(Semi) Scientific way to explain fast space travel

Question

I am looking for a (if only semi or even the tiniest bit) scientifically explainable way to make a spaceship travel fast enough that interplanetary travel (take the average distance between any 2 of the planets in our solar system for an example if you will) only takes either hours at it's shortest and a couple of days at most.

Is there any way i can handle this that does not end with having to resort to sci-fi like engines, wormholes, black holes or the good old classic of having a Hyperdrive?

Answer 1

Disclaimer: I am completely disregarding time and space dilation in this post because those only start to make a difference for napkin calculations when you really approach light speed, and I am dealing with fractions of up to 2% of it. Some pedant calculations would move the travel times a few fractions of a second up or down at most. I feel that is unnecessary here.

If you accelerate a ship at 1G (~9.8m/s²), you will reach approximately 0.1% of light speed in about eight and a half hours. 1G is cool because it feels natural to us. This is all napkin calculations, i.e.: me playing with rounded values and the browser's console:

The closest approximation between Mars and Earth is about 40 million kilometers. At 0.1% light speed you can clear that distance in 95 hours.

Notice that this means crashing on Mars with enough speed to restart global tectonism on it. You need to decelerate to orbital speed in order to avoid that. A more sane approach, then, would be to accelerate for half of the way, and decelerate for the other half.

If your average speed during the trip is 1% of the speed of light, then you can clear the distance in nine hours and a half. You will have to accelerate much faster than 1G, so this will be extremely uncomfortable for humans - maybe get everyone in stasis during the trip. And by stasis I mean take a page from people who feel sick in airplanes today. Strap the passengers in and feed them dramamine, you can't feel uncomfortable if you are in a coma.

Anyway, the usual formula here is:

$$ average \space velocity = \frac{initial \space velocity + final \space velocity}{2}$$

Consider you want to go from zero to max speed until half the trip, then turn thrusters around and decelerate for the other half, we are actually looking for:

$$1\%\space lightspeed = \frac{max \space speed}{2}$$

You have to reach 2% lightspeed with 20 million kilometers to go (because, remember, you will turn the thrusters around once you get to that speed). The formula is:

$$s = s_0vt + \frac{at^2}{2}$$

Since we start at zero, it becomes:

$$20 \space billion \space meters = \frac{at^2}{2}$$

$$40 \space billion \space meters = at^2$$

For, say, four hours of acceleration (14,400 seconds):

$$a = \frac{4 \times 10^{10}m}{1.44 \times 10^{4} s^2} = \frac{4 \times 10^6m}{1.44 s^2}$$

That is approximately 2.7x10⁶ m/s², or about 275.5G. I am pretty sure that is NOT survivable even for the hardiest materials we know of. So either accept a best flight time of about four to five work days from here to Mars or handwave things away.

Getting from the Earth's surface to outer space and reaching a fraction of the speed of light on rockets alone does not require any handwaving. New Horizons is leaving the solar system at 0.005% the speed of light relative to us - no wormholes, no Alcubierre drive involved. But in order to make it realistic, your spacecraft would have to rely on scifi thrusters and fuel in order to be able to carry passengers. A realistic design, even for near future, would be too large to build, even if it were built in orbit. There is no way around that without some heavy handwaving.

Answer 2

You don't have to exceed light speed

The good news:

1. You don't need to go faster than light, light can go from one end of the solar system to the other in less than a day.
2. If you're going that fast, you can pretty much ignore orbital mechanics (at least until you slow down).

The bad news:

1. Speeding up and slowing down takes a lot of energy.
2. Speeding up and slowing down too fast can hurt people in the ship.
3. Right now, we only know how to travel quickly in space by hurling a reaction mass in the opposite direction.

One Possible Solution

You have some handwavium technology that can reduce the mass of the ship and crew (so small that it is essentially 0) Then relatively little energy is required to move the ship to say 0.5c, and it can be done very quickly. The people on board also have their mass reduced, so they're not affected by any significant g-forces. Then, when the ship is nearer to the destination, the ship slows down, then the mass reduction device is turned off.

Answer 3

A nuclear powered, supercritical steam rocket is probably your best bet.

However, you're dealing with some serious engineering problems here, not the least of which is reaction mass.

These kinds of travel paths are known as brachistochrone paths (for shortest time-distance), and require a lot of acceleration.

The good thing though is this acceleration helps you as you can use it to generate your artificial gravity.

For example, if we want to fly to Mars at Earth gravity, that's 1G acceleration, and per the equations of motion, this would allow you to reach Mars in about 1.3 days when the two planets are at their closest (roughly 55 million km).

This all sounds great, but it's highly unlikely you'd be able to maintain 1G of acceleration for the entire ride. The reason being that to maintain 1G of acceleration on your spacecraft for the entire hour-long trip, you would need a massive amount of fuel. This comes from the thrust-to-weight ratio of your rocket, as well as its specific impulse. The specific impulse is essentially the amount of time you can fire your engines for 1 kg of propellant. So if you determine you need to fire your thrusters for 100,000 seconds (roughly a day), and you can carry 1 million kg of propellant, you'd need an I_sp of 1x10⁶/100000 = 10. 10 is an extremely low I_sp,

You'd have to calculate whether 1 million kg of propellant (1000 metric tons) can actually accelerate your ship at 1G. The problem is, to move your ship, you also have to move that 1 million kg of propellant. This is the tyranny of the rocket equation.

Another problem is that high efficiency rocket engines (i.e. high I_sp) also tend to have low thrust-to-weight ratios, and so low acceleration. This isn't always true, and in a nuclear supercritical steam rocket, it's least likely to be true, but it's still a difficult trade-off.

But there is a silver lining here: the time it takes to travel a certain distance at a certain acceleration is the square root of that distance divided by the acceleration, so if you half the acceleration, you don't necessarily double the travel time.

For example, if we instead accelerate at Moon gravity (1/6th that of Earth), the travel time only jumps to 3 days. This also pushes the I_sp we need down, and thus can lower our fuel needs.

If we run through our equation again, 1x10⁶/260000 = 4, so our I_sp requirements have dropped significantly. If your Isp is higher, you can get away with carrying less fuel, so let's fix our supercritical steam rocket's I_sp at 800. This is very high efficiency, but it's not outside of the theoretical rating for a nuclear supercritical rocket.

Roughly 88% of a rocket's mass is used up getting it to low Earth orbit, however, if we assume our spaceship is already in LEO, our mass requirements drop. So let's say we can get away with 60%. Let's see how our rocket does now.

Per the rocket equation, to get from Earth to Mars in 3 days requires a change in velocity of 55x10⁹ / 2.6x10⁵ = 211 km/s.

Plugging this into the rocket equation for our mass fraction of 60%, we get a delta V of 3.3 km/s (!!). this is a factor of 100 too small, so what do we do?

If our delta-v and our I_sp are fixed, the only thing we can play with is the mass fraction. So what mass fraction allows us to accelerate at 1/6 G for 3 days?

Per the rocket equation, assuming our empty spacecraft weighs about 100 metric tons (roughly the mass of the ISS), we need roughly 10 TRILLION metric tons of propellant. So we're screwed. We have to increase our I^sp, thus making our rocket more efficient, or we have to lighten the rocket, or we have to relax our flight time. At this point, we basically have to give up on using the acceleration to create our artificial gravity.

So what can we do? Well, we can already put a space probe on Mars in about 3 months flight time, so let's assume we have more technology than a basic boring chemical rocket and go from there. What if we want to get to Mars in a month?

Well this gives us a required average acceleration of 1.5x10^-6 m/s² and a delta-V of 418 m/s. Really though, the only thing that matters here is the delta-V, because we can essentially use any acceleration that gets us to the desired delta-V, then coast along for the rest of the ride (the only reason we wanted to maintain a specific acceleration before was for artificial gravity).

Plugging through the rocket equation, this gives us a much more manageable reaction mass requirement of 5480 kg of propellant, or about 5 metric tons.

So what's going on here? Why the huge disparity? Well, the rocket equation has a natural log in it which turns into an exponent when we try to solve backwards.

This means that the shorter you want your travel time to be, the higher your acceleration needs to be, which in turn makes your reaction mass requirements rise exponentially.

Note that this is a gross oversimplification of the math involved. If you actually want to use brachistochrone paths, your velocity isn't constant for the whole flight, which affects your travel time and delta-V calculations.

However, by using the linked equations you can play around with the math and get good rules of thumb for how long travel should take in your universe. Want things to go faster or slower? Play with the I_sp without losing thrust-to-mass ratio. Want to see if you can get a reasonable travel time for a given I_sp? Then fix the impulse and delta-V to see if you can get a reasonable reaction mass requirement.