Length of a Shadow Cast on a Sphere

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.

• I'm not sure if it would be shorter, but, since you have extracted a right-triangle with an altitude $R\sin\alpha$, you could also use the Euclid's theorem. Jun 17, 2020 at 5:26

Your labels for $$\alpha,\theta$$
Solve for $$x,$$ by eliminating/ plugging in for $$y$$
$$(x-(H+R))^2 +y^2= R^2,\; y = x \tan \theta \;;$$ Plug into $$\alpha= \tan^{-1}\frac{ x \tan \theta}{R}, s = R \alpha .$$