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To see Hawaii from Texas how high in altitude would you have to be? I know the curvature of the earth has a big role in this

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TL;DR: It depends on what one means by Hawaii, what one means by Texas, and whether one is concerned with atmospheric effects. The answer is somewhere between less than 2400 km and over 12600 km.

A note regarding the edit

This answer is a significant change from the previous answer where I focused solely on Honolulu and El Paso. I also made a mistake in that previous version of the answer; I used the longitude of El Paso where I should have used its latitude.

Assumptions

I'll be

  • Assuming a spherical Earth with a radius of 6371 km
  • Completely ignoring atmospheric effects
  • Assuming "seeing" means the altitude at which the point in question is barely visible on the limb of the Earth as seen from that altitude.

Locations

The Hawaiian Archipelago is a very long chain of islands, stretching from Hawaii Island to Kure Atoll.I'll be using the following locations in the Hawaii Archipelago :

Location Coordinates
Mauna Kea 19°49'14.4"N 155°28'5.0"W
Honolulu 21°18'25"N 157°51'30"W
Ni'ihau 21°54'0"N 160°10'0"W
Midway Atoll 28°12'27"N 177°21'0"W
Kure Atoll 28°25'0"N 178°20'0"W

The peak of Mauna Kea on Hawaii Island is about 4.2 km above sea level. I'll be using that elevation for Mauna Kea; for all other locations I'll be using sea level.

I'll be using the following locations in Texas:

Location Coordinates
El Paso 31°45'33"N 106°29'19"W
Sweetwater 32°28'5"N 100°24'26"W
Texarkana 33°26'14"N 94°4'3"W

EL Paso is about as far west as one can get in Texas; it's closer to Los Angeles than it is to Texarkana. I'm guessing Texarkana is the furthest location in Texas from the Hawaiian islands. Sweetwater is about halfway between El Paso and Texarkana.

The elevation of the Texas locales is irrelevant as I'll be reporting height above sea level (height above 6371 km).

Mathematics

"Barely seeing" a point at sea level from another point at elevation requires calculating the angular separation between the two points via

$$\cos\theta = \sin\phi_1\sin\phi_2 + \cos\phi_1\cos\phi_2\cos(\lambda_1-\lambda_2) \tag{1}$$ where $\theta$ is the angular separation, $\phi_1$ and $\phi_2$ are the latitudes of the two points, and $\lambda_1$ and $\lambda_2$ are the longitudes of the two points.

Since we're just barely seeing the point in Hawaii, the angle between the line segment from the center of the Earth to the point in Hawaii and the line segment from the point in Hawaii to the observation point (the point above the location in Texas) is 90°. This makes the triangle formed by the center of the Earth, the point in Hawaii, and the observation point a right triangle

Denoting $R=6371~\text{km}$ as the radius of the spherical Earth, $h$ as the height above sea level the Earth at the point in question, and $\theta$ as calculated via equation (1), this means $$\cos\theta = \frac R{R+h} \tag{2}$$ or $$h = R\left(\frac1{\cos\theta}-1\right) \tag{3}$$

For locations in Hawaii at sea level (all but Mauna Kea), equations (1) and (3) suffice to find the requisite altitude. A couple of extra steps are needed for Mauna Kea: Use equation (2) to the angular separation at which a point 4.2 km above sea level just becomes visible from sea level. (The answer is 2.0799° with the spherical Earth assumption, which corresponds to about 231.2 km on the surface of the Earth.) This reduces the altitude at the Texas site as all that is needed to be able to see that point at sea level.

Results

The following table portrays the angular separation between the points in Hawaii and the points in Texas (first line in each table cell) and the altitude above sea level for the point in Texas needed to barely see the point in Hawaii (the second line in each cell). For Mauna Kea, the table also reports in parentheses the reduced angular distance (the angular distance via equation (1) less 2.0799°).

El Paso Sweetwater Texarkana
Mauna Kea 45.294° (43.214°)
2371 km
50.462° (48.382°)
3222 km
55.762° (53.682°)
4386 km
Honolulu 46.702°
2919 km
51.833°
3939 km
57.072°
5349 km
Niʻihau 48.424°
3230 km
53.525°
4346 km
58.720°
5899 km
Midway Atoll 60.367°
6514 km
65.057°
8736 km
69.728°
12017 km
Kure Atoll 61.086°
6806 km
65.752°
9142 km
70.394°
12616 km
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    $\begingroup$ Hawaii's highest point is about 4km though, so you can estimate a bit less. Still, probably negligible in spite of other effects. $\endgroup$
    – yo'
    Feb 5 at 14:09
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    $\begingroup$ @yo' I took that into account in my significant update to the answer. I made a mistake in that answer, and it needed correction. In correcting the answer, I took the opportunity to address you concern. $\endgroup$ Feb 7 at 13:52
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    $\begingroup$ +1 for going "above" and beyond with the table. It is amazing to know that you need to be 12000 km above eastern-most Texas to see western-most Hawaii. $\endgroup$
    – JohnHoltz
    Feb 7 at 16:28
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    $\begingroup$ Oh look, star.nesdis.noaa.gov/goes/… ... 22300 miles up = 35900 km... and there's Hawaii and Texas, and some extra because you're about 3 times higher :) $\endgroup$ Feb 8 at 5:09
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    $\begingroup$ @JeopardyTempest At some point increased elevation attains less and less increased coverage. The maximum angular separation at 12600 km is about 70.4°. From geostationary altitude (e.g., GOES-West), it's about 81.3°. From about 1.5 million kilometers up (e.g., DSCOVR), it's about 89.8°. Even from an infinite distance, it's 90°. Point of diminishing returns. $\endgroup$ Feb 8 at 7:31

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