Surface area of intersection of sphere and off-angle cone (Footprint of satellite) A satellite has a telescope looking down at a section of the Earth, so its field of view is a cone with some angle φ. The telescope can slew any direction θ (non-perpendicular to Earth), and doing so will change the surface area in its field of view. What is the equation for that surface area? You can assume that the earth is a sphere of radius $r$ and the satellite is $h$ above the ground.
Here is an illustration of the parameters as well as the surface area in light blue.

 A: I do not have an answer, but I got interested in the problem so I thought I might share a couple of pictures I drew with Mathematica.
Fix some cartesian coordinate system in $\mathbb{R}^3$, centered at the center of the sphere. After some rotations, you can assume that the satellite is looking down a line parallel to the $z$-axis and perpendicular to the $x,y$-plane. I will call $\vartheta$ the angle between this line and the segment joining the satellite with the center of the sphere, this is $\pi/2$ minus the original $\vartheta$ of the question. So $\vartheta$ and the height $h$ give us the position $v$ of the satellite, while its "view cone" has amplitude $\varphi$.
A point $p$ on the sphere can be seen from the satellite if and only if these two conditions hold:

*

*the vectors $p-v$ and $(0,0,-1)$ must form an angle smaller than $\varphi$;

*the point $p$ is not "over the orizon" with respect to $v$, this is equivalent to saying that the distance between $p$ and $v$ should be less than $\sqrt{(r+h)^2-r^2}$, where $r$ is the radius of the sphere.

With this in place, now it is quite easy to draw a picture of the viewed area. I will leave the Mathematica code below if you want to play around a bit.
As for the area, I have just been able to compute some values. I do not know if it admits an analytic expression, this would be interesting! Anyway, I assume that $r=1$, $\varphi=\pi/6$ and $h=0.5$, and see how the area changes when the "tilting" $\vartheta$ becomes more pronounced, from $0$ to $1$ radians. I think there should be an analytic expression for the value of $\vartheta$ which realizes the maximal area, it would be interesting to see!
Fig.1 : the satellite is at the tip of the cone, which represents its "field of view" (it should be infinitely tall, but it made for an ugly picture). The meshed yellow region is the part of the surface which can be seen from the satellite.
Fig.2 : Area on vertical axis, $\vartheta$ on the horizontal one, for $\varphi=\pi/6$ and $h=r/2$, $r=1$.

$\vartheta$ on the horizontal one, for $\varphi=\pi/6$ and $h=r/2$, $r=1$." />
Some Mathematica code to draw the viewed area:
vert[h_, \[Theta]_] := {0, (1 + h)*Sin[\[Theta]], (1 + h)*Cos[\[Theta]]}
c0[h_, \[Theta]_] := {0, (1 + h)*Sin[\[Theta]], 0}
angle[h_, \[Theta]_, u_, v_] := ArcCos[((1 + h)*Cos[\[Theta]] - Sin[v])/Sqrt[Abs[Cos[u]*Cos[v]]^2 + (-((1 + h)*Cos[\[Theta]]) + Sin[v])^2 + (Cos[v]*Sin[u] - (1 + h)*Sin[\[Theta]])^2]]
    Manipulate[Show[
Graphics3D[{Opacity[0.6], Sphere[], Green, Opacity[0.3], Cone[{c0[h, \[Theta]], vert[h, \[Theta]]},(1 + h)*Cos[\[Theta]]*Tan[\[Phi]]]}, Boxed -> False],
ParametricPlot3D[{Cos[v]*Cos[u], Cos[v]*Sin[u], Sin[v]},{u, 0, Pi}, {v, 0, 2*Pi}, RegionFunction -> Function[{x, y, z, u, v}, angle[h, \[Theta], u, v] <= \[Phi] && Norm[{Cos[v]*Cos[u], Cos[v]*Sin[u], Sin[v]} - vert[h, \[Theta]]]^2 <= h^2 + 2*h], PlotPoints -> 100]],{{h, 0.3}, 0, 1}, {{\[Theta], Pi/4}, 0, Pi/2}, {{\[Phi], Pi/4}, 0, Pi/2}]

Some Mathematica code to compute the area (not particularly efficient):
    vert[h_, \[Theta]_] := {0, (1 + h)*Sin[\[Theta]], (1 + h)*Cos[\[Theta]]}
ViewedArea[h_, \[Theta]_, \[Phi]_] := ImplicitRegion[x^2 + y^2 + z^2 == 1 && ArcCos[(-z + (1 + h)*Cos[\[Theta]])/Sqrt[x^2 + (z - (1 + h)*Cos[\[Theta]])^2 + (y - (1 + h)*Sin[\[Theta]])^2]] <= \[Phi] && 
    Norm[{x, y, z} - vert[h, \[Theta]]]^2 <= h^2 + 2*h, {x, y, z}]

