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The most energetic proton ever detected in the cosmic rays coming to Earth from space had an astounding kinetic energy of 3.0xx10^(20)eV (enough energy to warm a teaspoon of water by a few degrees). What were the proton's Lorentz factor gamma and speed v (both relative to the ground-based detector)? |
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Answer» Solution :KEY IDEAS (1) The proton.s LORENTZ factor `gamma` relates its total energy E to its mass energy `mc^(2) (E=gamma mc^(2))`. (2) The proton.s total energy is the sum of its mass energy `mc^(2)` and its (given) kinetic energy K. Calculations: Putting these ideas together we have `gamma=(E )/(mc^(2))=(mc^(2)+K)/(mc^(2))=1+(K)/(mc^(2)). "" `(36-58) the proton.s mass energy `mc^(2)` is 938 MEV. substituting this and the given kinetic energy into Eq. 36-58, we obtain `gamma=1+(3.0xx10^(20)eV)/(938xx10^(6)eV)` `=3.198xx10^(11)~~3.2xx10^(11)`. This computed value for `gamma` is so large that we cannot use the definition of `gamma` to find v. Try it, your calculator will tell you that `beta` is effectively equal to 1 and thus that v is effectively equal to c. Actually, v is almost c, but we want a more accurate answer, which we can obtain by first solving Eq. for `1-beta`. To BEGIN we write `gamma=(1)/(sqrt(1-beta^(2)))=(1)/(sqrt((1-beta)(1+beta)))~~(1)/(sqrt(2(1-beta)))`, where we have used the FACT that `beta` is so close to unity that `1+beta` is very close to 2. (We can round off the sum of two very close numbers but not their difference.) The velocity we seek is contained in the `1-beta` term. Solving for `1-beta` then yields `1-beta=(1)/(2gamma^(2))=(1)/((2)(3.198xx10^(11))^(2))` `=4.9xx10^(-24)~~5xx10^(-24)`. Thus, `beta=1-5xx10^(-24)` and, since `v=betac`, `v~~`0.999 999 999 999 999 999 999 995c. |
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