I am trying to reconcile two different calculations of fuel for a 1g constant acceleration/deceleration trip to Alpha Centauri.
One found here (http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html) indicates that a 1 kg net payload needs about 38 kg of fuel assuming 100% efficiency, so 40 kg ship consisting of a 1 kg motor and 38+ kg fuel could make the trip.
The other (http://nathangeffen.webfactional.com/spacetravel/spacetravel.php) says that a 40 kg ship (payload+fuel) making a 1g 4.3 light-year journey with a fuel conversation rate of 1 (which I assume means 100% efficiency) would need 177 kg of fuel on the trip, which is impossible unless it gets fuel during the trip from somewhere.
The first site indicates that with enough fuel, you can have a 1g constant acc/dec trip to any distance, while the second says you cannot even go 1 light-year at 1g acc/dec without your fuel exceeding the mass of your spacecraft. Are these based on different underlying assumptions somewhere? Or am I mistaken in believing that they apply to the same situation? Or is one just wrong?
Best Answer
The maximum velocity $v$, observer time $t$, and traveler proper time $T$ calculated on the "Space travel calculator" site are identical to those on John Baez's relativistic rocket page, but the energy requirement and fuel mass calculations are botched.
If you check the explanation notes provided, the energy requirement is calculated as $e = 2mc^2(\gamma -1)$, where $\gamma = 1/\sqrt{1-v^2/c^2}$ is calculated for the maximum velocity $v$ at the trip midpoint and the mass $m$ is the total initial mass, including both payload and fuel. This is supposed to account for the kinetic energy acquired/dissipated during both acceleration and deceleration, but fails to take into account that at the same time the fuel mass decreases constantly until completely depleted by the end of the trip. With this erroneous assumption in place, the fuel requirement is then calculated as $M = e/rc^2$, where $r$ is the fuel conversion rate or efficiency. So for unit fuel efficiency $r=1$, the fuel mass is incorrectly estimated as $$ M = 2m(\gamma -1) $$ where $m$ technically still includes the initial fuel load!
In contrast, the Baez page does account for fuel consumption while applying energy and momentum conservation, and finds that the required fuel (M) to payload (m) mass ratio is $$ \frac{M}{m} = e^{\frac{aT}{c}} - 1 $$ where $a$ is the constant acceleration in the traveler's frame and $T$ is the total trip proper time. As an exercise, you may want to check that the above expression can be cast in the much simpler form $$ \frac{M}{m} = \frac{2 \frac{v}{c}}{1 - \frac{v}{c}} $$ where $v$ is again the maximum velocity at trip midpoint (acceleration and deceleration legs being assumed of practically identical durations).