Amount of fuel for accelerating a spaceship

Question

Suppose I have a spaceship and fire up its engines, accelerating to 10000 km / h. Now I turn off my engine and see I have consumed 10 % of my fuel (90 % left).

How much fuel does it take to accelerate the starship to 20000 km / h, assuming I don't hit a rock and can keep my speed at 10000 km / h without expending any fuel whatsoever?

Will I need another 10 % of my starting fuel, or do I require more fuel, i.e. energy use linear or non linear? I am under the impression that the closer I get to lightspeed, the more energy I need...?

Answer 1

Will I need another 1 % of my starting fuel, or do I require more fuel, i.e. energy use linear or non linear?

Ignoring relativistic effects, you require less fuel to go from 10000 km/hour to 20000 km/hour as opposed to going from 0 km/hour to 10000 km/hour. While 20000 km/hour sounds fast, it's actually quite slow. That's not even enough to get you from the ground into orbit about the Earth, which requires a Δv somewhere between 34000 km/hour to 40000 km/hour. That 20000 km/hour is also very slow compared to the speed of light, which is a bit over 1000000000 km/hr. Relativistic effects start appearing at about 1% of the speed of light, and aren't significant until about 10% of the speed of light.

I'll generously assuming your rocket initially was 90% fuel by mass. (Making a rocket whose mass is initially 85% fuel is hard.) I'll also assume that your rocket is in empty space, well removed from any gravitating bodies. In that case, the ideal rocket equation applies: $$\Delta v = v_e \ln\left(\frac{m_i}{m_f}\right)$$ Your rocket attained a Δv of 10000 km/hour after burning 10% of the initial fuel. This means your rocket has an exhaust velocity of 29.45 km/s (106033 km/hr), which puts it in the class of a typical ion propulsion engine.

If the rocket burns all of the fuel while accelerating in a straight line, the final speed would be 244148.9 km/hour. This is more than twice the value naively obtained by multiplying your 10000 km/hour by ten.

Answer 2

This is governed by the Tsiolkovsky rocket equation

$$\Delta V = ln\left( \frac{M_{i}}{M_{f}} \right)v_e$$

where $\Delta V$ is your total change in velocity, $v_e$ is the effective exhaust velocity of your engine, $M_{i}$ is the mass before the burn, and $M_{f}$ is the mass at the end of the burn.

Here's the issue - it takes a minimum mass ratio to reach a particular $\Delta V$ for a given effective exhaust velocity, as given by

$$\frac{M_i}{M_f} = e^\left(\frac{\Delta V}{v_e}\right)$$

Because of the exponential term, the mass ratios get ugly as you aim for higher $\Delta V$. For anything we're flying today, they get stupid huge for reaching speeds beyond tiny fractions of $c$. For that reason, I'm starting with speeds on the order of 0.0001 $c$. This would all break down at relativistic speeds, but we're not going to get anywhere near relativistic speeds with conventional reaction drives.

If you're using an ion engine like the Dawn spacecraft ($v_e$ ~ 30380 m/s), then the mass ratio required to reach 0.0001 $c$ (~30000 m/s) is 2.685; for every kilogram of mass you want to accelerate, you must expend ~1.7 kg of propellant¹. This is the most efficient engine in use today; the SSME $v_e$ tops out at ~4410 m/s in a vaccuum, giving us a mass ratio of a little over 900 (899 kg of propellant for each kg of final mass).

This means we have to work backwards; if we want to do two burns, we have to work out how much propellant we need to reserve for the second burn first, and then factor that into the calculation for the first burn. We're assuming magical propellant tanks with no mass.

For the second burn, we're accelerating the just the dry mass of the spacecraft by 30000 m/s. If our spacecraft masses 1000 kg and our mass ratio is ~2.7, that means we need to reserve 1700 kg of propellant for that burn.

For the first burn, we're accelerating the spacecraft plus the propellant needed for the second burn. Our $\Delta V$ is the same, so our mass ratio is the same, so we need 1.7 * (1000 + 1700) = 4590 additional kilograms of propellant, for a total starting mass of 8290 kg.

I mentioned that the mass ratios get ugly as your $\Delta V$ goes up. If you want to accelerate by 0.001 $c$ (300000 m/s), the mass ratio for the Dawn engine jumps up to ~19437; for every kg of mass you want to accelerate, you have to expend over nineteen thousand kg of propellant.

We're not going to reach anything near light speed using any reaction drive that has to carry its own propellant. You either need to gather propellant mass as you fly (Bussard ramjets), use solar or laser sails, or use a true reactionless drive like the Star Trek warp drive.

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1. The mass ratio is propellant mass plus final mass divided by final mass. A mass ratio of 2 means we have 1 kg of propellant for each kg of final mass, a mass ratio of 3 means we have 2 kg of propellant for each kg of final mass, etc. So we subtract 1 from the mass ratio to figure out how much propellant we use.

Answer 3

The answers using the Tsiolkovsky rocket equation are fine, but they may not clarify the underlying physics. It's simpler if we take the case of $\Delta V\ll v_e$. In that approximation, the mass of the ship is constant. The question would then be whether the fuel used is proportional to the rocket's final momentum (which goes linearly with $\Delta V$), or its final kinetic energy (which goes like $\Delta V^2$). It seems like a paradox, because in a rocket, the fuel is providing both the momentum (reaction mass) and the kinetic energy.

The resolution of the paradox is that in the final state, there is not just kinetic energy in the rocket. There is also kinetic energy in the fuel. So it's erroneous to assume that we have to use up fuel-energy in proportion to $\Delta V^2$. The fuel used is proportional to $\Delta V$.

Another way to see this is that we can pick a frame of reference that's momentarily at rest relative to the ship at some point in time after it's already been accelerating. In that frame, the efficiency of the motor has to be the same as the efficiency when the motor starts from rest, because in that frame, the rocket is momentarily at rest.