- If this is true that today's rockets use half the fuel they used to?
No. One might imagine that 60+ years of development must have produced
large gains, but chemical rocket performance is fundamentally limited by
the amount of energy in the chemical fuels, and the 1960s engines were
already getting at least 2/3 of the maximum theoretically possible
performance (see comparison table below).
Now,
ion thruster technology
has advanced greatly, but those can't be used to reach orbit.
- What metrics would one compare to determine the fuel efficiency and
how to compare them?
The usual primary metric is
specific impulse.
Effective exhaust velocity
But specific impulse is a somewhat unintuitive quantity to understand,
so let's start with
effective exhaust velocity,
which is the average speed of an exhaust particle (in the backward
direction). For example, the
Rocketdyne F-1 engines
used in the first stage of the
Saturn V (the Apollo rocket)
have an effective exhaust velocity of 2.58 km/s at sea level.
What does 2.58 km/s mean in terms of rocket performance? It means if
you build a rocket whose weight is about 63% fuel, and you fire the
engine in deep space until the fuel runs out, the rocket will now be
going 2.58 km/s faster in whatever direction it was pointing:
What if the fuel is not 63% of the weight? Use the
Tsiolkovsky rocket equation:
$$\Delta v = v_e \mathrm{ln} \frac{m_0}{m_f}$$
where $\Delta v$ is how much your velocity changes, $v_e$ is the
effective exhaust velocity, $m_0$ is the initial mass of rocket plus
fuel, and $m_f$ is the final mass of the empty rocket. I started with
63% $= \left( \frac{e-1}{e} \right)$ because then $\frac{m_0}{m_f}$ is $e$, whose
natural log is 1, meaning $\Delta v = v_e$.
Notice that it doesn't matter how long the burn takes, nor the thrust of
the engine, the final change in velocity is the same. That's the magic
of the rocket equation!
So, what is change in velocity, $\Delta v$, good for? In the solar
system there are two main uses for $\Delta v$: launching from the surface
to achieve orbit, and transferring from one orbit to another. The
article
Delta-v budget has some
examples, but the most relevant to Apollo is the $\Delta v$ to get into
low Earth orbit from a sea level launch, which is (very roughly) around
10 km/s. That breaks down as about 8 km/s of required velocity to stay
in orbit (any slower and you'll come back down) and 2 km/s spent lifting
the rocket against gravity and pushing through the air on the way up.
The bottom line is, for any given mission, you need a certain amount of
$\Delta v$. And while you can get more $\Delta v$ by increasing the
proportion of fuel, that gets diminishing returns very quickly due to
the natural log in the rocket equation. On the other hand, any increase
in $v_e$ translates directly to a proportional increase in $\Delta v$,
which means more mission without sacrificing payload.
Comparisons
So let's take a quick comparison of $v_e$ for the F-1 and the
SpaceX Merlin engine.
This is a relatively fair comparison because both burn
RP-1 (refined kerosene) and
liquid oxygen in a
gas-generator cycle.
These characteristics are good for a first stage due to high energy
density per unit volume and high thrust, although other fuels have
better $v_e$.
F-1 2.58 km/s (sea level)
Merlin 2.77 km/s (sea level)
F-1 2.98 km/s (vacuum) 65% of max
Merlin 3.05 km/s (vacuum) 66% of max
Theoretical max 4.61 km/s (vacuum)
The
theoretical maximum
is based on the total chemical energy in the fuel.
I speculate that the better $v_e$ for the Merlin has more to do with its
smaller size, thus making it easier to achieve stable, efficient
combustion, than with technology improvements aimed at performance.
Specific impulse
Finally then, what is specific impulse? It's obtained from $v_e$ by
dividing by the gravitational acceleration on Earth:
$$ I_{sp} = \frac{v_e}{g} $$
where $g$ is usually
standard gravity, or
about $9.81 \frac{m}{s^2}$. The resulting quantity has units of seconds. For
example, for the F-1 at sea level, $I_{sp} = 263 s$.
What is the physical significance of $I_{sp}$? Well, consider our
rocket from before with 63% fuel by mass. Suppose we start the rocket
while it is sitting on the pad, let it just barely lift off, then hover
just off the pad until it runs out of fuel (this assumes we can
arbitrarily throttle the engine without affecting its performance, which
is not realistic, but ignore that). $I_{sp}$ is how long it will hover.
That is because, for every second of hovering, we consume 9.81 m/s of
$\Delta v$ in order to overcome gravitational acceleration accumulated
during that second. After $I_{sp}$ seconds, all of our $\Delta v$ is
gone.
Fuel types
The question mentioned fuel types. This answer is already too long,
but I'll just briefly mention that different
fuel types do have
different performance characteristics, but they also come with other
tradeoffs, and which is best is highly dependent on the mission
objectives. For example, the Saturn V used RP-1/LOX in its first
stage for high thrust and energy density per volume, but
LH2/LOX in its second
and third stages for better energy density per unit mass and $v_e$,
while the Apollo spacecraft (command/service and lunar modules) used
hypergolics
for reliability and storability.
