1100_Physics ProblemsTechnical Physics

1100_Physics ProblemsTechnical Physics - Chapter 39 P39.18...

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Chapter 39 441 P39.18 The orbital speed of the Earth is as described by Fm a = : Gm m r mv r SE E 2 2 = v Gm r S == ×⋅ × × 6 67 10 1 99 10 1496 10 298 10 11 30 11 4 .. . . Nm kg kg m ms 22 ej e j . The maximum frequency received by the extraterrestrials is ff vc obs source Hz Hz = + × −× × 1 1 57 0 10 1 2 98 10 3 00 10 1 2 98 10 3 00 10 57 005 66 10 6 48 6 . . . The minimum frequency received is obs source Hz Hz = + × × 1 1 57 0 10 1 2 98 10 3 00 10 1 2 98 10 3 00 10 56 994 34 10 6 6 . . . The difference, which lets them figure out the speed of our planet, is 57 005 66 56 994 34 10 1 13 10 64 . = × bg Hz Hz . P39.19 (a) Let f c be the frequency as seen by the car. Thus, cv c = + source and, if f is the frequency of the reflected wave, c = + . Combining gives = + source af . (b) Using the above result, fc v f c v −=
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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