# 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 at 19:46

## 1 Answer

At 99% the speed of light the behaviour would be almost completely determined by special relativity. The scenario is well-investigated for synchrotrons. In principle a synchrotron or a storage ring, e.g. around the equator of Earth, could be built.

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 $$f_o=f_s\cdot\sqrt{(1+v/c)/(1-v/c)}=f_s\cdot\sqrt{0.01/1.99}=0.070888\cdot f_s.$$

By this we now have calculated the observed frequencies for three positions of the circling particle/radio to give some idea about the oscillation of the observed frequency.

More details about the relativistic transverse Doppler shift, see e.g. Ives-Stilwell experiment. Close to the experiment with the observer in the center of the circling particles are Mössbauer rotor experiments. In this cases ions or atomic nuclei emitting or absorbing at known wavelengths are used as "radios".

This paper describes a slower version of the transverse Doppler shift, as observed by using the GPS satellites moving with just 4 km/s. In this slow case the gravitational frequency shift, as predicted by general relativity to be induced by the gravitational field of Earth, plays a relevant role, relative to the (in this case) small transverse Doppler shift. Here the GPS satellites are the moving radios.

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