hw1ans

# hw1ans - Phys 341 Homework#1 Solutions 1(a From dimensional...

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Phys 341: Homework #1 Solutions 1. (a) From dimensional analysis: D ( GMT 2 ) 1 / 3 = [ M - 1 L 3 T - 2 × M × T 2 ] 1 / 3 = [ L ] (b) From dimensional analysis, the gravitational potential energy is E ∼ - GM 2 R ∼ - GM 2 ( GMT 2 ) 1 / 3 ∼ - G 2 / 3 M 5 / 3 T - 2 / 3 Taking the derivative with respect to time (and dropping numerical factors of order unity), we have dE dt G 2 / 3 M 5 / 3 T - 2 / 3 dT dt (c) Now we assume that the change in energy is due entirely to gravitational radiation, so we set dE/dt equal to the power we estimated. (With a minus sign, since radiating power causes E to decrease.) Then solving for dT/dt we get: dT dt ∼ - G - 2 / 3 M - 5 / 3 T 2 / 3 × Mc 3 D ± GM c 2 D ² n ∼ - ± GM c 3 T ² (2 n - 3) / 3 where I used the result from part (a) to replace D , and then I collected all the terms. Including the numerical prefactor then yields dT dt ∼ - 24 , 000 ± GM c 3 T ² (2 n - 3) / 3 (d) For M M ± and T = 27906 . 98 s, we have GM c 3 T [6 . 67 × 10 - 8 cm 3 g - 1 s - 2 ][1 . 99 × 10 33 g] [3 × 10 10 cm s - 1 ] 3 [27906 . 98 s] 1 . 76 × 10 - 10 I then get the following estimates of dT/dt for di±erent values of n : n dT/dt 1 - 4 . 3 × 10 7 2 - 1 . 3 × 10 1 3 - 4 . 2 × 10 - 6 4 - 1 . 3 × 10 - 12 5 - 4 . 2 × 10 - 29 6 - 1 . 3 × 10 - 25 Clearly n = 4 is the exponent that gives something close to the observed value of dT/dt . (Our estimate does not precisely equal the observed value mainly because the two bodies are actually neutron stars that a slightly more massive than the Sun.)

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## This note was uploaded on 09/08/2008 for the course PHYS 341 taught by Professor Keeton during the Spring '08 term at Rutgers.

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hw1ans - Phys 341 Homework#1 Solutions 1(a From dimensional...

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