2
$\begingroup$

For a given distance of travel, your fuel consumption can be calculated by running hours instead of distance. This is common when you're traveling by boat for instance. You don't typically say "I have enough fuel to travel 10 miles" instead you would say "I have enough fuel to run for 1 hour".

So here's my question. In a hypothetical light-speed spaceship, what would your consumption look like? For instance if you, as the passenger were riding for one year at light speed, and your rocket consumed 10,000 speed crystals (sc) in order to achieve this your fuel efficiency would be 10,000 speed crystals/year or about 1.14sc/hour.

What would the outside observers be able to say about your efficiency? Assuming the earth bound observer ages 7.2 years, would they observe an efficiency of 10Ksc/7.2 years or about 0.15sc/hour?

Whether you observe from inside or outside the spaceship, the consumption would be 10,000sc/light-year. But which "efficiency rating" is correct?

$\endgroup$
3
  • $\begingroup$ I am quite certain that the fuel consumption of ships is given as so much fuel per hour at so many knots; and that the maximum distance that a boat can travel with a given amount of fuel depends very much on the speed at which it travels. (And the fuel consumption of the fictional ship would be given in speed crystal kilograms per hour. One tiny crystal and one big crystal are not the same.) $\endgroup$
    – AlexP
    Dec 25, 2021 at 0:04
  • 1
    $\begingroup$ Asking about relativistic effects for a question with a "faster-than-light" tag is unanswerable, because the physics model that includes relativity specifically rules out the possibility of faster than light travel or even travelling at lightspeed itself. So the only answer that can be given is "whatever your techno-magic allows". If you were asking about a ship going at .99 c then a meaningful answer could be given. $\endgroup$ Dec 26, 2021 at 15:52
  • $\begingroup$ I suppose either measurement could be useful. The crew of the ship would probably work in terms of how they experience the passage of time, but they might have sometimes calculated it from the reference of their home planet when planning the trip. $\endgroup$ Aug 14 at 4:54

3 Answers 3

8
$\begingroup$

Space is not a lake.

The thing about travel on Earth is you have to keep expending energy to overcome drag - drag from air resistance, water resistance and gravity. You use fuel constantly to keep moving. Stuff on Earth that stops expending energy comes to a halt.

In space there is no drag. Once you are up to speed you stay at speed. If I accelerate to 0.9c I will stay at 0.9c indefinitely unless I expend energy to decelerate or hit something. Dividing your fuel consumption over the duration of travel does not make sense if you only expend energy at the beginning and end of your trip.

Probably energy expenditure for space travel will be in terms of m (your mass) and v (the velocity at which you intend to travel), and will budget fuel for initial acceleration of your mass and subsequent deceleration of your mass minus mass of fuel you expended earlier.

$\endgroup$
6
  • 2
    $\begingroup$ For lower speeds the "no drag" rule is true, but once a ship reaches a significant fraction of light speed it will experience drag from relativistic collisions with interstellar dust, especially in denser regions of the galaxy. Even at 1/3c a ship would need serious shielding and point defense to avoid significant damage from dust, so at 0.9c you would definitely have to consume fuel to maintain speed and power active shielding. $\endgroup$ Dec 24, 2021 at 22:41
  • 1
    $\begingroup$ "unless I expend energy to decelerate" You may not even need it, if you don't mind landing on destination at 0.9c. $\endgroup$ Dec 26, 2021 at 2:18
  • $\begingroup$ so there will probably be a running fuel cost to operate the ship and an separate calculable/advertised fuel cost for departure and arrival. $\endgroup$ Dec 27, 2021 at 1:15
  • 3
    $\begingroup$ @Whey_Isolate while technically true, those costs are utterly dwarfed by the energy cost of accelerating the ship in the first place, and then decelerating it at its destination. Interstellar drag is only really significant if you're trying to capture and use the interstellar medium for energy or propulsion, as in a Bussard ramjet. $\endgroup$ Dec 27, 2021 at 18:12
  • $\begingroup$ @PostlimFort There would be a time cost for the ship itself, whether it’s moving or not. If you borrowed a billion dollars for a ship that will last 30 years, you’d owe the bank nearly a quarter million dollars per day. Energy for lights and life support is a free but useful byproduct of the reactor idling. $\endgroup$
    – StephenS
    Dec 27, 2021 at 18:54
0
$\begingroup$

The efficiency should be relative to the speed of the engine. Though the rest of space is moving at one rate of time, the engine is experiencing another, and thus consuming fuel at its own rate of time.

$\endgroup$
0
$\begingroup$

Fuel consumption is mostly consequential to changing velocity (acceleration).

Your fuel consumption unit, then, is gee-hour per speed crystal.

Because relativistic mass increases with speed, the amount of energy required to obtain a single-unit increase in velocity becomes infinite near the speed of light.

Applied to this fuel efficiency measurement, you should expect to get your most efficiency gee-hours per speed crystal when you are at rest. Near the speed of light, fuel efficiency is poor, approaching zero as the speed of light is reached.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .