# Time dilation on an object circling earth

How would a receiver on Earth hear a radio transmission from an object circling earth at 99% of the speed of light for 24 hours. The transmission from the object circulating would be non-stop.

Because time dilation would occur, would the transmission be slowed to the receiver? The object would travel so near earth that the signal would not have any delay reaching earth.

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There's no way any object with non-zero rest-mass would be orbiting the Earth at that velocity, it would shoot it straight off since it would have orbital energy way past one that could be sustained in orbit due to gravitational attraction of the planet. But if it was (strictly from mathematical point of view), then its wavelength would oscillate as a sine wave from close to Planck length to near infinity with an average at the frequency of transmission. –  TildalWave Jan 25 '14 at 19:46

At 99% the speed of light the frequency $f_s$ of the circling object should occur red-shifted by a factor of a little more than 7 to an observer in the center of the circle due to the transverse relativistic Doppler effect: $$f_o = f_s/\gamma=f_s\cdot\sqrt{1-v^2/c^2}=f_s\cdot\sqrt{1-0.99^2}=f_s\cdot\sqrt{0.0199}=0.141067 f_s.$$ For an observer immediately near the ring, the frequency $f_o$ is the same as for the observer at the center for the signal emitted, when the particle was diametral on the other side of the ring. When approaching the observer along the lign of sight, the signal of the particle is blue-shifted to $$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{1.99/0.01}=f_s\cdot\sqrt{199}=14.1067\cdot f_s.$$ When leaving the observer along the lign of sight, the signal of the particle is red-shifted to $$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{0.01/1.99}=0.070888\cdot f_s.$$