# Why does the moon’s transit appear so inclined in this STEREO-B video?

Question: Why does the moon’s transit appear so inclined in this STEREO-B video?

This video of the Moon transiting the Sun on Feb 25 2007 is from STEREO-B during its solar orbital insertion.

In the video, the plane of the transit appears to be 24* inclination to the Sun’s rotational axis. The inclination of the Moon’s orbit to the ecliptic is 5.14*. The inclination of the Sun’s axis is 7.25* to the ecliptic. I can’t find the orbital inclination of STEREO-B, but it is usually shown coplanar with the ecliptic.

5.14+7.25 is a long way from 24. I assume the discrepancy is because the point of view is moving, but I can’t picture the relative motion.

By the Sun’s rotation, North is up. The waxing gibbous Moon is moving to the right. STEREO-B is moving to the left. Wouldn’t Stereo-B’s motion make the transit appear less inclined, rather than more?

• cannot find the heliocentric orbit inclination, but they were in a 182 × 403,810 kilometers) at a 28.5-degree Earthcentric inclination until ejected (by encounters with the moon). With apogee way past the moon, and inclination well above the moon's, the encounter would also have been from well "above" or "below" the moon, further increasing the off-ecliptic component of STEREO-B's movement. Dec 9, 2021 at 21:18
• @cutekitty-pleasestopbarking This NASA site stereo.gsfc.nasa.gov/orbit.shtml describes the launch of STEREO B as having two gravity assists from the moon, 6 weeks apart. The video was taken over a month after the second assist, when STEREO B was in its heliocentric orbit, about 1,000,000 miles from Earth. It should have acquired its permanent orbital inclination by that time. Dec 9, 2021 at 22:36
• @Woody there's a GIF in Does the arrow in this STEREO trajectory animation point heliocentric prograde, or towards the Sun? This is a very cool question! +1
– uhoh
Dec 10, 2021 at 1:40
• From Horizons, STEREO-B's inclination to the ecliptic on 2007-Feb-25 was quite low: 3.152843735218575E-01, i.e., just over 0.3°. I used this batch file: SOF MAKE_EPHEM=YES COMMAND=-235 EPHEM_TYPE=ELEMENTS CENTER='500@10' START_TIME='2007-Feb-20' STOP_TIME='2007-Feb-27' STEP_SIZE='1 DAYS' REF_SYSTEM='ICRF' REF_PLANE='ECLIPTIC' OUT_UNITS='KM-S' ELM_LABELS='YES' TP_TYPE='ABSOLUTE' CSV_FORMAT='YES' OBJ_DATA='YES'  Dec 10, 2021 at 1:59
• @Ohoh. I saw the same animation on another site (didn't save the link). The text described the arrow as pointing towards the Sun. Arrow movement correlates with time from launch to gravity assist. Dec 10, 2021 at 4:29

We're looking at two moving objects from a moving platform. Over the timespan from 2007-Feb-25 7:30 to 2007-Feb-25 18:00, the Moon's ecliptic latitude is changing faster than its ecliptic longitude. But we also need to subtract the Sun's movement to account for the effect shown in those video frames.

Here are a couple of diagrams produced using JPL Horizons, using a 30 minute time step. The Sun is red, the Moon is blue. The small dots show the positions, the large circles show the positions and apparent angular sizes of the bodies at the start and end of the time period.

The Moon is moving up & to the left, the Sun is moving to the right. Note that the Moon's path is even steeper than what the video shows. We're using an equirectangular projection of the celestial sphere, but the angular size of the viewing region is small, so the inevitable distortion is small.

If we subtract the Sun's motion, we get this:

The changes in ecliptic longitude and latitude, in arc-minutes, are:

Body Long Lat
Sun 26.21 0.14
Moon -4.95 12.17
Relative -31.17 12.03

So the angle of the Moon's path against the Sun is $$\arctan\left(\frac{12.03}{31.17}\right)\approx21.1°$$.

Here's a crude animated version:

Incidentally, the Moon was moving on a fairly curved trajectory (relative to STEREO-B) around that time.

Here's a plot covering two days, using a 1 hour time step.

Here's a link to the Sage / Python script I used to make the static plots.

• The diagrams, if I have it right, are in Geocentric Ecliptic Coordinates which has its origin at the center of the Earth. The Earth plays no part in the geometry of this video. The point of view is from STEREO B which has a different orbital plane, position and velocity. Therefore the “STEREO Celestial Coordinates” (if you chose to use them) should use a different origin and “Vernal Equinox”. An alternative would be to use Heliocentric Ecliptic Coordinates which is much more conventional and would clearly show the relative angular positions of the Moon and STEREO. Dec 10, 2021 at 22:25
• @Woody The diagrams are from the POV of STEREO-B. Yes, they use the ecliptic plane, but they are not geocentric. Please take a look at the Reference Frames section of the Horizons docs, ssd.jpl.nasa.gov/horizons/manual.html#frames particularly, the Ecliptic of Standard Epoch subsection. Dec 10, 2021 at 22:38
• Great. That makes much more sense. But why does the animated version show the Moon’s transit from 4 o’clock to 10 o’clock while the video shows 8 o’clock to 2 o’clock? The Sun’s rotation shows North is “up” in the video, so it wasn’t posted inverted. Dec 10, 2021 at 23:23
• @Woody I assume that the image was originally inverted, and it's simply been rotated to put the Sun's north pole at the top. (I doubt that the optics in the satellite would include a prism or mirror to undo the inversion). Note that the Sun's longitude is increasing, as we'd expect from the satellite being in a prograde orbit. Dec 11, 2021 at 0:19
• I just noticed that Horizons returns the angular diameter of the bodies, not the angular radius, so my Sun & Moon circles are twice as big as they should be. Oops. Fortunately, that doesn't affect the trajectory stuff. May 22, 2023 at 4:51