Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy enigma in uniformly moving reference frames (I hope this doesn't double post)

  1. Jul 25, 2007 #1
    A man I met the other day found out I am studying physics and asked me a question about energy, namely kinetic energy. Below in the box is an edited version of what he postulated to me.

    He seems to want to determine his speed compared to a universal reference frame, which is silly, but I haven't been able to reason out an interesting question that occurred to me in listening to him...



    Imagine you are at rest in a reference frame (S). You have a spaceship with 16 units of potential chemical energy (E_chem). You use E_chem to accelerate your spaceship to a velocity (v) of 5.66. Your spaceship now has 16 units of kinetic energy (K).

    In a completely different scenario, you are at rest in reference frame (S') which is moving at a velocity of 4 (V') relative to the reference frame of the first situation, S. in S' you use 8 units of E_chem to accelerate your ship to a velocity of 4 relative to S'. Given that you were moving initially with respect to S with a velocity of 4 and a kinetic energy of 8, and seeing that your final velocity is now 8 compared to S after using 8 units of E_chem (V' + v), it seems, by comparing the scenarios that kinetic energy is linearly related to velocity.

    but K = .5*m*v^2.

    Okay, either I just haven't done energy equations in too long, or I'm missing something, but something is wrong here. care to explain?



    ==========================================================
    ==========================================================
    0) I am in a space ship.

    1) I am at rest relative to a buoy that I will take to be my reference point.

    2) I accelerate the first time, using 8 units of energy, and measure a
    velocity of 4 relative to the buoy.

    3)I launch another buoy (whose mass is negligible relative to the mass of my space ship) so that it is at rest with respect to me (and thus moving at a velocity of 4 relative to the first buoy).

    4) I accelerate the second time, using 8 units of energy and measure a velocity of 1.66 relative to the second buoy, and 5.66 to the first.

    Finally, in view of the above, it seems to that I can distinguish
    between being at rest, and being in motion. The method (given in the
    box below) is simple, although it *differs* from the modified
    scenario just explained above... however, were the method given in the
    box below to fail, it seems to me that the failure would force a
    contradiction to what was said above.

    ----------------------------------------------------------------------
    I am adrift in space, so very far away from everything that I have no
    reference points whatsoever. What do I do to determine my speed? Simple.
    I launch a *single* bouy (whose mass is negligible relative to the mass of my
    space ship), so that it is at rest relative to myself,
    and then accelerate using 8 units of energy. Next, I measure my
    velocity relative to the bouy!
    ----------------------------------------------------------------------

    If I happened to be at rest, my velocity relative to the bouy will be
    different that it would have been if I happened to be in motion...
    ==========================================================
    ==========================================================
     
  2. jcsd
  3. Jul 26, 2007 #2

    rcgldr

    User Avatar
    Homework Helper

    Kinetic energy is relative to a frame of reference. It's the difference in the veolcity of an object versus the velocity of a frame of reference that determinse the kinetic energy.

    The issue with step 4 in the box is that 8 units of energy are relative to the fisrt buoy, not the second. The rate of work done is power, which can also be considered to be force times speed. In your example, the speed was relative to the first buoy.

    In the case of a real rocket, what is happening is that rocket expells a small part of it's own mass at a very high velocity. The frame of reference I prefer to use here is the rocket itself. All the work is done on the fuel. The increase in kinetic energy of the fuel relative to the rocket can be measured, so there is a way to state the power of a rocket in reguard to the rate of work done to accelerate the spent fuel.

    If I use the rockets initial velocity as a frame of reference, then the rocket produces constant thrust independent of the rocket's velocity from it's initial velocity (and acceleration increases as the rocket's mass is decreasing). Effectively the power of the rocket increases with speed. The issue is that this is ignoring the kinetic energy of the fuel which is moving at a high speed opposite that of the rocket's acceleration. The total energy is the energy gained from the expansion of burnt fuel, and is distributed between the kinetic energies of the fuel and rocket.

    The mass initially expelled will always have a very high velocity in the opposite direction of the rockets initial acceleration, but eventually the rocket will achieve a forwards speed that some of the fuel will end up with a velocity in the same direction as the rockets acceleration.
     
  4. Jul 26, 2007 #3
    I'm having trouble putting into words exactly what I am thinking, but after reading the previous posts, this may help...

    It seems that dependent on your choice of reference frames, you will gain a different amount of velocity using the same amount of energy.

    In reference frame S where you are moving at v=4 (K=8), 8 units of energy will let you attain a final v=5.66 (K=16). However if you are using reference frame S', which is moving at V'=4 relative to S, and assuming you are at initially at rest within S' (in S' your K=0; in S your K=8), using 8 units of energy will give you a final v=4 relative to S' (therefore in S' your K=8). Consequently, applying your knowledge about how fast you were moving relative to S and your new velocity in S', your final v=8 in S (V' + v), but this means that you have K=32 in S, which is impossible because the system only had 16 units of energy total.

    I think that I have been scrutinizing this problem too closely and have fallen into an obvious trap dealing with comparing the two reference frames, but I can't seem to find the error because it has become a matter of course in my approach to this problem. I hope this states my situation more clearly.
     
  5. Jul 26, 2007 #4

    rcgldr

    User Avatar
    Homework Helper

    In your scenario, if 16 units of chemical energy accelerate the rocket to 5.66, then the initial mass of the rocket was greater by 16 units of fuel, and the mass of the rocket after acceleration is now 1 (based on your statement that 1/2 m (5.66)^2 = 16.

    What is unknown is the mass of the fuel per unit of chemical energy.

    During the acceleration, the onboard fuel (as well as the rocket) is getting an increase in kinetic energy, which increases the total energy of the onboard fuel (kinetic and chemical) relative to the initial reference point (the first buoy).

    Also, after using 8 units of chemical energy, the velocity of the remaining fuel and rocket will be less than 4. Total energy will be:

    8 = 1/2 (mass of rocket + mass of fuel with 8 units of chemical energy) v^2

    Since mass of rocket is 1, then

    8 = 1/2 (1 + mass of fuel with 8 units of chemical energy) v^2
     
  6. Jul 26, 2007 #5
    I was assuming, like in an elementary physics problem, that the mass of fuel relative to the rocket was negligible, and for all practical calculations, the mass of the rocket could be considered constant. However this would not satisfy the conservation of momentum, which may be contributing to what I am overlooking.

    Either way, given the above scenario (with one object at rest relative to you and the other moving) do you think you would achieve the greater or lesser change in velocity (1.66 vs 4). Intuitively I would choose 1.66.
     
  7. Jul 26, 2007 #6

    rcgldr

    User Avatar
    Homework Helper

    Neither. The larger velocity change would be less than 4, and the smaller velocity change would be greater than 1.66.
     
  8. Jul 26, 2007 #7

    Ivan Seeking

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The problem is that you are using energy in S' in order to calculate the speed in S. What you actually measure is velocity and the energy spent in your frame of reference. So you expend 8 units of energy in your frame can call this the S' frame. You measure the S buoy moving away at Vs=-4. Now, in the S' frame of reference you expend another 8 units of energy, which means that your velocity wrt to the buoy in S' is 4. If we assume that the velocities are small enough to ignore relativistic effects, then you would measure S moving at a velocity of -8 wrt the S" frame. From this you would calculate that you have an energy of 32 in the S frame; that is, if you assume that you still have a mass of 1 in the S frame [again if we ignore relativity]..

    You are implicitly assuming that (Vs2 + Vs'2) = (Vs + Vs')2. But the energy equation doesn't allow us to add like this. You have to take the final velocity wrt a given frame of reference, and then calculate the energy in that frame.

    This resolves your paradox as well. Given your example, no matter where you start, if you define a buoy to be at rest and expend 8 units of energy, the buoy will be moving away from you with a velocity of 4.
     
  9. Jul 26, 2007 #8
    Aha! That is the problem.

    I thought that I was getting confused when comparing the reference frames and their respective energies, but I couldn't put my finger on exactly where I was making the bad assumption.

    Thanks Ivan!

    PS: Mods, I would close this thread to future posts, but don't know how.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Energy enigma in uniformly moving reference frames (I hope this doesn't double post)
Loading...