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Matter & Antimatter (2)

  1. Feb 3, 2007 #1
    I am having some trouble on the following problem, can someone help me? Any help is greatly appreciated!:smile:

    1) A spaceship has a mass of 20000kg. It has a matter-anti-matter engine which can transform the annihilation energy into kinetic energy of the spaceship with an efficiency of 5%. How many grams of matter and anti-matter do you need to take the ship completely out of the influence of the earth's gravitational force (achieving escape velocity)?

    My Guess:
    The escape velocity on the surface of the earth is 11200m/s
    Let M be mass of rocket
    m be mass of antimatter
    K be kinetic energy

    K=(gamma-1)Mc^2 (then I substitute the values 11200m/s, etc. into this formula)
    I get K=1.26x10^12 J
    Now set K=0.05*(2mc^2) and solve for m gives me the answer, AM I RIGHT? Is this the correct way to solve this problem?? I am really not sure...

    Also, I have some trouble understanding the scenario, so I am really not sure if I have done the calculations correctly:

    Is the engine turned off at the moment it leaves the earth's surface or is it on all the way (i.e. annihilation never stops?) ?

    Is the spaceship continuously losing mass due to the annihilation reaction? If so, how can I take this into account in my calculations?

    For the matter/antimatter, how can I find the kinetic energy before the collision of the matter/anti-matter?

    By the way, the total (relativistic) energy is defined to be equal to kinetic energy + rest energy. But how come there is no potential energy? For example, why is the gravitational potential energy not taken into account for the equation of total (relativistic) energy?

    Thanks for your help!
  2. jcsd
  3. Feb 4, 2007 #2
    Can anyone help me? Thank you!
  4. Feb 4, 2007 #3


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    Your approach is correct. Except I don't understand the '2' in K=(2*m*c^2). And it is overkill to use the relativistic form to find the kinetic energy. You can just use (1/2)*m*v^2. Earth escape velocity is hardly relativistic. As to some of your other questions,
    the most efficient use of energy would be to accelerate all at once (though probably not very comfortable). Otherwise the answer would depend on the details of how you accelerated. To see this, imagine you just hovered for a long time. You'd be burning fuel, but not getting any where. But the problem really just says there is 5% efficiency in converting energy to KE so you don't have to worry about the details. You are continually losing mass (just like a real rocket) so this should be accounted for. But if you look at the final answer, you'll realize this correction is insignificant (unlike a real rocket). I don't understand at all the question about KE before collision of matter-antimatter. Finally, potential energy is a property of a system - not of a single object. You can speak of the potential energy of a rock 1 km above the earth, but you could equally well say that is the potential energy of the earth relative to the rock. You can't really assign it a definite place. Does that help?
    Last edited: Feb 4, 2007
  5. Feb 4, 2007 #4
    The "2" appears because there are 2 particles: matter & antimatter, and m is the mass of one of them.

    So is the engine turned off at the moment it leaves the earth's surface and no further annihilation occurs? (i.e. the spaceship keeps going upward just by its inertia?)

    And is the the process of mass lost (ldecreasing from 20000kg) finished BEFORE the spaceship is launched? (i.e. maintaining a constant mass as soon as it moves?)

    Should I again assume ZERO kinetic energy of the matter/anti-matter before the collision? But if they are not moving, how can they collide?

    The classical definition of total energy is kinetic + potential energy. I don't understand where the potential energy is accounted for in the total relativistic energy formula. (For example, if an object changes its height, its gravitational potential energy changes...but how come the term "potential energy" isn't appearing in the total relativistic energy equation?)

    Thanks for your help!
  6. Feb 4, 2007 #5


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    The 2 is wrong. The amount of energy depends only on the total mass annihilated (which would already include both matter and antimatter). As the problem is stated, the details of the trip history don't matter. You are only told that when the ship has reached escape velocity the energy debt is the total mass destroyed less an efficiency factor. The mass lost by the 20000kg ship is much less than a gram (as you'll see once you've solved for it). Negigible. KE of the fuel has nothing to do with the problem. Potential energy is a property of a system, not an individual object. Wait, I think I already said that. Think about it!
  7. Feb 4, 2007 #6
    I'm very sorry, I don't quite get it...

    In the other post, I also used 2mc^2 as the intrinsic energy to do a similar calculation.
    Please see:

    If I let "m" be the mass of ONLY the matter and let "m" be the mass of ONLY the antimatter, shouldn't the total intrinsic energy be mc^2+mc^2=2mc^2?

    Say for example, if I get the answer for "m" to be m=12g, then I can say 12g of matter AND 12g of antimatter is needed for the spaceship to achieve escape velocity, right? (An EQUAL amount of matter and antimatter annilhilate, yes?)
  8. Feb 5, 2007 #7


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    Ok. You are correct. Whether the 2 is there or not simply depends on what 'm' means. If 'm' means total mass (matter and antimatter) then no 2. If 'm' means 'matter only (not antimatter) then there is a 2. Just be clear what m is when you state the answer. EQUAL, right.
  9. Feb 5, 2007 #8
    I see!

    By the way, thanks for your help!
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