Interstellar Travel Kinematics

2026-04-05

You want to take a rocket to another star system a distance D away.

The Tsiolkovsky rocket equation shows that your delta-v depends on your engine's exhaust velocity V_e and your mass ratio f

$$ \large{} \Delta v = {V_e} \ln f \tag{1} $$

That's the ratio of your fully-loaded "wet" mass M₀ to your "dry" mass when all your fuel is spent, M_f

$$ \large{} f = \frac{{M_0}}{{M_f}} \tag{2} $$

You only benefit in the logarithm of this ratio because every bit of additional fuel must accelerate itself along with the fuel you already had.

Delta-v is your total change in velocity during your mission. For a rendezvous, the optimal trajectory is to accelerate at a constant rate to the midpoint and then decelerate at the same rate until you stop at your destination. Therefore your maximum speed V_max, which you reach at the midpoint, is half your total delta-v.

The time it takes to arrive, T, is given by your average speed V

$$ \large{} T = \frac{D}{V} \tag{3} $$

And this average speed is half your V_max and thus one quarter your delta-v.

Great, ready to calculate T. Just substitute and you've got it

$$ \large{} T = \frac{4 D}{{V_e} \ln f} \tag{4} $$

Well, not quite. You have to be able to accelerate fast enough. Under constant acceleration a, your average speed to the midpoint is always V_max/2, but there's no guarantee it'll be one quarter the delta-v from the rocket equation. There's a second constraint on it

$$ \large{} {V_{max}} = \sqrt{a D} \tag{5} $$

Acceleration, meanwhile, depends on engine power

$$ \large{} a = \frac{2 P}{m {V_e}} \tag{6} $$

Engine power is all kinds of important in terms of cooling etc. You may not be able to make an engine powerful enough to live up to Tsiolkovsky's expectations. How can you tell? Your power:weight ratio at launch has to be

$$ \large{} \frac{{P_{max}}}{{M_0}} \geq \frac{{V_e}^3 \ln^2 f}{8 D} \tag{7} $$

You can see that increasing distance actually helps here, because it gives you more time to accelerate and use your available delta-v. And exhaust velocity hurts, because it increases delta-v and power.

If your engine is powerful enough, the travel time really is given by Eq. 4. If it's not... one option is to blend an inert propellant into your exhaust. Adding inert propellant doesn't change engine power, but it increases thrust (and decreases exhaust velocity) by the square root of the blending ratio.

Calculating the travel time is a little more involved, but it turns out that blending doesn't hurt as much as you'd think.

Previously, our M₀ = 100,000 tonne ship with 300 GW of engine power got to Proxima b in 184 years. To do it 'full Tsiolkovsky' requires

$$ \large{} {P_{max}} = \frac{100,000 \text{ t} \cdot (0.04 \text{ c})^3 \cdot \ln^2(20)}{34 \text{ ly}} = 4.8 \text{ TW} $$

of engine power and arrives in

$$ \large{} T = \frac{17 \text{ ly}}{0.04 \text{ c} \cdot \ln(20)} = 142 \text{ years} $$

Sixteen times the max engine power for 77% of the travel time. And instead of 77,000 tonne of water, all 95,000 tonne of propellant has to be uranium. Maybe not a great deal anyway.

Happy Easter!