# Is it physically possible to see mountains from 500 km away?

For less extreme, but very reliable, observations, consider some listed by Commander C. L. Garner of the Coast and Geodetic Survey in 1933 [...] He also credits the 1911 sighting of the Fairweather Mountains in Alaska [highest point is Mt. Fairweather 15 325 ft or 4670 m] from the ship Explorer from the Gulf of Alaska, 330 miles [531 km] away."

Earth’s mean average radius is 3959 miles by a right angled tangent of 330 miles gives a hypotenuse extending well above the surface of the earth of sqrt(39592 + 3302) = 3973 miles. Subtracting the radius 3959 miles gives the drop in the curvature of the earth from the beginning point where the Explorer made the sighting: 13.7 miles, or 72 492 ft.

72 492 ft minus Mt. Fairweather’s height 15 325 ft gives 57 167 ft to spare for a whole mountain range to disappear beneath the curvature of the earth.

What could account for this sighting 'through' the curve of the earth?

• I'm voting to close this question as off-topic because this has no relation to astronomy. – HDE 226868 Sep 13 '15 at 22:28
• Refraction, in this case probably anomalous refraction – Conrad Turner Sep 14 '15 at 3:51
• Hi Gerald, this seems to better fit in the Earth Science site, one of their mods has agreed it is more likely to fit in there so I will migrate this over for you now. – RhysW Sep 14 '15 at 17:20
• You've missed refraction, which will give a (smallish) range increase, and also that the viewpoint on the ship will not be at sea level, but probably some tens of metres above it. That will give a substantial increase in visual range, but I don't know whether it's enough. – Semidiurnal Simon Sep 15 '15 at 19:32
• You also have about an 8% increase is visible distance just due to normal atmospheric refraction. with a drastic temprature change the refraction can get a lot higher. en.wikipedia.org/wiki/Atmospheric_refraction – John Mar 16 '17 at 19:55