Tuesday, June 21, 2011

Interplanetary Flight Times.

Too often when we talk about future interplanetary flight we assume the trajectories used will be Hohman transfer orbits, elliptical orbits that intersect of both the orbit of departure and the destination tangentially, in the case of Mars this involves flight times of around 250 days. But even with todays chemical rockets the mass ratios for flight times of 150 days may be optimal for cost, and with thermal systems using hydrogen as the propellant flight times of just 70 days to Mars may not only be reasonable, but sensible, with hydrogen propellant and exhaust velocities of 10km/s a delta v of 6.5km/s is achievable with a mass ratio of only 2. All it requires is a thermal energy source to heat the propellant, aerobraking at Mars, and a propellant source on Mars for the return.


  1. Is there anything other than nuclear thermal that could work for this?

  2. I wrote this in irritation to the popularity of various nuclear propulsion schemes that have been mooted, my intent was to argue that solar thermal propulsion was a far more sensible option, I ended up doing so on this thread:

    I think the point Proponent makes in #8 reply about the advantages of traditional chemical rockets are interesting, though there's the issue of getting the far greater mass into space if you want the same delta V as can be achieved with STP.

  3. Hello Andrew,

    How about putting a fuel tank in one of these quick orbits and making a rendez-vous in space for a fuel transfer, similar to the Aerial propellant transfer proposition? Would this allow for slowing at Mars without requiring aerobraking and reduce the overall project cost and time? Could an even faster orbit be used, with a stopping system more powerful than aerobreaking?
    The fuel tank might be boosted to its orbit using a high ISP solar propulsion system and very little fuel (relatively speaking).

    Do you know if these quick orbits can swing back towards Earth in case of failure, like the moon missions?

    On the question of orbits, I’ve realized that a high G moon surface rotovator probably requires either secure orbital capture or a very robust orbital injection system, since the orbit of a ‘package’ will return directly to the rotovator area in case of failure to capture, possibly hitting the moon’s surface. I naively thought you could launch a ‘package’ into a nice circular orbit, but that’s not how it works!


    Michel Lamontagne

  4. Hi Michel, the arrival velocity at Mars is going to be even higher than the peak velocity relative to Earth, so it may be that the determining factor for flight times will be just how much velocity can be safely shed using aerobraking at Mars. Carrying extra fuel for powered deceleration is an option, but doing so quickly pushes up the total mass for little gain. I'm a fan of aerobraking because it gives a pretty impressive return on mass, for example the shuttle uses it to shed about 8km/s by employing a heat shield that only makes up about 10% of its weight, and the shield isn't even used up! Rocket braking would take about 5 times the shuttles orbital weight in fuel.

    These quick trajectories will take you at least to the outer solar system without a check on velocity. As Earth's orbital velocity around the sun is 29.8km/s, and solar escape velocity from 1AU from the sun is 42.1km/s, and we're more or less adding whatever velocity we're giving to our ship onto Earths orbital velocity (29.8+12=41.8km/s).

    I naively thought you could launch a ‘package’ into a nice circular orbit, but that’s not how it works!

    One of the first ScFi stories (H. G. Well?)about space flight was about a cannon used to launch a projectile into space that went once around Earth and came back and destroyed the cannon to the demise of the gun crew. Though it wouldn't quite work like that as, with the Earths rotation, the cannon would have moved out of the way while the projectile was in flight, mind you, the Moon rotates far more slowly...

  5. Martin J Sallberg
    Useful Casimir effect for cheap spacelaunches.
    The Casimir effect is traditionally demonstrated by placing two thin parallel plates mere micrometers apart in a vacuum and letting them slam together. The effect is due to vacuum energy. It can in principle be used to modify the vacuum for cheap spacelaunches and efficient space travel, but that requires preventing the plates from slamming together, so that the Casimir effect remains. That can be done by repulsive magnetic fields or by mechanically holding the plates in the edges (only in the edges, to keep the space between them). Another possibility is to abandon the parallel plates altogether and use microchannels or other microscopic holes instead. Anyone is free to build it, I am not going to claim any patent or money.

  6. Dear mr Sallberg,

    I suggest the following experiment: Get a glass of water and carefully put a bit of water on a flat surface. The water will gather into a mostly circular shape and rise up to make a bubble, beating gravity, under the influence of Van der Walls forces, which are similar to Casimir forces. Can a space drive be made out of this?

    No, because even if you do this a million times, the drop will never move across the surface, since the forces are balanced all around the drop. In the same way, a force between two plates (perpendicular) is not a force across two plates (parallel), and will not produce movement. There is no reaction to this force.

    The trick in the Casimir drive is to postulate that the Casimir effect between two plates, which can be positive or negative, can be exchanged across two sets of plates, giving a net flow of energy and therefore a net force. The problem is that, as far as I know, this exchange trick has never been done, and may be impossible.

    The second way to use the Casamir effect as a drive would be to extract the vaccum energy and then use it to power a conventional drive. However, this is basically what a nuclear reactor does, although from a more conventional source. There is no guarantee than an eventual Casimir effect energy source would be lighter than a nuclear reactor. Remember that the cost of a nuclear reactor is not really the fuel, which is plentiful and rather cheap, but the heat dissipation system and the heat to energy conversion system. What would be the energy losses from a Casimir energy source? If the system is inefficient, it might actually make a worse drive that a nuclear of even a chemical one.


    Michel Lamontagne

  7. One problem with solar thermal, which mitigates its utility, is the low thrust/mass ratio. Low thrust trajectories typically require larger delta-vees than high-thrust, due to gravity losses.

  8. Hmmm, if you're gonna use chemical you might need a way to connect a bunch of propellant modules together. Which is the theme of my latest project. I put it on my blog if you're interested.