<p><a href="https://i.sstatic.net/gCxfE.png" rel="noreferrer"><img src="https://i.sstatic.net/gCxfE.png" alt="enter image description here"></a></p>

<p>It might be easier to use <a href="https://en.wikipedia.org/wiki/Similarity_(geometry)#Similar_triangles" rel="noreferrer">similar triangles</a>. </p>

<p><span class="math-container">$$\frac{Rd}{b} = \frac{Rs}{a}$$</span></p>

<p><span class="math-container">$Rs$</span> is 696,000 km and <span class="math-container">$Rd$</span> is 5 km. <span class="math-container">$a$</span> is about 150,000,000 km and c is 80 km.</p>

<p>The the smallest triangle is used next:</p>

<p><span class="math-container">$$\frac{r}{b-c} = \frac{Rs}{a}$$</span></p>

<p>Solve for r and you get</p>

<p><span class="math-container">$$r = Rd - Rs \frac{c}{a} = \text{4.63 km}.$$</span></p>

<p>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.</p>

<p><span class="math-container">$Rs/a$</span> 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. </p>

CC BY-SA 4.0
2019-10-03T17:22:54.007Z
This is linear optics, which should be fine.  It would be interesting to calculate the Spot of Arago for various   $R_d$  .
Carl Witthoft
CC BY-SA 4.0
2019-10-03T17:28:55.057Z
I considered this, but it doesn't generalize quite as nicely. The one that breaks down first is that c is not going to stay at 80km for larger discs. It also doesn't work close to the Sun, but for this specific case this is indeed all you need.
SE - stop firing the good guys
CC BY-SA 4.0
2019-10-03T18:34:37.540Z
A priori, you cannot know that $a$ is about 150 million km - for example, what if the disk had the same diameter as the Sun? Instead, $(R_d-r) : (R_s-R_d)  = c : (a-b)$ and $a-b$ is the distance from Sun to disk (so *that* is approximately 150 million km)
Hagen von Eitzen
CC BY-SA 4.0
2019-10-03T20:29:59.140Z
Thank you very much for your reply. Your information was very useful to me. They confirmed my hypothesis.

Do you think an orbit of less than 80 km could be used to place such a disk? What could be the lowest orbit, in your opinion,  low altitude circular view, on which such a disk could be placed?
Ion Corbu
CC BY-SA 4.0
2019-10-03T20:30:23.067Z
To keep such a disk in orbit would be two solutions:

1. The disk has an orbital speed of about 21,000 km / hour. That is, it rotates around the Earth in about 90 minutes. Possible solution but that does not interest me.
Ion Corbu
CC BY-SA 4.0
2019-10-03T20:30:41.857Z
2. To have propellants (thermochemical, electrical, electromagnetic, ionic. Photonic) that would be the force required to keep such a disk in orbit (80 km) above a fixed point. To counteract gravitational force. And to prevent the disc from falling into orbit and its disintegration. The specific weight of such a disc could be about 10kg / m2
Ion Corbu
CC BY-SA 4.0
2019-10-04T01:40:31.377Z
That's a different question, but things at 80 km will not stay in orbit very long due to atmospheric drag. If you'd like to explore that, please ask a separate question.
uhoh
CC BY-SA 4.0
2019-10-03T11:47:52.743Z
2019-10-03T12:38:27.927Z
2019-10-03T12:38:27.927Z
uhoh
<p>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.</p>

<p>First, we need to find the angle of the shadow cone peak.</p>

<p>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:</p>

<p><span class="math-container">$$v = \sin^{-1}\left(\frac{r_{sun}}{r_{peak}}\right)$$</span></p>

<p>We don't have <span class="math-container">$r_{peak}$</span>, 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.</p>

<p>To correct the angle, we can calculate a new <span class="math-container">$r_{peak}$</span>:</p>

<p><span class="math-container">$$r_{peak} = r_{earth-orbit} - r_{LEO} + \frac{r_{disk}}{\tan(v)}$$</span></p>

<p>Using the new <span class="math-container">$r_{peak}$</span> to calculate a new <span class="math-container">$v$</span> should converge very quickly to the new angle.</p>

<p>I get <span class="math-container">$v = 0.004651$</span></p>

<p>Now, we need to find how this shadow cone projects on the Earth.</p>

<p>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, <span class="math-container">$d$</span>.</p>

<p><span class="math-container">$$r_{umbra} = r_{disk} - d\tan(v)$$</span></p>

<p>Again, we don't exactly have <span class="math-container">$d$</span>, but the orbital altitude is very close. But we can use the <span class="math-container">$r_{umbra}$</span> estimate we obtained to get a better <span class="math-container">$d$</span>:</p>

<p><span class="math-container">$$d = r_{LEO} - \sqrt{r_{earth}^2 - r_{umbra}^2}$$</span></p>

<p>And yet again, the values should converge very quickly.</p>

<p>I get <span class="math-container">$r_{umbra} = 4.628 km$</span></p>

<p>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 <span class="math-container">$4.628 km$</span></p>

CC BY-SA 4.0
2019-10-03T09:53:02.443Z
This appears to assume that the Sun and the disk are directly  overhead. When this is not true, which will be most of the time, the shadow will be something more like an ellipse, but, because the shadow will be further from the disk, the region of full shadow may be smaller or non-existent.
Steve Linton
CC BY-SA 4.0
2019-10-03T10:13:06.083Z
It does indeed assume that, as that's what the OP appears to assume in their diagram.
SE - stop firing the good guys
CC BY-SA 4.0
2019-10-03T20:28:12.867Z
Thank you very much for your reply. Your information was very useful to me. They confirmed my hypothesis.

Do you think an orbit of less than 80 km could be used to place such a disk? What could be the lowest orbit, in your opinion,  low altitude circular view, on which such a disk could be placed?
Ion Corbu
CC BY-SA 4.0
2019-10-03T20:28:51.610Z
To keep such a disk in orbit would be two solutions:

1. The disk has an orbital speed of about 21,000 km / hour. That is, it rotates around the Earth in about 90 minutes. Possible solution but that does not interest me.
Ion Corbu
CC BY-SA 4.0
2019-10-03T20:29:19.927Z
2. To have propellants (thermochemical, electrical, electromagnetic, ionic. Photonic) that would be the force required to keep such a disk in orbit (80 km) above a fixed point. To counteract gravitational force. And to prevent the disc from falling into orbit and its disintegration. The specific weight of such a disc could be about 10kg / m2
Ion Corbu
CC BY-SA 4.0
2019-10-03T08:08:45.047Z
2019-10-03T08:08:45.047Z
SE - stop firing the good guys
<p>Let's see how far we get without a calculator (so possibly only approximately).</p>

<p>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 <span class="math-container">$10\,\text{km}$</span>.
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: <span class="math-container">$\frac12^\circ\cdot\frac\pi{180^\circ}\approx 0.009$</span>. Multiply with the <span class="math-container">$80\,\text{km}$</span> altitude to arrive at <span class="math-container">$\approx 700\,\text{m}$</span> as the thickness of the penumbra annulus, which has the original disk boundary in its middle, i.e., the central shadow is <span class="math-container">$\approx 10\,\text{km}-700\,\text{m}$</span> wide and the penumbra to its outer edge is <span class="math-container">$\approx 10\,\text{km}+700\,\text{m}$</span> wide.</p>

CC BY-SA 4.0
2019-10-03T20:23:22.677Z
2019-10-03T20:23:22.677Z
Hagen von Eitzen
<p>How to calculate the shadow on earth of a large orbital disk (low orbit)?<a href="https://i.sstatic.net/sZZ1w.png" rel="noreferrer"><img src="https://i.sstatic.net/sZZ1w.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-10-03T04:43:50.497Z
For someone who wishes to explain the mathematics, I've done the approximate calculations for half of the problem: the umbra on the Earth's surface should be ~9.25 km in diameter, assuming the curvature of the Earth is negligible (which is an assumption I really shouldn't be making.)
Cloudy7
<amateur-observing>
Calculate the shadow on earth of a large orbital disk
Ion Corbu
# Calculate the shadow on earth of a large orbital disk

(Asked by: username_0 on 2019-10-03)

How to calculate the shadow on earth of a large orbital disk (low orbit)?[![enter image description here](https://i.sstatic.net/sZZ1w.png)](https://i.sstatic.net/sZZ1w.png)

**Tags:** amateur-observing

> **Comments:**
> **username_1**: For someone who wishes to explain the mathematics, I've done the approximate calculations for half of the problem: the umbra on the Earth's surface should be ~9.25 km in diameter, assuming the curvature of the Earth is negligible (which is an assumption I really shouldn't be making.)  

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## 3 answers

### Answer by username_2 on 2019-10-03. Score: 9

[![enter image description here](https://i.sstatic.net/gCxfE.png)](https://i.sstatic.net/gCxfE.png)

It might be easier to use [similar triangles](https://en.wikipedia.org/wiki/Similarity_(geometry)#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.

> **Comments:**
> **username_3**: This is linear optics, which should be fine.  It would be interesting to calculate the Spot of Arago for various   $R_d$  .  
> **username_4**: I considered this, but it doesn't generalize quite as nicely. The one that breaks down first is that c is not going to stay at 80km for larger discs. It also doesn't work close to the Sun, but for this specific case this is indeed all you need.  
> **username_5**: A priori, you cannot know that $a$ is about 150 million km - for example, what if the disk had the same diameter as the Sun? Instead, $(R_d-r) : (R_s-R_d)  = c : (a-b)$ and $a-b$ is the distance from Sun to disk (so *that* is approximately 150 million km)  
> **username_0**: Thank you very much for your reply. Your information was very useful to me. They confirmed my hypothesis.  Do you think an orbit of less than 80 km could be used to place such a disk? What could be the lowest orbit, in your opinion,  low altitude circular view, on which such a disk could be placed?  
> **username_0**: To keep such a disk in orbit would be two solutions:  1. The disk has an orbital speed of about 21,000 km / hour. That is, it rotates around the Earth in about 90 minutes. Possible solution but that does not interest me.  
> **username_0**: 2. To have propellants (thermochemical, electrical, electromagnetic, ionic. Photonic) that would be the force required to keep such a disk in orbit (80 km) above a fixed point. To counteract gravitational force. And to prevent the disc from falling into orbit and its disintegration. The specific weight of such a disc could be about 10kg / m2  
> **username_2**: That's a different question, but things at 80 km will not stay in orbit very long due to atmospheric drag. If you'd like to explore that, please ask a separate question.  

---

### Answer by username_4 on 2019-10-03. Score: 5

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 $4.628 km$

> **Comments:**
> **username_6**: This appears to assume that the Sun and the disk are directly  overhead. When this is not true, which will be most of the time, the shadow will be something more like an ellipse, but, because the shadow will be further from the disk, the region of full shadow may be smaller or non-existent.  
> **username_4**: It does indeed assume that, as that's what the OP appears to assume in their diagram.  
> **username_0**: Thank you very much for your reply. Your information was very useful to me. They confirmed my hypothesis.  Do you think an orbit of less than 80 km could be used to place such a disk? What could be the lowest orbit, in your opinion,  low altitude circular view, on which such a disk could be placed?  
> **username_0**: To keep such a disk in orbit would be two solutions:  1. The disk has an orbital speed of about 21,000 km / hour. That is, it rotates around the Earth in about 90 minutes. Possible solution but that does not interest me.  
> **username_0**: 2. To have propellants (thermochemical, electrical, electromagnetic, ionic. Photonic) that would be the force required to keep such a disk in orbit (80 km) above a fixed point. To counteract gravitational force. And to prevent the disc from falling into orbit and its disintegration. The specific weight of such a disc could be about 10kg / m2  

---

### Answer by username_5 on 2019-10-03. Score: 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.


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