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.enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ Jun 17, 2020 at 5:26

1 Answer 1


enter image description here


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 $$ \alpha= \tan^{-1}\frac{ x \tan \theta}{R}, s = R \alpha . $$


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