How to calculate curvature of the Earth and its effect on a 25-meter footprint? I am working on a problem that considers a remote sensing instrument hovering over the Earth with a 25 meter footprint. I'd like to see the effects of Earth's curvature on this footprint even if it's minuscule.
The Earth curves about 8cm per 1 km or $0.00007858km/1km$. I tried using this ratio and converted it to meters: $0.00007848km/1000m$ = $0.0000007848km$ of curvature per meter
$0.00000007848m \cdot 25m$
$ = 0.000001962km$ of curvature in footprint.
My advisor said this doesn't make sense to solve curvature with a linear solution since it does not behave linearly. I think he's referring to the fact that the 'footprint' of a laser shining onto the surface curves in both directions since it's a circle. It doesn't just curve in one dimension.
What would've been the correct steps to approach this problem or represent it?
It sounds like I would just add another dimension to it and scale it to something like the following:
$T(x,y)=(0.000001962m)y^2 \cdot (0.000001962m)x^2$ if x and y are in meters.
 A: I don't think your advisor is referring to the circular footprint of the laser at all.
Instead, your advisor is telling you that the expression "$8$ cm per $1$ km"
is highly misleading. If the Earth were a perfectly smooth ellipsoid and you could construct a line perfectly tangent to the surface of the Earth at some point, the surface of the Earth would "drop" approximately $8$ cm below that line when you looked at a point $1$ km from the point of tangency.
But to call this "$8$ cm per $1$ km" implies that if you went to a point $2$ km from the point of tangency, the surface of the Earth would have dropped about $16$ cm below the line. Instead, the surface will have dropped approximately $31$ cm.
You can find a few more examples at Earth Curvature Calculator.
The deflection of the Earth's surface due to curvature is approximately proportional to the square of the distance over distances of a few kilometers or less;
if you double the distance the deflection quadruples,
and if you multiply the distance by $10$ the deflection is multiplied by $100.$
The deflection at just $25$ meters from the tangent point is much, much less than you thought.
And if $25$ meters is the diameter of you footprint rather than the radius, the deflection is reduced again by a factor of four.
You can actually work out the deflection yourself by first principles using the Pythagorean Theorem. Construct a right triangle whose hypotenuse is the radius of the Earth, and one of whose legs is the radius of the your "footprint" on the surface of the Earth. The length of the other leg is the Earth's radius minus the deflection of the part of the surface covered by the footprint.
