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B Measurements within our solar system with relativity in mind

  1. Nov 9, 2016 #1
    ok I know that anybody in the heavens will bend spacetime with the gravity it has. What I was wondering was how do we measure around them like from the earth to mars. When mars in now quit on the opposite side of the sun ( so we can see it at sunset). do we measure the parabolic curve in spacetime that light takes from the sun's gravity or do we take into account the curve and make adjustments to get a straight line through the system for distances between the two planets. Also to go even feather do we take into account the curve around the planets as well???

    sorry I didn't know where to post the thread and thank you ahead of time for any and all feed back
  2. jcsd
  3. Nov 9, 2016 #2


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    Distances within the solar system can be approximated as being measured along straight lines for almost all purposes. If you want to include GR, then I believe you would need to measure or calculate the curvature along the path, though I admit I have little idea how scientists do this.
  4. Nov 9, 2016 #3


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    It depends what your application is. You can certainly talk about the "straight line" distance from Earth to Mars. But if what you want to know is "how long will it take my radio message to reach Mars" (or you're using radar to locate a spacecraft) then you'd need to worry about the curved distance.
  5. Nov 9, 2016 #4
    do we or is there any kind of map of the curvatures either for the solar system or bigger like the local stars may be even our side of the milky way?
    I know I have seen maps for the magnetic fields for the milky way. Is there any for gravity?
    I know that we can do the math for each individual object and maybe list them but that's not really the same feel you get with a grid or map either in 2d or 3d
  6. Nov 9, 2016 #5
    wow this is the first time i have try to google something and came up with no hits lol
  7. Nov 9, 2016 #6

    Fervent Freyja

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    Distances can be derived by measuring the parallax. The greatest elongation point, or the point where the body is furthest from the sun in orbit, can be used with the principle of triangulation to find the distance. There are other variations of the method, it depends on what body it is and what baseline is being used.
  8. Nov 9, 2016 #7
    Distances to certain object in the solar system are directly measurable with sonar. The moon, Venus, NEOs, and I think even Mars is reachable.

    Stretching of spacetime within the solar system is pretty minimal anyway. Newton's laws describe the orbits of all planets very accurately with the exception of Mercury and even that isn't terribly far off.
  9. Nov 9, 2016 #8
    great info for finding out how far away something is in space and I approached it too was hoping for some kind of map or grid that showed the cravacher of spacetime around the stars and more massive planets that one could see the path that light would take from an outside perspective
  10. Nov 10, 2016 #9


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    I hope you meant radar. Otherwise that's a really, really good sonar! :-p
  11. Nov 10, 2016 #10


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    Well, part of the problem is that Einstein's field equations for GR are extremely complicated and nonlinear, and solving them requires a lot of time and computational power. I doubt a map of the curvature of nearby space even exists. Though I could be mistaken.
  12. Nov 10, 2016 #11


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    The curvature of the light pave is really, really small even when it comes to light passing near the Sun. Let's take the Mars on the other side of the Sun example. we will assume that Mars is 1 deg away from the center of the Sun's disk. If you compare Mars' actual position in its orbit against its apparent position due to the light path curving, you get a shift that works out to bit just less than 1/2 of Mars' own radius. (The atmospheric distortion caused by Mars being near the horizon will be much larger than this.
    As far as the planets go, Let's assume that Saturn is passing near Jupiter in line of sight, for simplicity's sake we assume this happens when Jupiter and Earth are on exactly the same side of the Sun. Saturn appears to pass 1/4 of a degree from Jupiter's center. How far far off would our line of sight estimate of Saturn's position be compared to its actual position? Less than 2 km.
  13. Nov 10, 2016 #12
    Yes, Newtonian dynamics does give good accuracy in the Solar System but the orbit of Mercury being "off" at all is what inspired Einstein to write his most famous work and Eddington to try and prove Albert's work.

    When it comes to the original question, when they are planing to send a probe, they do measure the distance as a line and of course they tend to find when the line (trajectory) is the shortest to preserve fuel etc. Sometimes they use the Hohmann transfer orbit. If you are interested in the subject, the link provided is for Orbital Mechanics and should be a good source. Just in case you don't like it, here's NASA to the rescue, enjoy.
  14. Nov 10, 2016 #13


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    For a 2-d surface, curvature is represented by a single 3-d number, it takes more numbers to describe the curvature of 3 dimensional space, and even more (20 unique numbers, as I recall - 21 numbers with one constraint resulting in 20 so-called degrees of freedom) to describe the curvature of a 4-d space-time. So it's hard to draw a map of the curvature, you'd need 20 4-dimensional maps.

    There is something that can be regarded as a space-time map of the solar system, this is a mathematical formula called the metric. The interpretation of the metric as a map is given by Misner in his paper "Precis of General Relativity", https://arxiv.org/abs/gr-qc/9508043. The paper though discusses a different, but related problem. Rather than discussing how to make a map of the universe, the paper talks about the issues of how we make a map of the Earth via the global positioning satellite system, GPS. The basic problems are similar though - how do we go from physical observations of signals, to some sort of overall conceptual representation of time and space?

    The relevant quote that discusses interpreting a metric as a sort of space-time map is given in the following short quote from the much longer paper.

    Also important is the following quote about the role of the metric:

    Moving onto the greater question of the "map" (metric) of the solar system, and beyond. Astronomers do need such a map to organize their experimental observations, so that they can go from observations via optical and radio telescopes, to some sort of stellar coordinates. I believe that modern star coordinates are based on the International Celestial Reference System, the ICRS, https://en.wikipedia.org/wiki/International_Celestial_Reference_System.

    The details are quite technical, though. You can find a very terse explanation in the IAU resolutions, for instance http://syrte.obspm.fr/IAU_resolutions/Resol-UAI.htm, the IAU 2000 resolutions. These have been tweaked at least once, the basic IAU resolutions from 2000 were modified in 2006, https://www.iau.org/news/pressreleases/detail/iau0603/. I'm not sure how current the IAU 2006 resolutions are, to be honest.

    The resolutions are so terse that there are papers explaining them in more detail to the astronomers, with PHD degrees in the field. I'm not aware of any really good descriptions written at the lay level, and I don't recall offhand which explanatory papers I saw (though I recall seeing them).

    Another related topic you might find interesting is star catalogues, see for instance https://en.wikipedia.org/wiki/Star_catalogue. Hipparcos is one of the more modern catalogues, though with only 118,218 stars, it is listed as a specialized catalogue in the wiki article.
  15. Nov 11, 2016 #14
    thank you for the info lol it makes very good reading :partytime:
  16. Nov 11, 2016 #15


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    Spacetime curvature due to gravity is very feeble for low mass, low density objects like planets and stars. Only massive compact bodies, like neutron stars and black holes, have sufficient effect to be of any practical significance and only in close proximity to such bodies. We can safely ignore curvature under ordinary circumstances.
  17. Nov 11, 2016 #16


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    I think you mean space curvature* rather than spacetime curvature. Spacetime curvature can be ignored only in circumstances where the tidal effects of gravity are negligible.

    *under a suitable choice of decomposition of spacetime into space and time.
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