NavList:

I did this using JPL Horizons as a data source [because it exists! :) ...it's also relatively easy to use]. I first generated an hourly table of RA and Dec for Mars and then for Jupiter on 24 August. Hmmm... seems that Antoine is testing us! I see an approximate three-hour discrepancy. Then I ran it at 10-minute intervals and finally at 1-minute. The RA and Dec, with decimal degrees chosen as an output option, is completely analogous to longitude and latitude in an ocean-sailing scenario. We have two vessels travelling on paths that are nearly straight lines for some short enough interval of time, and from those lons and lats we can calculate time and separation as they pass each other.

How to calculate the distance from Jupiter to Mars? I decided to treat this as a "plane sailing" problem in traditional navigation (with a spreadsheet to avoid pain). To do that, we can treat differences in Dec just like differences in latitude on the ground, and it's fair to call these differences "dy" --that is, they are simple offsets in cartesian coordinates. The differences in longitude, as with a comparable problem on the ground (or at sea) have to be multiplied by a longitude scaling factor given by cos(lat) where the lat can be any reasonable mean or "average" latitude for the case at hand. I chose a factor of 0.926. So I take the differences in longitude (diffs of RA in degrees for Mars and Jupiter) at each minute of UT and multiply (scale) those differences by 0.926. Now they're on the same scale as the Dec differences, and we can call those scaled differences "dx". Finally the separation distance is Pythagorean: square root of the sum of the squares... sep =√(dx² + dy²).

In my spreadsheet I made a simple graph of the resulting separation in seconds of arc over the course of several minutes. During a span of five minutes, the separation changes by only a small fraction, a few hundredths, of a second of arc. Thus the actual minimum separation distance and the time of that minimum separation should be treated as approximate at that level: a few minutes of time and a fraction of a second of arc at best.

My result: 1103.317 seconds of arc or 18.38862 minutes of arc at about 14:53:10 UT. See attached images from my little spreadsheet.

Why no conversion to TT?? Anyone who wants it can do that conversion, but in any case the difference is a known quantity at the level of accuracy of the problem, and there's no reason to prefer TT over UT here. Plus NavList readers know what UT is, of course, while TT is another jargon barrier. So... I'm sticking with UT.

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA

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