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Speed, as we know, doesn't exist without first having a reference point. We then say that the reference point isn't moving at all, and speed is then measured in relation to the reference point.

What is the standard reference point when determining the speed of objects in astronomy? Earth?

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Any chance to get an accept on this one? –  e-sushi Dec 17 '13 at 1:58
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As Einstein realized and like you correctly state, you indeed can't measure the speed of an object by itself since it has to be measured relative to something else.

As a logic result, if you ask the question "How fast is X moving?", you will have to specify that you want the speed with respect to another object because motion cannot be measured without a reference point.

Some examples:

  • If you ask how fast Earth is moving with respect to its own axis, the center of the Earth will be your reference point.
  • If you want to know how fast the Earth orbits the Sun, the Sun will be your reference point.
  • If you want to know how fast the Moon's distance from Earth increases, Earth will be your reference point.
  • If you want to know the speed of our solar system in the milky way galaxy, the center of the Milky Way will be your reference point.

"standard" reference point

If no reference point is given, it could be assumed that the "standard reference point" is the location of the observer. In the realms of astronomy, that would be the location of the astronomer.

It should be noted that the astronomer could well be on an space mission outside Earth's atmosphere, or he/she could be using a telescope in Earth's orbit. Therefore, "Earth" can not generally be assumed to be a standard reference point for astronomy because, depending on the precision of the measurements and the location of (for example) the telescope, assuming "Earth" may result in inexact measurements.

EDIT

As @TidalWave correctly commented, there's also the International Celestial Reference Frame (ICRS) which can help you find reference points, calculate distances, etc. according to a celestial reference system, which has been adopted by the International Astronomical Union (IAU) for high-precision positional astronomy. The origin of ICRS resides at the barycenter of the solar system.

Wrapping it up:

If no reference point is defined and if we can assume that the astronomer works according to the rules and definitions of the International Astronomical Union (which would be the regular, if not ideal case), the International Celestial Reference Frame provides you with a "standard reference point" at it's center (which is the barycenter of the solar system).

In rare cases where IAU compliance can not be assumed (something that "might" happen in amateur realms), it has to be assumed that the reference point is the point of observation.

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The last part is a bit off considering IAU adopted standards, plus it's impossible to determine one's speed by moving with the reference point. You might also wanna mention the International Celestial Reference Frame (ICRS) to also answer the last part of the question. It's all there... ;) –  TildalWave Oct 11 '13 at 2:33
    
Well, when making measurements of the cosmic microwave background radiation, one has to subtract off the dipole moments associated with the movement of the satellite in orbit around the Earth, the Earth's motion due to it revolving around the sun, and the motion due to the sun about the center of the galaxy. I would imagine that this would always need to be a step whenever high accuracy observations (from telescopes in space) are made of distant objects. See: ned.ipac.caltech.edu/level5/March05/Scott/Scott2.html –  astromax Oct 11 '13 at 2:44
    
@astromax Exactly. That's why I said it's not correct to assume Earth as a standard reference point. –  e-sushi Oct 11 '13 at 2:51
    
@astromax True that as well, though for academic reasons, we can boil it all down to determining Inertial Frame of Reference, which might of course be different depending on yours and your object of observations common parameters that would describe time and space homogeneously, isotropically, and in a time-independent manner for both. So yes, with respect to background radiation, what you say is absolutely correct. –  TildalWave Oct 11 '13 at 2:55
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