I know that the Earth isn't a sphere, not even an ellipsoid, but for my measurements, its an acceptable approximation. Assuming I have a coordinate(lat,lon) and a distance( e.g.: 1000km ), what is the surface area on the earth with that distance radius?


Just a small amplification of the answer by Ross Millikan. Use the same notation as the article he linked to. I take it that your $1000$ km is the surface of the Earth distance from the center $C$ of your circle to the furthest points $P$ from the center. In the picture linked to, $C$ is the top of the sphere, and $P$ is any point on outer edge of the bottom of the cap.

Assume that this surface of the Earth distance is $d$, and that is is $\le$ $1/4$ of the circumference of the Earth (that's not necessary, but it makes visualization easier). Let the radius of the Earth be $r$.

Then the angle $\theta$ subtended by the arc $CP$ at the centre of the Earth is given by $$\theta=\frac{d}{r}.\tag{$\ast$}$$ The "$h$" in the linked picture is given by $h=r-r\cos\theta$. The surface area is $2\pi rh$, which is $$2\pi r^2(1-\cos\theta).\tag{$\ast\ast$}$$ Compute $\theta$ using $(\ast)$, and then use $(\ast\ast)$ to find the surface area.

  • $\begingroup$ Thank you for the detailed answer. I was missing the (*) equation. $\endgroup$ – p1100i Apr 23 '12 at 13:53
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    $\begingroup$ Actually this answer is incorrect. θ = (d / r), in radians, or θ = (d * 360 / (2 πr)), in degrees. $\endgroup$ – Hans Brende Feb 22 '16 at 3:17
  • $\begingroup$ Unless, of course, you are speaking in turns ;) $\endgroup$ – Hans Brende Feb 24 '16 at 5:55
  • $\begingroup$ @HansBrende: Thank you. I will look at it again, tomorrow perhaps. $\endgroup$ – André Nicolas Feb 24 '16 at 6:05
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    $\begingroup$ @HansBrende: Thank you for spotting the incorrect $\theta$. Fixed. $\endgroup$ – André Nicolas Mar 1 '16 at 6:12

You could look at Spherical Cap for the formula. If you treat the earth as a sphere, the coordinates of the center do not matter, just the radius and the radius of the earth.

  • $\begingroup$ Thx for the reply, if there would be a distributed solution option I would give the half for u, but since Andrés answer was more detailed... $\endgroup$ – p1100i Apr 23 '12 at 13:55

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