Thruster May Shorten Mars Trip (from six months to a week!)

Photonics.com News ^ | 9/7/07

[DungeonMaster]

The bad news is it works just like the flux capaciter and requires 1.26 gigawatts to run.

[PAR35]

2. A lot of braking power.

No, it wouldn't. They said that they would get you there in a week. Nothing about you being in any shape to ever come back. Just keep on accelerating until impact.

[JamesP81]

Anybody know how long a human can function under sustained multiple g-loads?

It wouldn't be a heavy g-load. The thruster produces 35 micronewtons. 1 micronewton is 1,000,000th of a newton. The g-load would be very low.

[JamesP81]

So unless we can develop some kind of sophisticated radar that can allow our spacecraft to see these objects millions of miles away and automatically make slight corrections in course to avoid these tiny objects or unless we develop an exterior to the spacecraft that is impervious to collisions at high rates of speed, we are going to continue to be limited in how fast we can go in space.

There are only three solutions to the collision problem: deflector shields, sophisticated armor plating, or highly accurate and reliable weapons (probably lasers or particle beams) that can incinerate the objects. The weapons solution is likely unworkable; we're talking about having a computer system with a radar that can detect a tiny pebble from a million miles out, target, and fire on it before you run over it, correcting for lag imposed by the lightspeed barrier on your radar.

[KevinDavis]

[burzum]

If you wanted to start and stop at rest you would need to have your thrusters on for half the trip in one direction and then reverse direction for the second half.

In the simple case of a constant force (assuming a constant mass):

x(t) = x_0 + v_0t + 0.5 * a t² (x_0 and v_0 are assumed to be zero)
a = 2*x(halfway)/t² = 2*50e9 m/(3.5 days)² = 1.09 m/s²

This is about 1/9th of a g. If they had the capability to have 1 g of acceleration then the transit time would only be 56 hours (1/3 of a week).

[burzum]

Oh, I should also mention that since power equals force times velocity, it probably wouldn’t be feasible to be able to power the thrusters after the beginning of the trip so the initial forces will have to be much higher.

[burzum]

Oh, I should also mention that since power equals force times velocity, it probably wouldn’t be feasible to be able to power the thrusters after the beginning of the trip so the initial forces will have to be much higher.

Err, I mean that the thrusters will have the same power applied, but will lose thrust the faster you go, so your initial thrust has to be much higher than the lower thrust (using the same power) later in the trip (because KE = 1/2 mv² and you have to pay 4 times the energy for every doubling of velocity).

[DuncanWaring]

Ummmmm ... no.

Whatever mass you're kicking out of the back of the vehicle is accelerated relative to the vehicle. No matter how fast the vehicle is going, if the mass-flow-rate and the delta-v is the same, the thrust produced is the same and the energy cost is the same.

[toddlintown]

Once you go warp, you’ll never look back.

[Old Student]

“That was just his desktop prototype. He says it’s scalable.
But I have a lot of doubts about it myself. We see too many of these too-good-to-be-true scams.

But... IF it’s real... OH, BOY!”

Maybe, just MAYBE, I’ll get to visit Mars before I die, after all.

Old Student <————CROSSES FINGERS REAL HARD!

[Old Student]

“If it was built by using “off the shelf parts” then it’s not patentable, IIRC.......”

Which means anyone who could read the blueprints could build one. (Given the cash, of course.)

Power to the People! ;)

I can see a LOT of people taking off on their own. Also a lot of debris fluttering down...

[burzum]

You do not conserve energy in your case. KE = 1/2 mv². The higher the velocity, the higher the cost per incremental change in velocity. Since, P = F * v, your rocket would become much more powerful the faster it goes if it had a constant force. Does it make sense to you that an object that hasn’t changed its mass becomes more powerful the faster it goes when it is only powered by a constant power source?

[burzum]

Oh, and before you mention it, an accelerating body is, by definition, not in an inertial reference frame. So your Galilean transformation arguments do not apply.

[anymouse]

Feel like weighing in? ;)

[r9etb]

35 µN? That ain’t a lot of cookies....................

That depends, actually -- space is funny that way. We used to model continuous thrust forces of 0.05 lb for the Shuttle (which weighs 225,000 lb). That's a tiny acceleration, but very noticeable over the course of an orbit.

If you can keep even this small thruster firing for long periods of time, it adds up to a lot. Of course, it really all boils down to power: how much does it take to get this mouse-fart thrust?

[r9etb]

if photons have no mass, how can they provide acceleration in a vacuum?

They have no rest mass. But they have a small mass by virtue of the fact that they have energy:

E_photon = mc², so m = E_photon/c²

And they have an exhaust velocity = c.

Thrust = mass flow rate * exhaust velocity. So if you have some number N photons per second, then

T = NE_photon/c² * c = NE_photon/c.

I don't rightly know how you'd count photons, but it occurs to me that NE_photon is the total photon energy flux, so you don't really need to count photons, you just need to know the photon energy output by the thrust "beam".

[r9etb]

If my back of the napkin calculations are right, and If he can scale it up by a billion-fold (i.e., to 35,000 Newtons), a million kilogram ship (i.e., one of respectable size), starting at an initial velocity of zero, would travel about one kilometer towards Mars in a week’s time

Uh, no. a = 35e3/1e6 = 35e-3 m/sec².

It would take about 28.6 sec per meter per second -- so a day's thrust would give you around 3 km/sec of delta-V.

[BenLurkin]

Interesting!

[Frank_Discussion]

F=ma

F = 35 µN
m = Mass of your spacecraft

Solve for a!

(Hint, not a lot of G’s, but you don’t need a lot as long as the acceleration is constant for a long period of time...)