4
$\begingroup$

How precisely can the distance to the Sun be measured? Wikipedia says the distance to the Moon can be measured upto millimeter precision. But Wikipedia article on distance to the Sun says only about Astronomical Unit and nothing on the precision of the measurement of the distance to the Sun. I am certainly aware that we now use radar to measure the distance to the Sun, and I remember reading the precision in its measurement somewhere on the Internet, but I can't find it anymore. When I try to find the relevant information on the Internet, all I find is educational articles on things such as parallax, which is certainly superceded by radar measurements, or articles on lunar distance, which I am not looking for. Relevant references will highly be appreciated.

Edit: I just found out on Internet that we measure the distance to the Sun through distance to the Venus or Mercury. Anyway I want to know the precisions or errors in these measurements..

Improve this question
$\endgroup$
6
  • $\begingroup$ naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/… may or may not be helpful; note that we know the moon's orbit to within a meter, not within a millimeter (I trust NASA more than wp) $\endgroup$
    – user21
    Feb 17, 2020 at 16:06
  • $\begingroup$ Can't we measure the distance to moon by just using laser time-of-flight? Then we should know it very accurately. $\endgroup$
    – Natsfan
    Mar 13, 2020 at 17:39
  • 2
    $\begingroup$ No, you know the distance to the mirror. @jmh $\endgroup$
    – ProfRob
    Mar 13, 2020 at 19:56
  • $\begingroup$ Thanks. But won't laser light reflect off the surface of the moon and return to the earth with enough intensity to detect? Or do we have to use a mirror? $\endgroup$
    – Natsfan
    Mar 13, 2020 at 20:43
  • 1
    $\begingroup$ Re I am certainly aware that we now use radar to measure the distance to the Sun: The Sun is one of the solar system bodies for which radar measurements of distance are essentially impossible. We could ping Pluto with radar (which we cannot yet do) before we could ping the Sun. See for example this and this at physics.SE. $\endgroup$
    – David Hammen
    Oct 17, 2021 at 18:55

2 Answers 2

Reset to default
1
$\begingroup$

According to E. V. Pitjeva & E. M. Standish, it is (on average) +/- 3 meters. The measurement was made with over half a million observations of different types in 2008. Generally using the positions of planets, asteroids, etc, and spacecraft and modeling the gravitational effects of all bodies.

IAU 2012 Resolution B2 adopts the value of the AU to be 149,597,870,700m exactly. Mostly just to have an exact value to work with, not because the measurement is exact.

Improve this answer
$\endgroup$
0
$\begingroup$

This is a partial answer too long for a comment, but it may help to get the ball rolling...

Starting from this answer we can see that the standard gravitational parameter of the Sun used by JPL in their development ephemerides is 1.32712440040944E+20. While that doesn't mean we know the Earth's orbit to 1 part in $10^{14}$ it hints that it's believed to be known surprisingly well!

It's possible that something like this might shed some light on the subject, but it's not an easy question to answer. The Planetary and Lunar Ephemeris DE 430 and 431, IPN Progress Report 42-196 (February 2014) Part of the problem is that we don't know exactly where the solar system barycenter is with respect to the Sun because there may be bodies far away we haven't detected yet. That's only a small effect on the Sun-Earth orbit. If I had to guess, I'd say it's somewhere between 1 and 100 meters uncertainty between the Earth and the center of the Sun. Where the edge of the Sun falls is a horse of a different color!

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ The second paragraph is not correct. From the link you yourself provided, "The mass parameter of the Sun was defined by $GM_{\odot} = k^2$, where Gauss’s constant k = 0.01720209895 is a defined value. ... For DE430 and DE431, $GM_{\odot}$ has been set to $k^2$ since our current estimate is consistent with this value given the current value of the au." $\endgroup$
    – David Hammen
    Feb 16, 2020 at 14:01
  • $\begingroup$ @DavidHammen I'll have a look thanks! My "stack" is overflowing right now so it may take a day or so... $\endgroup$
    – uhoh
    Feb 17, 2020 at 3:38
  • $\begingroup$ FWIW, ssd.jpl.nasa.gov/api/… gives 132712440041.93938 $\rm{km^2/s^3}$ for $GM_\odot$, which equates exactly to $k^2$ using standard 53 bit double precision arithmetic. $\endgroup$
    – PM 2Ring
    Apr 2, 2022 at 23:04
  • $\begingroup$ High precision Gauss k from GM $\endgroup$
    – PM 2Ring
    Apr 2, 2022 at 23:04
  • 1
    $\begingroup$ Sorry, I don't really have a better answer either. $\endgroup$
    – PM 2Ring
    Apr 2, 2022 at 23:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .