I'm doing a project for a classical mechanics class about shadows cast on a sphere. Though the topic is physics-ish, the derivation is entirely math and geometry. Below is my derivation, but I'd appreciate a second pair of eyes to check it over. I've yet to really define the constraints on $\theta$ or whether to take the + or - sign of the $\pm$.
Meta: This post proposes an answer to the same question posed here but my diagram is different so don't take my setup to be the same.
In the left diagram, a vertical object of height $H$ sits normal to the surface of the sphere (reduced to a circle for simplicity). Incoming light casts a shadow (beneath $\lambda$) at an angle $\vartheta$ from the surface of the object. This shadow is a distance $\mathcal S = R\alpha$ from the base of the object, where $R$ is the radius of the sphere, and $\alpha$ is the angle from the base to the tip of the shadow, as measured from the center of the sphere.
The right diagram is a close-up of the triangle constructed in the left diagram.
From the two diagrams above it is evident that $R\cos\alpha+\varepsilon = R$.
$\implies \varepsilon = R - R\cos\alpha$
$\implies H + \varepsilon = H + R - R\cos\alpha$
Using Pythagoras on the rightmost figure we have,
$\begin{aligned}\lambda^2&=\left(H+R-R\cos\alpha\right)^2 + R^2\sin^2\alpha \\&= \left(H + R(1-\cos\alpha)\right)^2 + R^2\sin^2\alpha\\&= H^2 + 2HR(1-\cos\alpha) + R^2(1-\cos\alpha)^2 + R^2\sin^2\alpha\\&= H^2 + 2HR - 2HR\cos\alpha + R^2(1 - 2\cos\alpha + \cos^2\alpha) + R^2\sin^2\alpha\\&= H^2 + 2HR +R^2 -2HR\cos\alpha-2R^2\cos\alpha + R^2\cos^2\alpha+ R^2\sin^2\alpha\\&= H^2 + 2HR +R^2 - (2HR + 2R^2)\cos{\alpha} + R^2\\&= (H^2 + 2HR +2R^2) - (2HR + 2R^2)\cos\alpha\end{aligned}$
The rightmost figure also tells us that $\lambda\sin\vartheta = R\sin\alpha$.
$\implies \lambda^2 = \cfrac{R^2\sin^2\alpha}{\sin^2\vartheta}$ $= \cfrac{R^2}{\sin^2\vartheta}(1 - \cos^2{\alpha})$ $= \cfrac{R^2}{\sin^2\vartheta} - \cfrac{R^2}{\sin^2\vartheta}\cos^2\alpha$
Equating these two expressions for $\lambda^2$ gives,
$\cfrac{R^2}{\sin^2\vartheta} - \cfrac{R^2}{\sin^2\vartheta}\cos^2{\alpha}= (H^2 + 2HR +2R^2) - (2HR + 2R^2)\cos\alpha$
$\implies \cfrac{R^2}{\sin^2\vartheta}\cos^2\alpha- (2HR + 2R^2)\cos{\alpha} + \left(H^2 + 2HR +2R^2 - \cfrac{R^2}{\sin^2\vartheta}\right) = 0$
Noting that this result is a quadratic equation in terms of $\cos^2\alpha$, we can solve for $\cos\alpha$ using the quadratic formula:
$\cos\alpha = \cfrac{(2HR + 2R^2) \pm \sqrt{(2HR + 2R^2)^2 -4\cfrac{R^2}{\sin^2\vartheta}\left(H^2 + 2HR +2R^2 - \cfrac{R^2}{\sin^2\vartheta}\right)}}{2\cfrac{R^2}{\sin^2\vartheta}}$
Finally, using the arc-length identity $\mathcal S = R\alpha$, where $\alpha$ is measured in radians, we obtain the distance a shadow is cast as measured on a sphere (given the setup described in the figure above):
$\begin{aligned}\mathcal S&= R\arccos\left(\cfrac{(2HR + 2R^2) \pm \sqrt{(2HR + 2R^2)^2 -4\cfrac{R^2}{\sin^2\vartheta}\left(H^2 + 2HR +2R^2 - \cfrac{R^2}{\sin^2\vartheta}\right)}}{2\cfrac{R^2}{\sin^2\vartheta}}\right) \end{aligned}$ P.S. In case my professor sees this and thinks I just copied it rather than posting it, I'll state the following: The content of this post is original to J. White who's taking PHYS 331 in the Winter 2020 semester at McGill.
