4
$\begingroup$

To get more statistical significance for the suspected flyby-anomaly (or to refute it), it would be desirable to track as many hyperbolic earth flybys to within 1mm/s as we can get. Can we track natural objects approaching/leaving earth with this accuracy by long-term observations?

Laser tracking of manmade objects may be easier. Think of a swarm of passive reflectors, spring-emitted way before perigee of an earth gravity-assist, so they pass with different distance and leave in different directions. But I fear that we will never see such a maneuver with a significant plane change, because missions rarely ever leave the ecliptic. With the observations so far, in- and outbound declination seems significant, especially going from high declination back to low would possibly give insight. Will the effect be reversed?

I doubt we will ever see a mission dedicated to this effect, so studying it with natural objects would be useful.

|improve this question
$\endgroup$
  • 1
    $\begingroup$ @uhoh Yes, it sounds unreasonable if you remember that even tracking luna is done with a reflector. I was however not sure, what long-term integration of observations can do about this. $\endgroup$ – Andreas Nov 5 '16 at 14:52
  • 1
    $\begingroup$ One of OSIRIS-REx's major tasks is to carefully characterize the 3D mass distribution and to carefully map the reflectivity and emissivity of Bennu, to improve on the estimations of the non-gravitational effects that will affect it's orbit. For the experiment you are proposing, the natural bodies will be much low mass - because high mass NEOs are (luckily) infrequent! Low mass means the uncertainties in the non-gravitational effects will be magnified, and may overshadow any hope to extract meaningful data. So ambiguous reflector + uncertain force $\endgroup$ – uhoh Nov 5 '16 at 15:06
  • 1
    $\begingroup$ @uhoh Why wouldn't the anomaly be apparent for satellites in Earth's orbit? 1 mm/s is 31 km per year. $\endgroup$ – LocalFluff Nov 5 '16 at 15:20
  • 2
    $\begingroup$ For good accuracy, radar is useful. You can get a very accurate axial velocity measurement with Doppler radar (1 mm/s doesn't seem out of the question). Radial velocity depends on being able to correlate multiple measurements into a single track. $\endgroup$ – Hobbes Nov 5 '16 at 16:06
  • 2
    $\begingroup$ Here lpi.usra.edu/books/AsteroidsIII/pdf/3004.pdf are some results about 4179 Toutatis, a resolution as fine as 125 ns (19 m in range) and 8.3 mHz (0.15 mm/s in radial velocity) was achieved. $\endgroup$ – Uwe Nov 7 '16 at 15:47
5
$\begingroup$

Arecibo can measure speeds with an accuracy on the order of 1% in scanning mode (i.e. just observing asteroids as they pass through the field of view). Distances can be measured to 10-8 (10 ppb), I suspect this is done in a different, more accurate mode.

Emily Lakdawalla discusses this in her blog, inlcuding this graph that shows the position accuracy of one asteroid using optical observations only, or using Goldstone radar observations, with radar being 50 times more accurate:

Radar is much more accurate than optical observations only

This article quotes position accuracy of 10 m, and axial speed accuracy of 1 mm/s.

|improve this answer
$\endgroup$
  • $\begingroup$ Note the ratio of quoted position accuracy in the Space.com article (10 m) to distance (324,600 km) is about $3\times 10^{-8}$, right in line with the NASA article. $\endgroup$ – Chris Nov 21 '16 at 15:23
  • $\begingroup$ Optical gives position accuracy perpendicular to the line of sight while radar (one dish, doppler) gives position accuracy along the line of sight. Of course with multiple dishes you can get more transverse position information. That plot is position prediction accuracy, not position measurement accuracy. They are related, but one is not the same as the other. The Emily Lakdawalla blog link is very helpful - as they all are! $\endgroup$ – uhoh Nov 22 '16 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.