I'm asked to use "simple geometry" and appropriate approximations of astronomical distances and radii to explain why the shadow cast by Earth on Moon is only about 2.7 times of the Moon size, while in reality Earth is about 3.7 times as big. I'm really stuck with this problem, because there appears to be no way that I'm doing anything incorrectly, yet the results I've been getting are absurd.

My approach is this:

We can draw an isosceles triangle with the base of $2R_{Sun}$. Somewhere in this triangle there is Earth's diameter parallel to the base, as well as the Earth shadow's diameter parallel to the base.

Now, we need to find the height of the similar triangle, the line going from the Moon diameter to the vertex of the triangle, that is the median of this similar triangle. Let this be $y$. Once $y$ is found, we can find the radius of the Earth shadow on Moon; let it be $m$. The result will be the division of $R_{Earth}$ by $m$.

The constants as I defined for convenience are as follows:

$$d_{e-m}=380$$ $$d_{e-s}=1.5\times 10^5$$ $$R_{sun}=695.7$$ $$R_{earth} = 6.4269$$

Now, we can setup two equations, where $(d_{e-s}$ is the distance from Earth to Sun, $d_{e-m}$ is the distance from Earth to Moon, $h$ is the hypotenuse of the half of the isosceles triangle, and $x$ is the difference of $h$ and the hypotenuse of the similar triangle which is a half of the isosceles triangle, which has the base $R_{earth}$:

$$(d_{e-s}+d_{e-m}+y)^2 + R_{sun}^2 = h^2$$ $$R_{earth}^2+(d_{e-m}+y)^2=(h-x)^2$$

$x$ is found this way:


so $x=150002$.

Now, typing the above system of two equations into Wolfram Alpha I get two solutions, with $y$ being negative in every one of them. I've double-checked many times, and my setup is absolutely correct!

Please help!


1 Answer 1


We have $$ \frac{R_{sun}-R_{earth}}{d_{earth-sun}}=\frac{R_{earth}-R_{shadow}}{d_{earth-moon}}$$ hence $$\begin{align}R_{shadow}&= R_{earth}-\frac{R_{sun}-R_{earth}}{d_{earth-sun}}\cdot d_{earth-moon}\\ &=6371\,\text{km}-\frac{695700\,\text{km}-6371\,\text{km}}{149600000\,\text{km}}\cdot {384400\,\text{km}}\\ &\approx 4600\,\text{km}\\&\approx 0.72\cdot R_{earth}\end{align}$$ (and $\frac {2.7}{3.7}\approx 0.72$ as well).

  • $\begingroup$ Thanks. I didn't know about this relation. $\endgroup$
    – sequence
    Commented Jan 29, 2017 at 0:49

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