View Poll Results: Which ship will require more energy to maintain their speeds? Ship A will require more energy to maintain it's speed 0 0% Ship B will require more energy to maintain it's speed 0 0% They will require the same energy to maintain their speeds. 1 100.00% Voters: 1. You may not vote on this poll

# Thought experiment

by Harut82
Tags: experiment
 P: 19 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?
 Admin P: 23,716 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?
 P: 123 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.
P: 123
Thought experiment

 Quote by Borek 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.
 PF Gold P: 6,496 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.
P: 19
 Quote by cosmic dust 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.

 Quote by phinds 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.
P: 19
 Quote by cosmic dust 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.
P: 123
 Quote by Harut82 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)$\cdot$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 $\propto$ rate of energy gain = f(v) $\cdot$ v
Mentor
P: 17,534
 Quote by Harut82 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.
P: 19
 Quote by cosmic dust 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)$\cdot$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 $\propto$ rate of energy gain = f(v) $\cdot$ v
Please remember that my physics is very limited. ↑ I didn't get most of that.

 Quote by DaleSpam 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.

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