Thought Experiment: Fuel Consumption in Outer Space

In summary, the conversation between the two speakers revolves around a thought experiment about two spaceships of different masses traveling at the same speed in outer space. The question debated is which ship would require more energy to maintain its speed if they encounter air friction. While one speaker argues that the ship with more mass will need less energy due to its momentum, the other speaker argues that both ships will need the same amount of energy as the friction affects them equally. The experts in the conversation further explain that since the force and velocity are the same for both ships, the power loss and energy consumption will also be the same. They also suggest calculating the work done by the drag force to better understand the concept.

Which ship will require more energy to maintain their speeds?

  • Ship A will require more energy to maintain it's speed

    Votes: 0 0.0%
  • Ship B will require more energy to maintain it's speed

    Votes: 0 0.0%
  • They will require the same energy to maintain their speeds.

    Votes: 1 100.0%

  • Total voters
    1
  • #1
Harut82
19
0
My brother and I were discussing fuel consumption which lead to this thought experiment.

Let's say you have two identically shaped spaceships of different masses traveling in outer space where they are not effected by any forces. No wind drag, no friction, no gravity, etc.

Let's assume spaceship A has a mass of 1,000 Kg ans spaceship B has a mass of 10,000 kg. (it's filled with heavy stuff) Both are traveling side by side at 200 km/hr.

Both ships will travel side by side at 200 km/hr forever unless they encounter other forces. Now if these spaceships suddenly hit a patch of air and experienced air friction similar to air pressure on earth. Which ship would need more energy to keep its speed constant at 200 km/hr.

My brother argues that because ship B has more momentum the wind drag will affect it less and therefore need less energy to maintain it's speed.
I'm arguing that the friction affects both ships equally with a certain force and the will both require an opposite and equal force to maintain their speed.

What are your thoughts?
 
Last edited:
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  • #2
There is nothing requiring a poll here, as the answer is clear.

Assume drag is constant and path through the air is identical for both ships. Can you calculate work done by drag on the ship? Work done is equivalent to the change of the ship kinetic energy. To keep the speed constant you need to replenish the energy. Is there a difference in the energy lost by both ships?
 
  • #3
I agree... The force depends on the velocity and shape, so they are the same, and since the spaceships are co-traveling, they lose energy with the same rate since:

Power = f v = rate of energy loss

Your brother is partly right, since more mass means less deceleration, but also more mass means more energy (per lost velocity) required to keep velocity constant. These two factors balance each other, since they are proportional to 1/m and m, respectively.
 
  • #4
Borek said:
There is nothing requiring a poll here, as the answer is clear.

Of course, this is an exactly solvable problem and we have the tools to do it. Polls are needed when there is not an objective answer, which of course is not the case of this problem.
 
  • #5
To just add to what has already been said, you should avoid the mindset that says that factual, mathematically analyzable situations, are amenable to a poll. Learn science instead.

EDIT: I see cosmicdust beat me to it. It's beginning to sound like we're piling on. Sorry.
 
  • #6
cosmic dust said:
Of course, this is an exactly solvable problem and we have the tools to do it. Polls are needed when there is not an objective answer, which of course is not the case of this problem.



phinds said:
To just add to what has already been said, you should avoid the mindset that says that factual, mathematically analyzable situations, are amenable to a poll. Learn science instead.

EDIT: I see cosmicdust beat me to it. It's beginning to sound like we're piling on. Sorry.

I agree. I'm not able to solve it because I studied Architecture. My science education is limited. That's why I posted here to get some answers from you smart people. The poll was added for fun.
 
  • #7
cosmic dust said:
I agree... The force depends on the velocity and shape, so they are the same, and since the spaceships are co-traveling, they lose energy with the same rate since:

Power = f v = rate of energy loss

Your brother is partly right, since more mass means less deceleration, but also more mass means more energy (per lost velocity) required to keep velocity constant. These two factors balance each other, since they are proportional to 1/m and m, respectively.

I did use that argument. If these two factors balance each other then we was fully wrong :)
He was arguing that these two factors don't necessarily balance each other.

I'm looking for ways to explain to him why he's wrong (or right). If anyone can show me some calculations or convincing arguments I would appreciate it.
 
  • #8
Harut82 said:
I'm looking for ways to explain to him why he's wrong (or right). If anyone can show me some calculations or convincing arguments I would appreciate it.

It's pretty simple and doesn't need futher arguing:

Force = f(v) (function of velocity, where the functional form depends on shape)

Power loss = f(v)[itex]\cdot[/itex]v (equall for both ships)

To keep velocity or kinetic energy constant, you have to give energy at the same rate of losing it. But:

Consumption rate [itex]\propto[/itex] rate of energy gain = f(v) [itex]\cdot[/itex] v
 
  • #9
Harut82 said:
If anyone can show me some calculations or convincing arguments I would appreciate it.
The easy way is simply to calculate the work done by the drag force. That is w=f.d and since f is the same for both ships and d is the same for both ships then w is the same for both ships.

The only possible confusion is that it may not be clear that f is the same in both cases. If that is not clear then look at the drag equation (http://en.wikipedia.org/wiki/Drag_equation). The terms that affect it are the density of the air, the speed, the drag coefficient (shape of the object), and the cross sectional area. Mass or density of the object are not relevant.
 
  • #10
cosmic dust said:
It's pretty simple and doesn't need futher arguing:

Force = f(v) (function of velocity, where the functional form depends on shape)

Power loss = f(v)[itex]\cdot[/itex]v (equall for both ships)

To keep velocity or kinetic energy constant, you have to give energy at the same rate of losing it. But:

Consumption rate [itex]\propto[/itex] rate of energy gain = f(v) [itex]\cdot[/itex] v

Please remember that my physics is very limited. ↑ I didn't get most of that.

DaleSpam said:
The easy way is simply to calculate the work done by the drag force. That is w=f.d and since f is the same for both ships and d is the same for both ships then w is the same for both ships.

The only possible confusion is that it may not be clear that f is the same in both cases. If that is not clear then look at the drag equation (http://en.wikipedia.org/wiki/Drag_equation). The terms that affect it are the density of the air, the speed, the drag coefficient (shape of the object), and the cross sectional area. Mass or density of the object are not relevant.

Makes perfect sense. What about the point cosmic dust pointed out?
How can I show that the below two factors balance each other out? Can anyone elaborate on that a little further?
Your brother is partly right, since more mass means less deceleration, but also more mass means more energy (per lost velocity) required to keep velocity constant. These two factors balance each other, since they are proportional to 1/m and m, respectively.
 

Related to Thought Experiment: Fuel Consumption in Outer Space

1. What is a thought experiment?

A thought experiment is a mental exercise used by scientists and philosophers to explore a concept or theory without actually conducting a physical experiment. It involves imagining a scenario and using logical reasoning to analyze the potential outcomes.

2. How does fuel consumption work in outer space?

Fuel consumption in outer space is different from that on Earth due to the absence of gravity and air resistance. In space, a spacecraft's engines must continuously fire to maintain a constant velocity, resulting in a gradual decrease of fuel over time.

3. What factors affect fuel consumption in outer space?

The main factors that affect fuel consumption in outer space include the mass of the spacecraft, the type and efficiency of the engines, and the distance and trajectory of the journey. Other factors such as gravitational forces and solar radiation can also play a role.

4. Can thought experiments accurately predict fuel consumption in outer space?

While thought experiments can provide valuable insights and predictions, they cannot accurately predict fuel consumption in outer space. This is because there are many variables and unknown factors that can affect fuel consumption in real-life situations.

5. How do scientists use thought experiments to study fuel consumption in outer space?

Scientists use thought experiments to explore different scenarios and analyze the effects of various factors on fuel consumption in outer space. This can help them develop theories and models to better understand and predict fuel usage in space travel.

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