Wikipedia:Reference desk/Archives/Mathematics/2025 July 3

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July 3

Geometry / atmospherics query

Tonight I can see some noctilucent clouds, which are typically 80 km high in the upper atmosphere. From my position, they're low down in the sky, between 2° and 5° above the north-north-east horizon. What's the horizontal distance to them (nearest, and furthest), taking into account the curvature of the Earth? Thanks! - MPF (talk) 00:52, 3 July 2025 (UTC)[reply]

We'll take kilometres as the unit, using ${\displaystyle R=6371}$  and ${\displaystyle h=80}$ . The intersection of Earth with the plane formed by the vertical and the line of sight can be approximated by a circle with centre ${\displaystyle (0,-R)}$  and radius ${\displaystyle R,}$  so that the observer is at the origin while the clouds lie on a concentric circle having a radius of ${\displaystyle R+h.}$  The ___location of the clouds satisfies the equation ${\displaystyle x^{2}+(y+R)^{2}=(R+h)^{2},}$  in which ${\displaystyle x}$  is the horizontal distance and ${\displaystyle y}$  the height above the horizontal. Letting ${\displaystyle \alpha }$  be the angle of the line of sight with the horizontal, and putting ${\displaystyle \gamma =\cos \alpha ,\sigma =\sin \alpha ,}$  the equation for the line of sight is given by ${\displaystyle x\sigma =y\gamma .}$  Eliminating ${\displaystyle y}$  gives us:

${\displaystyle x^{2}+2\sigma \gamma Rx-\gamma ^{2}h(2R+h)=0.}$ 

Solving this quadratic equation for ${\displaystyle x}$  gives the positive root

${\displaystyle x=-\sigma \gamma R+\gamma {\sqrt {\sigma ^{2}R^{2}+h(2R+h)}}.}$ 

Plugging in ${\displaystyle \alpha =5^{\circ }}$  results in ${\displaystyle x=}$  597 km, while ${\displaystyle \alpha =2^{\circ }}$  gives us ${\displaystyle x=}$  814 km. For ${\displaystyle \alpha =0^{\circ }}$  we get ${\displaystyle x=}$  1013 km.  ​‑‑Lambiam 13:02, 3 July 2025 (UTC)[reply]

@Lambiam brilliant, many thanks! - MPF (talk) 17:34, 3 July 2025 (UTC)[reply]