Home > 0877 > GR0877 Problem 55

GR0877 Problem 55

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (A) Applying relativistic Doppler effect formula: \displaystyle \frac{\lambda_0}{\lambda_{lab}} = \sqrt{\frac{1+\beta}{1-\beta}} \approx \frac{4}{3}. Thus, 16-16\beta = 9 + 9 \beta \implies \beta = 7/25 = 0.28.

Found a typo? Comment!

Jump to the problem

1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Advertisements
Categories: 0877 Tags: , , , ,
  1. Pedro
    07.11.2012 at 07:16

    I must be getting something wrong, but according to wikipedia (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect) the lambda terms in the doppler equation are actually reciprocal of what is presented on this page. My textbook also seems to be showing them the other way around. Weird.

  2. Pedro
    10.11.2012 at 05:36

    I got it. My bad. Never mind! Thanks.

  3. Azalee
    12.03.2013 at 04:18

    I used delta_lamba/lamba_0 = v/c to get 1/3 c which is close enough to 0.28.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: