<p>The Earth's radius is about 6371 km, while the moon is 1737.4 km. The distance between the two bodies is 384399 km. The formula for angular size (in radians) is <span class="math-container">$\delta = c$</span>. Letting <span class="math-container">$x$</span> be equal to the distance from Earth, we get the equation <span class="math-container">$2 \arctan\Big(\frac{12742}{2x}\Big)=2 \arctan\Big(\frac{3474.8}{2(384399-x)}\Big)$</span>. Solving this equation gives <span class="math-container">$\frac{4081676715}{13514}\approx 302033\ \text{km}$</span>. Plugging in this value, we get <span class="math-container">$2^\circ\ 25' \ 1.78''$</span> as the angular diameter of the Moon, compared to <span class="math-container">$31' \ 5''$</span> as viewed from Earth, which is 4.66583 times smaller.</p>

CC BY-SA 4.0
2021-12-07T15:53:52.870Z
…should there be 2 solutions to the equation? One on the line segment between their centers (this answer), and one on the line extending beyond the moon… the point of the cone defined by the circumferential circles of the two bodies (sorry, that's a bad descrition.)
Craig Constantine
CC BY-SA 4.0
2021-12-07T16:38:43.350Z
@CraigConstantine There's only one solution. See https://www.desmos.com/calculator/cmnvyngk8i
WarpPrime
CC BY-SA 4.0
2021-12-07T17:16:15.177Z
There ARE multiple answers to the general question "from where is space". Craig's suggestion is one of those locations when on the line passing through the Earth and Moon but not in between the Earth and Moon. You answered the specific question of "on the line between the Earth and Moon". Also, 6371 and 1737 km are the radius of the Earth and Moon, not the diameter.
JohnHoltz
CC BY-SA 4.0
2021-12-07T17:25:59.603Z
@CraigConstantine There is only one solution to fasterthanlight's equation, because we need a different equation for "observer-Moon-Earth" colinearity: $2 \arctan\Big(\frac{12742}{2(384399+x)}\Big)=2 \arctan\Big(\frac{3474.8}{2x}\Big)$.  I get 144,133 km.  Also there are an infinite number of non-colinear solutions.  Also they will appear the same size when an observer is sufficiently far away that the angular diameter of both bodies is smaller than the minimum angular resolution of the observer, i.e. they both appear as point sources.
Connor Garcia
CC BY-SA 4.0
2021-12-07T17:27:29.570Z
@fasterthanlight That plot shows negative angular size, which makes no sense - you need an absolute value in there. At very large distances the earth will have larger angular size since it is actually larger, but at distances just outside the moon's orbit the moon will of course appear larger. To continuously transition from one to the other (moon larger to earth larger), they must have equal angular size at some point outside the moon's orbit. Craig is correct, there is a second solution.
Nuclear Hoagie
CC BY-SA 4.0
2021-12-07T19:46:29.053Z
For spheres, you need to use arcsin, not arctan. See https://en.wikipedia.org/wiki/Angular_diameter
PM 2Ring
CC BY-SA 4.0
2021-12-07T20:10:59.880Z
The set of all points of equal angular size appear to be a three-dimensional ovoid [centered behind the moon.](https://www.geogebra.org/calculator/qxbr8c4s)
notovny
CC BY-SA 4.0
2021-12-07T20:45:31.577Z
@notovny Your comment would make an outstanding second answer to this question.
Connor Garcia
CC BY-SA 4.0
2021-12-07T21:07:05.237Z
Silly number of significant figures.
ProfRob
CC BY-SA 4.0
2021-12-07T22:17:54.310Z
@ConnorGarcia Perhaps if I've had some more time to look at the math; I'm not certain, but investigating the GeoGebra shape makes it look like the result I called an ovoid is at least very, very close to a being a sphere about 113 000 km in radius, centered on a point 415 000 km from Earth.
notovny
CC BY-SA 4.0
2021-12-07T13:55:11.423Z
2021-12-07T17:18:57.537Z
2021-12-07T17:18:57.537Z
WarpPrime
<p>When viewing a sphere of radius <span class="math-container">$r$</span> at a distance <span class="math-container">$d$</span> from the centre of the sphere, you <em>don't</em> see a circle of radius <span class="math-container">$r$</span>. The extreme lines of sight are tangents to the sphere, as this diagram illustrates.</p>
<p><a href="https://i.sstatic.net/Pyq0Q.png" rel="noreferrer"><img src="https://i.sstatic.net/Pyq0Q.png" alt="Tangents to two circles" /></a></p>
<p>A tangent to a sphere or circle makes a right angle to the radius at the point of tangency, so we have 4 similar right triangles. Let the radii of the circles be <span class="math-container">$r_1, r_2$</span> and the respective distances from their centres to the origin be <span class="math-container">$d_1, d_2$</span>, so the (centre to centre) distance between the two circles is <span class="math-container">$d=d_1+d_2$</span>. Then</p>
<p><span class="math-container">$$d_1 = \frac{d\cdot r_1}{r_1+r_2}$$</span>
<span class="math-container">$$d_2 = \frac{d\cdot r_2}{r_1+r_2}$$</span></p>
<p>and the angular diameter <span class="math-container">$\theta$</span> is given by</p>
<p><span class="math-container">$$\sin\left(\frac\theta2\right) = \frac{r_1}{d_1} = \frac{r_2}{d_2} = \frac{r_1+r_2}{d_1+d_2}$$</span></p>
<p>Using values <a href="https://en.wikipedia.org/wiki/Earth" rel="noreferrer">from</a> <a href="https://en.wikipedia.org/wiki/Moon" rel="noreferrer">Wikipedia</a> for the radii of the Earth and Moon, and their mean distance, we get</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: left;">Body</th>
<th style="text-align: right;">Radius</th>
<th style="text-align: right;">Distance</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: left;">Moon</td>
<td style="text-align: right;">1737.4</td>
<td style="text-align: right;">82365.8</td>
</tr>
<tr>
<td style="text-align: left;">Earth</td>
<td style="text-align: right;">6371.0</td>
<td style="text-align: right;">302033.2</td>
</tr>
<tr>
<td style="text-align: left;">Sum</td>
<td style="text-align: right;">8108.4</td>
<td style="text-align: right;">384399.0</td>
</tr>
</tbody>
</table>
</div>
<p>with the angular diameter <span class="math-container">$\theta\approx 2.41734°\approx2°25'$</span> or <span class="math-container">$145$</span> arc-minutes, which is almost <span class="math-container">$4.6$</span> times larger than the Moon's mean angular diameter as seen from Earth's surface, which is <span class="math-container">$31.7'$</span>, although it ranges from <span class="math-container">$29.3'$</span> to <span class="math-container">$34.1'$</span>.</p>
<p>The Earth is approximately an ellipsoid, and there are several ways to define <a href="https://.wikipedia.org/wiki/Earth_radius" rel="noreferrer">the radius of the Earth</a>. In this answer, I'm using the arithmetic mean radius. Another option that makes sense in this context (but not mentioned on that page) is to use a geometric mean radius <span class="math-container">$\approx6367$</span> km. A circle of that radius has the same area as a cross-section of the the Earth, in a plane containing the poles, perpendicular to the equator.</p>
<hr />
<p>There's another viewing point behind the Moon, where the Moon just eclipses the Earth. Once again, we get similar right triangles.</p>
<p><a href="https://i.sstatic.net/Q6GDI.png" rel="noreferrer"><img src="https://i.sstatic.net/Q6GDI.png" alt="Diagram of Moon eclipsing Earth" /></a></p>
<p>Let <span class="math-container">$s$</span> be the distance from the viewing point to the Moon's centre, and once again <span class="math-container">$d$</span> is the distance from the Earth to the Moon. We have</p>
<p><span class="math-container">$$\sin\left(\frac\theta2\right) = \frac{r_1}s = \frac{r_2}{d+s}$$</span>
So
<span class="math-container">$$r_1\cdot d + r_1\cdot s = r_2\cdot s$$</span>
Hence
<span class="math-container">$$s = \frac{d\cdot r_1}{r_2-r_1}$$</span></p>
<p>Plugging in the values for the Earth and Moon, we get <span class="math-container">$s=144133.0$</span> km, and <span class="math-container">$\theta\approx1.38134°\approx1°23'$</span></p>
<hr />
<p>Those two viewpoints are on the diameter of a circle (actually a sphere) of radius</p>
<p><span class="math-container">$$q=\frac{d_1\cdot d_2}{d_2-d_1}$$</span></p>
<p>I'll add a derivation for that below.</p>
<p>At all points on that circle the Moon and Earth have equal angular size.</p>
<p><a href="https://i.sstatic.net/RaG39.gif" rel="noreferrer"><img src="https://i.sstatic.net/RaG39.gif" alt="Equal angular size anim" /></a></p>
<p>As in the previous diagrams, the Moon &amp; Earth circles are blue. The radii are approximately in their correct ratio in this diagram, but the distance between them is (of course) radically reduced, which magnifies the angular size. The large purple circle is the circle of radius <span class="math-container">$q$</span>, the small pale purple circle makes it a bit easier to see that the angles are equal.</p>
<p>Let <span class="math-container">$M=(-d_1, 0)$</span> be the centre of the Moon and <span class="math-container">$E=(d_2, 0)$</span> be the centre of the Earth, as in the top diagram. We want to find points <span class="math-container">$P=(x,y)$</span> such that
<span class="math-container">$$\sin(\theta/2)=r_1/PM=r_2/PE$$</span></p>
<p>That is,
<span class="math-container">$$d_2^2((x+d_1)^2+y^2)=d_1^2((x-d_2)^2+y^2)$$</span>
<span class="math-container">$$d_2^2(x^2+2d_1x+d_1^2+y^2)=d_1^2(x^2-2d_2x+d_2^2+y^2)$$</span>
<span class="math-container">$$d_2^2x^2+2d_1d_2^2x+d_1^2d_2^2+d_2^2y^2=d_1^2x^2-2d_1^2d_2x+d_1^2d_2^2+d_1^2y^2$$</span>
<span class="math-container">$$(d_2^2-d_1^2)x^2+2d_1d_2(d_1+d_2)x+(d_2^2-d_1^2)y^2=0$$</span>
<span class="math-container">$$x^2+2\left(\frac{d_1d_2}{d_2-d_1}\right)x+y^2=0$$</span>
<span class="math-container">$$\left(x+\frac{d_1d_2}{d_2-d_1}\right)^2+y^2=\left(\frac{d_1d_2}{d_2-d_1}\right)^2$$</span>
Let
<span class="math-container">$$q=\frac{d_1d_2}{d_2-d_1}$$</span>
Thus
<span class="math-container">$$(x+q)^2+y^2=q^2$$</span>
which is a circle centred at <span class="math-container">$(-q, 0)$</span> with radius <span class="math-container">$q$</span>.</p>
<p>Note that
<span class="math-container">$$\frac1q = \frac1{d_1}-\frac1{d_2}$$</span>
If <span class="math-container">$d_1=d_2$</span> then <span class="math-container">$q$</span> goes to infinity, and the circle degenerates to the vertical line <span class="math-container">$x=0$</span>, i.e., the Y axis.</p>
<p>Using the previous values of <span class="math-container">$d_1$</span> &amp; <span class="math-container">$d_2$</span>, <span class="math-container">$q\approx113249.4$</span> km.</p>
<hr />
<p>I should mention that these calculations assume the the Earth and Moon are perfect spheres, separated by a constant distance. In reality, none of those things are true, so the true angular sizes of the Earth and Moon are a little different to what I've calculated above.</p>
<p>As ProfRob says, the Earth isn't a perfect sphere. It's slightly flattened at the poles, with a <a href="https://en.wikipedia.org/wiki/Earth_ellipsoid" rel="noreferrer">flattening</a> factor of <span class="math-container">$f\approx 1/298.25642$</span>. The Moon is also flattened, but much less than the Earth (<span class="math-container">$f\approx 1/830$</span>), due to its much slower rate of axial rotation.</p>
<p>Also, the orbit of the Moon &amp; Earth about their barycentre is moderately eccentric, with a mean value of <span class="math-container">$\varepsilon\approx 0.0549$</span>, and the eccentricity changes depending on the distance to the Sun.</p>
<p>Here's a daily plot of the Earth-Moon distance for 2020, produced using Horizons.</p>
<p><a href="https://i.sstatic.net/NXJ2h.png" rel="noreferrer"><img src="https://i.sstatic.net/NXJ2h.png" alt="Earth-Moon distance, 2020" /></a></p>

CC BY-SA 4.0
2021-12-08T07:02:25.980Z
The difference in the result over just using the radius of a circle is only in the 4th significant figure. But the Earth's radius varies from 6378km to 6357km. So I think if you want quote that 4th significant figure you should discuss more carefully what radii you are using for the Earth and Moon.
ProfRob
CC BY-SA 4.0
2021-12-08T15:25:15.270Z
So the set of points  of equal angular size is truly a sphere? Cool. Mucking about with three-point circles and five-point conics on the intersection in Geogebra definitely made me think it was, and I wound up asking a question about that over on [Math Stack Exchange](https://math.stackexchange.com/questions/4327106/is-the-set-of-all-points-where-any-two-spheres-of-different-size-such-as-the-ea) .
notovny
CC BY-SA 4.0
2021-12-08T15:31:36.710Z
@notovny Well, it's a sphere if we pretend the Earth & Moon are spheres, with a fixed distance between them. ;) I really don't want to deal with that wobbly distance, and slicing ellipsoids with the plane containing their centres and the viewing point...
PM 2Ring
CC BY-SA 4.0
2021-12-08T15:48:22.460Z
@ProfRob True, for small angles, we can use the well-known approximation, $\theta\approx\sin\theta\approx\tan\theta$. The error from ignoring the flattening of the Earth (and of the Moon) is fairly small, the error from treating the Earth-Moon distance as constant is larger, although it just affects the absolute values of the viewing distances (and thus the angular diameter), not their ratio, which is determined by the radii.
PM 2Ring
CC BY-SA 4.0
2021-12-08T17:41:13.683Z
My point was, that the error associated with the variation in the Earth and Moon radii is in the 3rd significant figure and is more important than the $\theta \sim \tan \theta \sim \sin \theta$ approximation.
ProfRob
CC BY-SA 4.0
2021-12-08T17:47:06.403Z
@ProfRob Fair enough. But if we're calling trig functions anyway, we might as well use arcsin rather than arctan (assuming both are available).
PM 2Ring
CC BY-SA 4.0
2021-12-10T16:10:54.260Z
What a great answer.
Fattie
CC BY-SA 4.0
2021-12-10T16:12:40.920Z
@ProfRob *4th decimal* .. wait, is that true even in the "closest" solution, ie, when at the appropriate point on the midway line between the two?  (I guess the lunar astronauts would have seen this or near this point ...)
Fattie
CC BY-SA 4.0
2021-12-11T17:39:47.720Z
This is a great and informative answer. Let me ask about the presentation side: what software did you use to make the plots, especially the animated one?
lukeg
CC BY-SA 4.0
2021-12-11T17:54:54.503Z
Thank, @lukeg. I created the plots using [Sage](https://doc.sagemath.org/html/en/reference/plotting/sage/plot/plot.html), which is a mathematics software system built on top of Python. Sage uses matplotlib to do the actual plotting. You can play with some of my Sage / Python scripts linked in this answer: https://astronomy.stackexchange.com/a/47345/16685 Those scripts run on the SageMathCell server, so you can run them in your Web browser without installing anything. FWIW, I wrote these scripts on my phone. ;)
PM 2Ring
CC BY-SA 4.0
2021-12-11T17:59:52.777Z
@lukeg FWIW, Sage can also do interactive 3D plots & anims, eg https://astronomy.stackexchange.com/a/43350/16685 https://astronomy.stackexchange.com/a/28036/16685 https://astronomy.stackexchange.com/a/47554/16685
PM 2Ring
CC BY-SA 4.0
2021-12-07T21:25:50.843Z
2021-12-11T00:35:05.283Z
2021-12-11T00:35:05.283Z
PM 2Ring
<p>If I'm directly in between the Earth and Moon, what distance from the Earth would I have to be so that the Earth and Moon have the same apparent size?</p>
<p>How big would the moon appear compared to it's normal size?</p>

<the-moon><earth><angular-diameter>
Where in space would the Earth and Moon appear to be the same size?
cowlinator
# Where in space would the Earth and Moon appear to be the same size?

(Asked by: username_0 on 2021-12-07)

If I'm directly in between the Earth and Moon, what distance from the Earth would I have to be so that the Earth and Moon have the same apparent size?

How big would the moon appear compared to it's normal size?

**Tags:** the-moon, earth, angular-diameter

---

## 2 answers

### Answer by username_6 on 2021-12-07. Score: 69 | Accepted answer

When viewing a sphere of radius $r$ at a distance $d$ from the centre of the sphere, you *don't* see a circle of radius $r$. The extreme lines of sight are tangents to the sphere, as this diagram illustrates.

[![Tangents to two circles](https://i.sstatic.net/Pyq0Q.png)](https://i.sstatic.net/Pyq0Q.png)

A tangent to a sphere or circle makes a right angle to the radius at the point of tangency, so we have 4 similar right triangles. Let the radii of the circles be $r\_1, r\_2$ and the respective distances from their centres to the origin be $d\_1, d\_2$, so the (centre to centre) distance between the two circles is $d=d\_1+d\_2$. Then

$$d\_1 = \frac{d\cdot r\_1}{r\_1+r\_2}$$
$$d\_2 = \frac{d\cdot r\_2}{r\_1+r\_2}$$

and the angular diameter $\theta$ is given by

$$\sin\left(\frac\theta2\right) = \frac{r\_1}{d\_1} = \frac{r\_2}{d\_2} = \frac{r\_1+r\_2}{d\_1+d\_2}$$

Using values [from](https://en.wikipedia.org/wiki/Earth) [Wikipedia](https://en.wikipedia.org/wiki/Moon) for the radii of the Earth and Moon, and their mean distance, we get

| Body | Radius | Distance |
| --- | --- | --- |
| Moon | 1737.4 | 82365.8 |
| Earth | 6371.0 | 302033.2 |
| Sum | 8108.4 | 384399.0 |

with the angular diameter $\theta\approx 2.41734°\approx2°25'$ or $145$ arc-minutes, which is almost $4.6$ times larger than the Moon's mean angular diameter as seen from Earth's surface, which is $31.7'$, although it ranges from $29.3'$ to $34.1'$.

The Earth is approximately an ellipsoid, and there are several ways to define [the radius of the Earth](https://.wikipedia.org/wiki/Earth_radius). In this answer, I'm using the arithmetic mean radius. Another option that makes sense in this context (but not mentioned on that page) is to use a geometric mean radius $\approx6367$ km. A circle of that radius has the same area as a cross-section of the the Earth, in a plane containing the poles, perpendicular to the equator.

---

There's another viewing point behind the Moon, where the Moon just eclipses the Earth. Once again, we get similar right triangles.

[![Diagram of Moon eclipsing Earth](https://i.sstatic.net/Q6GDI.png)](https://i.sstatic.net/Q6GDI.png)

Let $s$ be the distance from the viewing point to the Moon's centre, and once again $d$ is the distance from the Earth to the Moon. We have

$$\sin\left(\frac\theta2\right) = \frac{r\_1}s = \frac{r\_2}{d+s}$$
So
$$r\_1\cdot d + r\_1\cdot s = r\_2\cdot s$$
Hence
$$s = \frac{d\cdot r\_1}{r\_2-r\_1}$$

Plugging in the values for the Earth and Moon, we get $s=144133.0$ km, and $\theta\approx1.38134°\approx1°23'$

---

Those two viewpoints are on the diameter of a circle (actually a sphere) of radius

$$q=\frac{d\_1\cdot d\_2}{d\_2-d\_1}$$

I'll add a derivation for that below.

At all points on that circle the Moon and Earth have equal angular size.

[![Equal angular size anim](https://i.sstatic.net/RaG39.gif)](https://i.sstatic.net/RaG39.gif)

As in the previous diagrams, the Moon & Earth circles are blue. The radii are approximately in their correct ratio in this diagram, but the distance between them is (of course) radically reduced, which magnifies the angular size. The large purple circle is the circle of radius $q$, the small pale purple circle makes it a bit easier to see that the angles are equal.

Let $M=(-d\_1, 0)$ be the centre of the Moon and $E=(d\_2, 0)$ be the centre of the Earth, as in the top diagram. We want to find points $P=(x,y)$ such that
$$\sin(\theta/2)=r\_1/PM=r\_2/PE$$

That is,
$$d\_2^2((x+d\_1)^2+y^2)=d\_1^2((x-d\_2)^2+y^2)$$
$$d\_2^2(x^2+2d\_1x+d\_1^2+y^2)=d\_1^2(x^2-2d\_2x+d\_2^2+y^2)$$
$$d\_2^2x^2+2d\_1d\_2^2x+d\_1^2d\_2^2+d\_2^2y^2=d\_1^2x^2-2d\_1^2d\_2x+d\_1^2d\_2^2+d\_1^2y^2$$
$$(d\_2^2-d\_1^2)x^2+2d\_1d\_2(d\_1+d\_2)x+(d\_2^2-d\_1^2)y^2=0$$
$$x^2+2\left(\frac{d\_1d\_2}{d\_2-d\_1}\right)x+y^2=0$$
$$\left(x+\frac{d\_1d\_2}{d\_2-d\_1}\right)^2+y^2=\left(\frac{d\_1d\_2}{d\_2-d\_1}\right)^2$$
Let
$$q=\frac{d\_1d\_2}{d\_2-d\_1}$$
Thus
$$(x+q)^2+y^2=q^2$$
which is a circle centred at $(-q, 0)$ with radius $q$.

Note that
$$\frac1q = \frac1{d\_1}-\frac1{d\_2}$$
If $d\_1=d\_2$ then $q$ goes to infinity, and the circle degenerates to the vertical line $x=0$, i.e., the Y axis.

Using the previous values of $d\_1$ & $d\_2$, $q\approx113249.4$ km.

---

I should mention that these calculations assume the the Earth and Moon are perfect spheres, separated by a constant distance. In reality, none of those things are true, so the true angular sizes of the Earth and Moon are a little different to what I've calculated above.

As ProfRob says, the Earth isn't a perfect sphere. It's slightly flattened at the poles, with a [flattening](https://en.wikipedia.org/wiki/Earth_ellipsoid) factor of $f\approx 1/298.25642$. The Moon is also flattened, but much less than the Earth ($f\approx 1/830$), due to its much slower rate of axial rotation.

Also, the orbit of the Moon & Earth about their barycentre is moderately eccentric, with a mean value of $\varepsilon\approx 0.0549$, and the eccentricity changes depending on the distance to the Sun.

Here's a daily plot of the Earth-Moon distance for 2020, produced using Horizons.

[![Earth-Moon distance, 2020](https://i.sstatic.net/NXJ2h.png)](https://i.sstatic.net/NXJ2h.png)

> **Comments:**
> **username_8**: The difference in the result over just using the radius of a circle is only in the 4th significant figure. But the Earth's radius varies from 6378km to 6357km. So I think if you want quote that 4th significant figure you should discuss more carefully what radii you are using for the Earth and Moon.  
> **username_7**: So the set of points  of equal angular size is truly a sphere? Cool. Mucking about with three-point circles and five-point conics on the intersection in Geogebra definitely made me think it was, and I wound up asking a question about that over on [Math Stack Exchange](https://math.stackexchange.com/questions/4327106/is-the-set-of-all-points-where-any-two-spheres-of-different-size-such-as-the-ea) .  
> **username_6**: @notovny Well, it's a sphere if we pretend the Earth & Moon are spheres, with a fixed distance between them. ;) I really don't want to deal with that wobbly distance, and slicing ellipsoids with the plane containing their centres and the viewing point...  
> **username_6**: @ProfRob True, for small angles, we can use the well-known approximation, $\theta\approx\sin\theta\approx\tan\theta$. The error from ignoring the flattening of the Earth (and of the Moon) is fairly small, the error from treating the Earth-Moon distance as constant is larger, although it just affects the absolute values of the viewing distances (and thus the angular diameter), not their ratio, which is determined by the radii.  
> **username_8**: My point was, that the error associated with the variation in the Earth and Moon radii is in the 3rd significant figure and is more important than the $\theta \sim \tan \theta \sim \sin \theta$ approximation.  
> **username_6**: @ProfRob Fair enough. But if we're calling trig functions anyway, we might as well use arcsin rather than arctan (assuming both are available).  
> **username_9**: What a great answer.  
> **username_9**: @ProfRob *4th decimal* .. wait, is that true even in the "closest" solution, ie, when at the appropriate point on the midway line between the two?  (I guess the lunar astronauts would have seen this or near this point ...)  
> **username_10**: This is a great and informative answer. Let me ask about the presentation side: what software did you use to make the plots, especially the animated one?  
> **username_6**: Thank, @lukeg. I created the plots using [Sage](https://doc.sagemath.org/html/en/reference/plotting/sage/plot/plot.html), which is a mathematics software system built on top of Python. Sage uses matplotlib to do the actual plotting. You can play with some of my Sage / Python scripts linked in this answer: https://astronomy.stackexchange.com/a/47345/16685 Those scripts run on the SageMathCell server, so you can run them in your Web browser without installing anything. FWIW, I wrote these scripts on my phone. ;)  
> **username_6**: @lukeg FWIW, Sage can also do interactive 3D plots & anims, eg https://astronomy.stackexchange.com/a/43350/16685 https://astronomy.stackexchange.com/a/28036/16685 https://astronomy.stackexchange.com/a/47554/16685  

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### Answer by username_1 on 2021-12-07. Score: 11

The Earth's radius is about 6371 km, while the moon is 1737.4 km. The distance between the two bodies is 384399 km. The formula for angular size (in radians) is $\delta = c$. Letting $x$ be equal to the distance from Earth, we get the equation $2 \arctan\Big(\frac{12742}{2x}\Big)=2 \arctan\Big(\frac{3474.8}{2(384399-x)}\Big)$. Solving this equation gives $\frac{4081676715}{13514}\approx 302033\ \text{km}$. Plugging in this value, we get $2^\circ\ 25' \ 1.78''$ as the angular diameter of the Moon, compared to $31' \ 5''$ as viewed from Earth, which is 4.66583 times smaller.

> **Comments:**
> **username_2**: …should there be 2 solutions to the equation? One on the line segment between their centers (this answer), and one on the line extending beyond the moon… the point of the cone defined by the circumferential circles of the two bodies (sorry, that's a bad descrition.)  
> **username_1**: @CraigConstantine There's only one solution. See https://www.desmos.com/calculator/cmnvyngk8i  
> **username_3**: There ARE multiple answers to the general question "from where is space". Craig's suggestion is one of those locations when on the line passing through the Earth and Moon but not in between the Earth and Moon. You answered the specific question of "on the line between the Earth and Moon". Also, 6371 and 1737 km are the radius of the Earth and Moon, not the diameter.  
> **username_4**: @CraigConstantine There is only one solution to fasterthanlight's equation, because we need a different equation for "observer-Moon-Earth" colinearity: $2 \arctan\Big(\frac{12742}{2(384399+x)}\Big)=2 \arctan\Big(\frac{3474.8}{2x}\Big)$.  I get 144,133 km.  Also there are an infinite number of non-colinear solutions.  Also they will appear the same size when an observer is sufficiently far away that the angular diameter of both bodies is smaller than the minimum angular resolution of the observer, i.e. they both appear as point sources.  
> **username_5**: @fasterthanlight That plot shows negative angular size, which makes no sense - you need an absolute value in there. At very large distances the earth will have larger angular size since it is actually larger, but at distances just outside the moon's orbit the moon will of course appear larger. To continuously transition from one to the other (moon larger to earth larger), they must have equal angular size at some point outside the moon's orbit. Craig is correct, there is a second solution.  
> **username_6**: For spheres, you need to use arcsin, not arctan. See https://en.wikipedia.org/wiki/Angular_diameter  
> **username_7**: The set of all points of equal angular size appear to be a three-dimensional ovoid [centered behind the moon.](https://www.geogebra.org/calculator/qxbr8c4s)  
> **username_4**: @notovny Your comment would make an outstanding second answer to this question.  
> **username_8**: Silly number of significant figures.  
> **username_7**: @ConnorGarcia Perhaps if I've had some more time to look at the math; I'm not certain, but investigating the GeoGebra shape makes it look like the result I called an ovoid is at least very, very close to a being a sphere about 113 000 km in radius, centered on a point 415 000 km from Earth.  

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