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?


2 Answers 2


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

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 Wikipedia 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. 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

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

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 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

  • 7
    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Commented Dec 8, 2021 at 7:02
  • 3
    $\begingroup$ 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 . $\endgroup$
    – notovny
    Commented Dec 8, 2021 at 15:25
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    $\begingroup$ @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... $\endgroup$
    – PM 2Ring
    Commented Dec 8, 2021 at 15:31
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    $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Commented Dec 8, 2021 at 15:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Commented Dec 8, 2021 at 17:41

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.

  • 4
    $\begingroup$ …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.) $\endgroup$ Commented Dec 7, 2021 at 15:53
  • 2
    $\begingroup$ @CraigConstantine There's only one solution. See desmos.com/calculator/cmnvyngk8i $\endgroup$
    – WarpPrime
    Commented Dec 7, 2021 at 16:38
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    $\begingroup$ 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. $\endgroup$
    – JohnHoltz
    Commented Dec 7, 2021 at 17:16
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    $\begingroup$ @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. $\endgroup$
    – Connor Garcia
    Commented Dec 7, 2021 at 17:25
  • 1
    $\begingroup$ For spheres, you need to use arcsin, not arctan. See en.wikipedia.org/wiki/Angular_diameter $\endgroup$
    – PM 2Ring
    Commented Dec 7, 2021 at 19:46

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