Faster than light isn't needed for interplanetary journeys. One gee acceleration is just fine. A micro-fusion device (see inertial confinement fusion)

http://en.wikipedia.org/wiki/Antimatter ... propulsion
http://en.wikipedia.org/wiki/Nuclear_fusion
Lithium deuteride, is a compact powder that when detonated by a tiny micro-fission device releases lots of energy;

21D + 63Li → 2 42He + 22.4 MeV

detonates and releases 645 gigajoules (about the same energy in 105 barrels of crude oil) per GRAM (about the mass of a penny) all without producing neutrons! Done in a magnetic nozzle this system is very interesting

http://mnx.pppl.gov/
Now jet energy is given by kinetic energy per unit mass

E = 1/2 m V^2

So we can rewrite the equation to figure out exhaust speed

V = SQRT(2*E/m) = SQRT(2*6.45e11/.001) = 25,400,000 m/s = 25,400 kilometers per second.

Now thrust is equal to

F = ma = mdot * V ---> F/2.54e7 = mdot

With F in newtons - to convert to kg divide by g0 = 9.802 m/s/s

F(kg)/2.49e8 = mass flow rate (kg/sec)

So, a nuclear fusion rocket burns about 4 micrograms of lithium deuteride per second produces 1 tonne of thrust.

Now constant acceleration to the moon - 360,000 km away achieves 59.4 km/sec - and doubles that to land softly on the lunar surface - while accelerating at 1 gee all along the way! The propellant fraction to achieve this is;

u = 1 - 1/exp(Vf/Ve) = 1 - 1/exp(120/25400) = 0.00471 ~ 0.5%

So, a 1 tonne vehicle carrying 10 kg - about 10 liters - of lithium deuteride is fully capable of flying to the moon at a constant 1 gee from Earth and flying back again at 1 gee - and have some added maneuvering room. Using advanced MEMs technology, this system would have a 1,000 to 1 thrust to weight ratio - so the propulsion system would weigh only 1 kg - and the 1 tonne system would mass only 5% - or 150 kg - and carry 849 kg payload. The trip would take 3 hours and 20 minutes.

Mars at 30 million km distant would boost at 1 gee for 15 million km - 542.3 km/sec - double that to slow down over the remaining 15 million km - and you have 1,084.6 km/sec. The propellant fraction needed to achieve this is;

u = 1 - 1/exp(vf/Ve) = 1 - 1/exp(1,084.6/25,400) = 0.04180 =~4.2%

Again for a 1 metric ton vehicle it must carry 84 kg - or 84 liters of LiD - the structure is still 150 kg - 234 kg in all - leaving 768 kg of payload capacity. This will take 1 day 6 hours and 45 minutes to complee the journey to mars at 1 gee.

This sort of spaceship would be quite capable. I envision a totally automated system. A sort of GOOGLE SOLAR SYSTEM that gives you your position on whatever planet you find yourself on - and you type in where you want to be - navigate on the software to precisely the location where you want to be - and press GO and *bam!* the software calculates;

1) lithium deuteride load to get you there and back

2) available payload

3) supply weight

4) payload weight (after subtracting your weight and the weight of others)

5) duration of voyage,

6) arrival date

7) return date.using duration at site and arrival date.

8) interesting sights along the way

http://www.chewednews.com/Pictures/winnebago.jpg
You can fly anywhere in the solar system out to Neptune and back in three weeks or less at 1 gee - and have the comfort of 1 gee all along the way.