Editing Horizon (section)

== Distance to the horizon {{anchor|Distance}}==

Ignoring the [[#Effect of atmospheric refraction|effect of atmospheric refraction]], distance to the true horizon from an observer close to the Earth's surface is about<ref name="ATYoungDistToHoriz">
{{cite news |url=http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html |title=Distance to the Horizon |work=Green Flash website (Sections: Astronomical Refraction, Horizon Grouping) |author=Young, Andrew T. |publisher=San Diego State University Department of Astronomy |access-date=April 16, 2011 |url-status=live |archive-url=https://web.archive.org/web/20031018020513/http://mintaka.sdsu.edu/GF/explain/atmos_refr/horizon.html |archive-date=October 18, 2003 }}</ref>

:<math>d \approx \sqrt{2\,h\,R} \,,</math>

where ''h'' is height above sea level and ''R'' is the [[Earth radius]].

The expression can be simplified as:

:<math>d \approx k \sqrt{h} \,,</math>

where the constant equals ''k''{{=}}{{val|3.57|u=km/m<sup>½</sup>}}{{=}}{{val|1.22|u=mi/ft<sup>½</sup>}}.
In this equation, Earth's surface is assumed to be perfectly spherical, with ''R'' equal to about {{Convert|6371|km}}.

===Examples===
Assuming no [[#Effect of atmospheric refraction|atmospheric refraction]] and a spherical Earth with radius R={{Convert|6371|km}}:
* For an observer standing on the ground with ''h'' = {{Convert|1.70|m}}, the horizon is at a distance of {{Convert|4.7|km}}.
* For an observer standing on the ground with ''h'' = {{Convert|2|m}}, the horizon is at a distance of {{Convert|5|km}}.
* For an observer standing on a hill or tower {{Convert|30|m}} above sea level, the horizon is at a distance of {{Convert|19.6|km}}.
* For an observer standing on a hill or tower {{Convert|100|m}} above sea level, the horizon is at a distance of {{Convert|36|km}}.
* For an observer standing on the roof of the [[Burj Khalifa]], {{Convert|828|m}} from ground, and about {{Convert|834|m}} above sea level, the horizon is at a distance of {{Convert|103|km}}.
* For an observer atop [[Mount Everest]] ({{Convert|8848|m}} in altitude), the horizon is at a distance of {{Convert|336|km}}.
* For an observer aboard a commercial passenger plane flying at a typical altitude of {{Convert|35000|ft}}, the horizon is at a distance of {{Convert|369|km}}.
* For a [[Lockheed U-2|U-2]] pilot, whilst flying at its service ceiling {{Convert|21000|m}}, the horizon is at a distance of {{Convert|517|km}}.

=== Other planets ===
On terrestrial planets and other solid celestial bodies with negligible atmospheric effects, the distance to the horizon for a "standard observer" varies as the square root of the planet's radius. Thus, the horizon on [[Mercury (planet)|Mercury]] is 62% as far away from the observer as it is on Earth, on [[Mars]] the figure is 73%, on the [[Moon]] the figure is 52%, on [[Mimas (moon)|Mimas]] the figure is 18%, and so on.

=== Derivation ===
[[File:CircleChordTangent.png|thumb|right|300px|Geometrical basis for calculating the distance to the horizon, tangent-secant theorem]]
[[File:GeometricDistanceToHorizon.png|thumb|right|300px|Geometrical distance to the horizon, Pythagorean theorem]]
[[File:Horizons.svg|thumb|right|300px|Three types of horizon]]

If the Earth is assumed to be a featureless sphere (rather than an [[Earth ellipsoid|oblate spheroid]]) with no atmospheric refraction, then the distance to the horizon can easily be calculated.<ref>{{cite web |url=http://blogs.discovermagazine.com/badastronomy/2009/01/15/how-far-away-is-the-horizon/#.WNp_MGczVEY |title=How far away is the horizon? |last=Plait |first=Phil |date=15 January 2009 |website=Discover |series=Bad Astronomy |publisher=Kalmbach Publishing Co. |access-date=2017-03-28 |url-status=live |archive-url=https://web.archive.org/web/20170329050848/http://blogs.discovermagazine.com/badastronomy/2009/01/15/how-far-away-is-the-horizon/#.WNp_MGczVEY |archive-date=29 March 2017 }}</ref>

The [[tangent-secant theorem]] states that
:<math>\mathrm{OC}^2 = \mathrm{OA} \times \mathrm{OB} \,.</math>
Make the following substitutions:
* ''d'' = OC = distance to the horizon
* ''D'' = AB = diameter of the Earth
* ''h'' = OB = height of the observer above sea level
* ''D+h'' = OA = diameter of the Earth plus height of the observer above sea level,

with ''d, D,'' and ''h'' all measured in the same units. The formula now becomes
:<math>d^2 = h(D+h)\,\!</math>
or
:<math>d = \sqrt{h(D+h)} =\sqrt{h(2R+h)}\,,</math>

where ''R'' is the radius of the Earth.

The same equation can also be derived using the [[Pythagorean theorem]].
At the horizon, the line of sight is a tangent to the Earth and is also perpendicular to Earth's radius. 
This sets up a right triangle, with the sum of the radius and the height as the hypotenuse. 
With

* ''d'' = distance to the horizon
* ''h'' = height of the observer above sea level
* ''R'' = radius of the Earth

referring to the second figure at the right leads to the following:

:<math>(R+h)^2 = R^2 + d^2 \,\!</math>
:<math>R^2 + 2Rh + h^2 = R^2 + d^2 \,\!</math>
:<math>d = \sqrt{h(2R + h)} \,.</math>

The exact formula above can be expanded as:

:<math>d = \sqrt{2Rh + h^2} \,,</math>

where ''R'' is the radius of the Earth (''R'' and ''h'' must be in the same units). For example,
if a satellite is at a height of 2000&nbsp;km, the distance to the horizon is {{convert|5430|km}};
neglecting the second term in parentheses would give a distance of {{Convert|5048|km}}, a 7% error.

=== Approximation ===
[[File:horizon_distance_graphs.svg|thumb|250px|Graphs of distances to the true horizon on Earth for a given height ''h''. ''s'' is along the surface of the Earth, ''d'' is the straight line distance, and ''~d'' is the approximate straight line distance assuming ''h'' &lt;&lt; the radius of the Earth, 6371 km. In [http://upload.wikimedia.org/wikipedia/commons/4/44/Horizon_distance_graphs.svg the SVG image], hover over a graph to highlight it.]]
<!-- [[File:How far away is the horizon.png|thumb|right|300px]] -->
If the observer is close to the surface of the Earth, then ''h'' is a negligible fraction of ''R'' and can be disregarded the term {{nowrap|(2''R'' + ''h'')}}, and the formula becomes-

:<math>d = \sqrt{2Rh} \,.</math>

Using kilometres for ''d'' and ''R'', and metres for ''h'', and taking the radius of the Earth as 6371&nbsp;km, the distance to the horizon is

:<math>d \approx \sqrt{2\cdot6371\cdot{h/1000}} \approx 3.570\sqrt{h} \,</math>.

Using [[imperial units]], with ''d'' and ''R'' in [[statute mile]]s (as commonly used on land), and ''h'' in feet, the distance to the horizon is

:<math>d \approx \sqrt{2\cdot3963\cdot{h/5280}} \approx \sqrt{1.5h} \approx 1.22 \sqrt{h} </math>.

If ''d'' is in [[nautical mile]]s, and ''h'' in feet, the constant factor is about 1.06, which is close enough to 1 that it is often ignored, giving:

:<math>d \approx \sqrt h </math>

These formulas may be used when ''h'' is much smaller than the radius of the Earth (6371&nbsp;km or 3959&nbsp;mi), including all views from any mountaintops, airplanes, or high-altitude balloons. With the constants as given, both the metric and imperial formulas are precise to within 1% (see the next section for how to obtain greater precision).
If ''h'' is significant with respect to ''R'', as with most satellites, then the approximation is no longer valid, and the exact formula is required.