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A space probe speeding towards the nearest star moves at $0.250c$ and sends radio information at a broadcast frequency of 1.00 GHz. What frequency is received on the Earth?
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$0.775 \textrm{ GHz}$
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OpenStax College Physics Solution, Chapter 28, Problem 25 (Problems & Exercises) (1:50)

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This is College Physics Answers with Shaun Dychko. A space probe is moving away from the Earth and for the purposes of this formula for relativistic Doppler effect, we take the velocity—<i>u</i>—to be positive when it is moving away from the observer. So when the source of the frequency is moving away from the observer, the velocity of that source is positive. So we have a velocity of positive 0.250<i>c</i>. And the source is emitting a frequency of 1 gigahertz. And the question is, what frequency will be observed on Earth? So we plug it in to this relativistic Doppler effect formula, which says we multiply the frequency of the source multiplied by the square root of 1 minus the velocity of the source divided by speed of light divided by 1 plus the velocity of the source divided by speed of light. So that's 1 gigahertz times square root of 1 minus 0.250<i>c</i> divided by <i>c</i> divided by 1 plus 0.250<i>c</i> divided by <i>c</i>. So I'm plugging in a positive number in place of <i>u</i>, in each case here. and this minuses from the formula. OK So that's 1 gigahertz times square root of 1 minus 0.250 divided by 1 plus 0.250, which is 0.775 gigahertz. And this is called red-shifted, as expected, and red-shifted means its wavelength has increased or, in other words, its frequency has decreased. When a source is moving away from an observer, the frequency it admits will be observed at a lower frequency than what was emitted by the source. And so 0.775 gigahertz is less than the 1 gigahertz originally emitted by the source.