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I live near Detroit, Michigan, USA. Let's say I'm viewing from 42.34768° N, 83.47610° W, altitude 718 feet (219 meters) above MSL (mean sea level.) (No, these are not my exact coordinates.)

I've been ready each night to view the KiNET-X rocket launch, from Wallops Island, Virginia, USA, which will wind up near Bermuda at an altitude of over 200 miles (322 km.) 32.3078° N, 64.7505° W.

Assuming the rocket winds up 200 miles over Bermuda, what is the degrees above or below the horizon that would be at, seen by my location?

And, how is that figured out? All news sites covering this are using the same graphic, saying it should be visible even farther from me, as far as Chicago. Maybe it will be visible during its journey, but maybe by the time it's over Bermuda it won't be visible. I unfortunately can't find its expected trajectory.

From my latitude and longitude to Bermuda's, https://www.nhc.noaa.gov/gccalc.shtml gives a distance of 1056 nautical miles (1956 km.)

Ultimately, what I'm trying to figure out is how close it will be to the horizon, so if it's very close, I can decide whether to drive somewhere with better visibility every night until it actually launches.

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  • $\begingroup$ I tried using Google Earth (desktop edition) to add a place at Bermuda, with 322 km altitude. That worked. Viewing from the altitude of my top story window, I can't see the pin marking the place. I increased the Bermuda marker's altitude to 1,000 km, and it still doesn't show it, so I think Google Earth just isn't handling it. If I look from the fake Bermuda 1,000 km marker, I easily see the marker for my home. If I look from the real Bermuda 322 km marker, I can see the marker for my home, but that doesn't tell me how close line of sight is to the horizon. $\endgroup$
    – user1902689
    May 12, 2021 at 21:02
  • $\begingroup$ If I get up quite high over my home, I easily see the Bermuda 322 km marker. As I zoom in to my home, getting to a lower altitude, there's a point where Google Earth just stops rendering the Bermuda 322 km marker, even when it's way above the horizon. $\endgroup$
    – user1902689
    May 12, 2021 at 21:08
  • $\begingroup$ I've tried visualizing ISS passes in Google Earth and have the same experience where it randomly decides to display or not display the markers $\endgroup$
    – BrendanLuke15
    May 13, 2021 at 17:44

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I played around with Google Earth Pro (it's free!) and was able to get the visualization and the math working.

Visualization:

I created a Line using the Ruler tool (with no regard for were the line was placed). Then I saved the line. Then find the line in your places and right click it and select 'Save Place As...' and save it as a .kml file (not .kmz).

Open the KML file using a text editor and edit the coordinates (near the bottom):

KML File

Important to note that the data is entered as degrees longitude, degrees latitude, elevation(units could be only meters or based on your settings).

Save this file and open it in Google Earth Pro. Find the KML file in your places and right click on it and open the properties dialog. GO to the altitude tab and use the drop down menu to select 'Absolute'. This will free the line from the surface. You can then position the camera to look 'down the line' and get a visual sense of the angle above the horizon (and obstructions with 3D buildings on).

Math:

I used Wikipedia's Ellipsoid page to turn a long/lat coordinate to an x,y,z cartesian vector (with the WGS-84 Ellipsoid as the base Earth shape). Note: use the last method shown in the above linked section (i.e., "Measuring angles directly to the surface of the ellipsoid, not to the circumscribed sphere", note that $a=b$ in these formulas).

Once you have the cartesian vectors you can find the angle above the horizon by defining two new vectors:

$V_1=2(\frac{X_{view}}{a^2} , \frac{Y_{view}}{a^2} , \frac{Z_{view}}{c^2})$, the vector normal to the ellipsoid at the viewing point, zenith/up, as described in this answer on the Math Stack Exchange

$V_2=(X_{rocket}-X_{view},Y_{rocket}-Y_{view},Z_{rocket}-Z_{view})$

The angle between these vectors will be the angle between the viewing altitude/elevation and zenith, so its complimentary angle (subtract this value form 90°) will be the altitude/elevation angle. I used the dot product:

$\theta=90°-cos^{-1}(\frac{V_1\cdot{V_2}}{|V_1||V_2|})=-0.0524°$

You will notice; however, from the Google Earth Pro visualization that it appears above the horizon. This alludes to the concept of the local horizon. Sometimes your horizon is more than 90° from zenith because of local topography.

I would suggest finding a better vantage point, but you'll likely need to go a long ways (using the given assumption of rocket location):

Location: Elevation Angle
Cleveland 1.6°
Detroit (top of GM building) 0.23°
Pittsburgh 3.5°
Washington, DC 7.3°
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    $\begingroup$ I cannot thank you enough!!! Such a detailed response! I used my exact GPS coordinates in Google Earth Pro, using a line like you say, and it works and shows the line above the horizon in it like you said. But, in it the view is completely obstructed by other houses and trees. Combined with your math, I know to have lower expectations, since I'm unfortunately not able to travel far for this. Bonus question, then! I wonder why the US Dept of State Geographer image that NASA + the media are sharing shows that Detroit and even Chicago should have a chance to see it. bit.ly/3uSWeeW $\endgroup$
    – user1902689
    May 15, 2021 at 19:58
  • $\begingroup$ Most importantly, their viewing is centered around Wallops, VA, the launch site. In retrospect, what they said is it "can reach altitudes more than 200 miles above Earth's surface", and they've said things on the live stream like they're checking weather and readiness with their observers on planes near Bermuda. Maybe it's releasing its barium vapor a lot closer to Wallops than I thought, and maybe they have observers in Bermuda that will watch it drift toward them. $\endgroup$
    – user1902689
    May 15, 2021 at 20:03
  • $\begingroup$ @user1902689 I've been reading this article, and interestingly the view circles are not concentric (implying a discernable ground track). I'm bout to open up a new question related to that. $\endgroup$
    – BrendanLuke15
    May 15, 2021 at 20:08

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