p44_050

# p44_050 - , is in large part due to the fact that, as the...

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50. (a) From E = NQ =( M sam / 4 m p ) Q we get the energy per kilogram of hydrogen consumed: E M sam = Q 4 m p = (26 . 2 MeV)(1 . 60 × 10 13 J / MeV) 4(1 . 67 × 10 27 kg) =6 . 3 × 10 14 J / kg . (b) Keeping in mind that a Watt is a Joule per second, the rate is dm dt = 3 . 9 × 10 26 W 6 . 3 × 10 14 J / kg =6 . 2 × 10 11 kg / s . This agrees with the computation shown in Sample Problem 44-5. (c) From the Einstein relation E = Mc 2 we get P = dE/dt = c 2 dM/dt ,or dM dt = P c 2 = 3 . 9 × 10 26 W (3 . 0 × 10 8 m / s) 2 =4 . 3 × 10 9 kg / s . This ±nding, that dm dt > dM dt
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Unformatted text preview: , is in large part due to the fact that, as the protons are consumed, their mass is mostly turned into alpha particles (helium), which remain in the Sun. (d) The time to lose 0 . 10% of its total mass is t = . 0010 M dM/dt = (0 . 0010)(2 . × 10 30 kg) (4 . 3 × 10 9 kg / s)(3 . 15 × 10 7 s / y) = 1 . 5 × 10 10 y ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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