Calculate the shadow on earth of a large orbital disk

Question

How to calculate the shadow on earth of a large orbital disk (low orbit)?

Answer 1

It might be easier to use similar triangles.

$$\frac{Rd}{b} = \frac{Rs}{a}$$

$Rs$ is 696,000 km and $Rd$ is 5 km. $a$ is about 150,000,000 km and c is 80 km.

The the smallest triangle is used next:

$$\frac{r}{b-c} = \frac{Rs}{a}$$

Solve for r and you get

$$r = Rd - Rs \frac{c}{a} = \text{4.63 km}.$$

Extra digits are not helpful because the exact diameter of the Sun depends on how you define it and the distance from the Sun to the Earth varies by almost +/-2 percent.

$Rs/a$ is about 0.00464 and that's also the half-angle of the Sun in radians. Convert it to degrees my multiplying by 180/pi and you get 0.266 degrees, or a quarter of a degree. The full diameter of the Sun is double that, or about a half degree.

Answer 2

Playing around with a computer algebra system, the problem actually has an exact solution, but it is ugly enough that a much more simple numerical approximation is more practical.

First, we need to find the angle of the shadow cone peak.

The peak, the centre of the Sun, and the tangent point on the Sun form a triangle with a right angle. Therefore, half the peak angle can be expressed as:

$$v = \sin^{-1}\left(\frac{r_{sun}}{r_{peak}}\right)$$

We don't have $r_{peak}$, but we have the distance from the Sun to the Earth, which is very close. This can give us a first estimate for the angle.

To correct the angle, we can calculate a new $r_{peak}$:

$$r_{peak} = r_{earth-orbit} - r_{LEO} + \frac{r_{disk}}{\tan(v)}$$

Using the new $r_{peak}$ to calculate a new $v$ should converge very quickly to the new angle.

I get $v = 0.004651$

Now, we need to find how this shadow cone projects on the Earth.

To calculate the radius of the projected disk, which has its centre slightly below the surface, we simply have to scale the disk by the cone steepness and distance between the disk, $d$.

$$r_{umbra} = r_{disk} - d\tan(v)$$

Again, we don't exactly have $d$, but the orbital altitude is very close. But we can use the $r_{umbra}$ estimate we obtained to get a better $d$:

$$d = r_{LEO} - \sqrt{r_{earth}^2 - r_{umbra}^2}$$

And yet again, the values should converge very quickly.

I get $r_{umbra} = 4.628 km$

To get the radius on the curve of the Earth instead, you can calculate the central angle and multiply with the circumference of the earth, but at 4 significant figures, the result is still $4.628 km$

Answer 3

Let's see how far we get without a calculator (so possibly only approximately).

Approximating the Sun as a point source and infinitely far away and the ground as flat, the shadow of the disk is a sharp disk of diameter $10\,\text{km}$. But the Sun is not a point source. From the top of our head, we may recall that the Sun (and the Moon) have an angular diameter of about half a degree. Converted to radians: $\frac12^\circ\cdot\frac\pi{180^\circ}\approx 0.009$. Multiply with the $80\,\text{km}$ altitude to arrive at $\approx 700\,\text{m}$ as the thickness of the penumbra annulus, which has the original disk boundary in its middle, i.e., the central shadow is $\approx 10\,\text{km}-700\,\text{m}$ wide and the penumbra to its outer edge is $\approx 10\,\text{km}+700\,\text{m}$ wide.