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B Energy required to change speed?

  1. May 25, 2016 #1
    Ok, this will be awkward but i somehow got lost in this thought,

    For low velocities, the formula for energy boils down to the classical E=1/2m*v^2

    So let's say two astronauts A and B are in empty space absent of gravity.

    B has a jetpack and accelerates to 5m/s which require E=1/2m*v^2= 1/2*100kg*(5m/s^2) = 1250 joule

    When B is finished accelerating, he finds another astronaut C right next to him (doesn't matter how he got there, he just is).

    B and C are moving at 5m/s relative to A. They can both consider themselves at rest according to SR.

    So B decides to accelerate again to 5m/s relative to C, requiring again 1250 joule.

    For such low velocities, we can add the velocities in the classical sense. So B would see A moving at 10m/s relative to himself with a total of 2500 joules used.

    But it requires 1/2*100kg*(10m/s)^2 = 5000 joules to accelerate to 10m/s. So where is my confusion?
     
  2. jcsd
  3. May 25, 2016 #2

    Ibix

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    Science Advisor

    You've accounted for the kinetic energy change of the astronaut, but not for the kinetic energy change of the exhaust mass. If you account for that, you'll find that the total change is frame invariant. The velocity of the second burn exhaust (in A's rest frame) is lower than that from the first since B was already moving when he accelerated the second time.

    It would be easier to consider two shots from a gun, with the recoil accelerating B, if you wish to check this.

    Edit: and the first burn had to accelerate not just B but also the reaction mass for the second burn, unless C provided a refuel.
     
    Last edited: May 25, 2016
  4. May 25, 2016 #3
    Hm, so i am picturing a cannonball+spring astronaut B carries at the beginning, The cannonball+spring weights 100kg, just as much as the astronaut. He pushes the spring together and releases it, so cannonball+spring fly in the other direction at 5m/s.
    If i am not mistaken, he had to use 2500 joules to achieve this as A would see him flying at 5m/s in one direction, while the cannonball+spring flies at 5m/s in the other direction. So 1250 joules + 1250 joules

    When he reaches C, C gives him another cannonball+spring. Same procedure, he had to use 2500 joules to get him and the cannonball fly at 5m/s relative to C in opposite directions.

    From the point of view of A, the 1st cannonball is flying at 5m/s relative to him, meaning 1250 joules energy difference to the initial state.
    The 2nd cannonball which formerly was flying at 5m/s away of him, is now at rest relative to him, so minus 1250 joules to the total of the system. A total of zero for the two cannonballs as far as energy of the system goes from A's point of view.

    So yes, i guess that would be 5000 joules for B moving at 10m/s.

    But what if the cannonballs were huge and massive, such that the energy passed onto the cannonballs was negligible small? In such a case it seems that B would have to just use the 1250 joules + a small extra to get himself to 5m/s in both cases and it would have cost him only 2500 joules + a tiny extra to get to 10m/s. So where is my thinking going bad again?
     
  5. May 25, 2016 #4

    jbriggs444

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    The idea that large cannonballs at a correspondingly low speed have negligible energy only works if the cannonballs are approximately at rest. Starting from rest, it is a good approximation for the first cannonball. You can make the cannonball big enough and slow enough that the energy expended on its recoil is negligible.

    Think, for instance of a runner pushing off the starting blocks at a race. The energy in the Earth's recoil is negligible.

    But for the second cannonball, the approximation fails. The cannonball is already moving appreciably. A small change in its velocity will not have a negligible impact on its energy. You can see this if you take the first derivative of ##\frac{1}{2}mv^2## with respect to momentum. Start by substututing p for mv and the formula for energy becomes ##\frac{1}{2}\frac{p^2}{m}##.

    Now differentiate with respect to p to get ##\frac{p}{m}##. This is the rate of change of energy in the cannonball as momentum is injected into it.

    For the first shot from rest, p=0 and injecting a little momentum into the cannonball results in negligible energy.
    For the second shot with the astronaut+2nd cannonball already moving, p is non-zero and injecting a little momentum into the cannonball results in non-negligible energy.
     
  6. May 25, 2016 #5

    Dale

    Staff: Mentor

    When you come up on these sorts of conundrums from "hand waving" arguments then the first thing you should do is actually work the math. Most of the time working the math will show you exactly where your thinking was going bad.

    Here, you would find the energy transferred into each object as a function of the energy in the spring, the two masses, and the initial velocity.
     
  7. May 25, 2016 #6

    Ibix

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    As @Dale says, it's always best to work through the maths. @jbriggs444's method is elegant, but if it wouldn't occur to you you can just grind it out long-hand. Say your astronaut has mass ##m## and each cannon ball has mass ##M## and that, from rest, the astronaut and cannon ball are accelerated to ##v## and ##V## respectively by the spring. Conservation of momentum relates ##v## and ##V## and, with that, you can easily work out the kinetic energy change of the astronaut and each cannon ball as seen from A's frame (or any other). Then you can add up the changes to get a total energy change, and even show that it is frame-independent if you want.
     
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