion Drives can achieve exhaust velocities of the right scale, but no power source yet available can supply the energy required. The reason why leads into the next couple of options so it’s worth exploring. For deep space missions the only working option for high-power is a nuclear fission reactor, since we’re yet to build a working nuclear fusion reactor. When a rocket’s thrust is limited by the power supply’s mass, then there’s a minimum power & minimum travel time trajectory with a specific acceleration/deceleration profile – it accelerates 1/3 the time, then cruises at constant speed 1/3 the time, then brakes 1/3 the time. The minimum Specific Power (Power per kilogram) is:

**P/M = (27/4)*S ^{2}*T^{-3}**

…where P/M is Power/Mass, S is displacement (distance traveled) and T is the total mission time to travel the displacement S. In units of AU and Years, the P/M becomes:

**P/M = 4.8*S ^{2}*T^{-3} W/kg**

However while the Average Speed is 474 km/s for a 6 year mission to 600 AU, the acceleration/deceleration must be accounted for. The Cruise Speed is thus 3/2 times higher, so the total Delta-Vee is 3 times the Average Speed. The optimal mass-ratio for the rocket is about 4.41, so the required Effective Exhaust Velocity is a bit over twice the Average Speed – in this case 958 km/s. As a result the energy efficiency is 0.323, meaning the required Specific Power for a rocket is:

**P/M = 14.9*S ^{2}*T^{-3} W/kg**

For a mission to 600 AU in 6 years a **Specific Power of 24,850 W/kg** is needed. But this is the ideal Jet-Power – the kinetic energy that actually goes into the forward thrust of the vehicle. Assuming the power source is 40% (40% drive and 10% payload) of the vehicle’s empty mass and the efficiency of the higher-powered multi-grid ion-drive is 80%, then the power source must produce **77,600 W/kg of power**. Every power source produces waste heat. For a fission power supply, the waste heat can only be expelled by a radiator. Thermodynamic efficiency is defined as the difference in temperature between the heat-source (reactor) and the heat-sink (radiator), divided by the temperature of the heat source:

**Thermal Efficiency = (T _{source} – T_{sink}) / T_{source}**

For a reactor with a radiator in space, the mass of that radiator is (usually) minimised when the efficiency is 25 % – so to maximise the Power/Mass ratio the reactor has to be really HOT. The heat of the reactor is carried away into a heat exchanger and then travels through the radiator to dump the waste heat to space. To minimise mass and moving parts so called Heat-Pipes can be used, which are conductive channels of certain alloys. Another option, which may prove highly effective given clever reactor designs, is to use high performance thermophotovoltaic (TPV) cells to convert high temperature thermal emissions directly into electrical power. High performance TPV’s have hit 40% efficiency at over 2,000 degrees C, which would also maximise the P/M ratio of the whole power system.

Pure Uranium-235, if perfectly fissioned (a Burn-Up Fraction of 1), releases 88 trillion joules (88 TJ) per kilogram. A jet-power of 24,850 W/kg sustained for 4 years is a total power output of 3.1 TJ/kg. Operating the Solar Lens Telescope payload won’t require such power levels, so we’ll assume it’s negligible fraction of the total output – a much lower power setting. So our fuel needs to be *at least* 3.6% Uranium-235. But there’s multipliers which increase the fraction required – not all the vehicle will be U-235.

**First**, the power-supply mass fraction and the ion-drive efficiency – a multiplier of 1/0.32. Therefore the fuel must be 11.1% U-235.

**Second,** there’s the thermodynamic efficiency. To minimise the radiator area (thus mass) required, it’s set at 25%. Therefore the U-235 is 45.6% of the power system mass. The Specific Power needed for the whole system is thus **310,625 W per kilogram**.

**The final limitation** I haven’t mentioned until now – the thermophysical properties of Uranium itself. Typically Uranium is in the form of Uranium Dioxide, which is 88% uranium by mass. When heated every material goes up in temperature by absorbing (or producing internally) a certain amount of heat – the so called Heat Capacity. The total amount of heat stored in a given amount of material is called the Enthalpy, but what matters to extracting heat from a mass of fissioning Uranium is the difference in Enthalpy between a Higher and a Lower temperature. Considering the whole of the reactor core and the radiator as a single unit, the Lower temperature will be the radiator temperature. The Higher will be the Core where it physically contacts the heat exchanger/radiator. Thanks to the Thermal efficiency relation we know that if the radiator is at 2,000 K, then the Core must be at least ~2,670 K. **The Enthalpy difference is 339 kilojoules per kilogram of Uranium Oxide core.** Extracting that heat difference every second maintains the temperature difference between the Source and the Sink to make Work (useful power) and that means a bare minimum of 91.6% of the specific mass of the whole power system must be very hot fissioning Uranium Dioxide core. Even if the Core is at melting point – about 3120 K – then the Enthalpy difference is 348 KJ/kg – 89.3% of the Power System is Core.

The trend is obvious. The power supply ends up being almost all fissioning Uranium, which is obviously adsurd.

**To conclude:** A fission powered mission to 600 AU will take longer than 6 years. As the Power required is proportional to the inverse cube of the mission time, the total energy required is proportional to the inverse square of the mission time. So a mission time of 12 years means the fraction of U-235 burn-up comes down to a more achievable 22.9% of the power supply’s total mass. A reactor core is more than just fissioning metal oxide. Small reactors have been designed with fuel fractions of 10%, but this is without radiators. A 5% core mass puts the system in range of a 24 year mission time, but that’s approaching near term Solar Sail performance.