E(A) = 5 GeV = 8 x 10^(-10) J We can figure out its gamma factor from 2 E(A) = gamma(A) * m * c Plugging in m = 9.11 x 10^(-31) kg, we find gamma(A) = 9771 We can then derive the velocity of proton A 1 v(A) = c * sqrt( 1 - ----------- ) gamma(A)^2 We can re-write the square-root term using the binomial expansion: 1 1 v(A) ~= c * ( 1 - --- * ----------- ) 2 gamma(A)^2 and the time it takes for proton A to reach the Earth: 1.5 x 10^(11) m t(A) = ------------------------------------ c * ( 1 - (1/2) * (1/gamma(A)^2) ) which we can re-write again using the binomial expansion as 1.5 x 10^(11) m 1 1 t(A) ~= ------------------ * ( 1 + --- * ------------- ) c 2 (1/gamma(A)^2) ) = 500.3 * ( 1 + 5.2 x 10^(-9) ) = 500.3 + 2.6 x 10^(-6) seconds --------------------- Now, let's look at the other electron, which has twice as much energy. E(B) = 5 GeV = 8 x 10^(-10) J We can figure out its gamma factor from 2 E(B) = gamma(B) * m * c Plugging in m = 9.11 x 10^(-31) kg, we find gamma(B) = 19541 We can then derive the velocity of proton B 1 v(B) = c * sqrt( 1 - ----------- ) gamma(B)^2 We can re-write the square-root term using the binomial expansion: 1 1 v(B) ~= c * ( 1 - --- * ----------- ) 2 gamma(B)^2 and the time it takes for proton B to reach the Earth: 1.5 x 10^(11) m t(B) = ------------------------------------ c * ( 1 - (1/2) * (1/gamma(B)^2) ) which we can re-write again using the binomial expansion as 1.5 x 10^(11) m 1 1 t(B) ~= ------------------ * ( 1 + --- * ------------- ) c 2 (1/gamma(B)^2) ) = 500.3 * ( 1 + 1.3 x 10^(-9) ) = 500.3 + 0.7 x 10^(-6) seconds Therefore, the high-energy electron B reaches the Earth VERY slightly earlier than the low-energy electron A t(A) - t(B) = [ 500.3 + 2.6x10^(-6) ] - [ 500.3 + 0.7x10^(-6) ] = 1.9 x 10^(-6) seconds But the difference is only about 2 micro-seconds for electrons, instead of about 6 seconds for protons.