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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online