A recent question on the Astronomy Stack Exchange asked "Where in space would the Earth and Moon appear to be the same size?". Though the question asked specifically about points on the line drawn between the Earth and the Moon, idle curiosity and the comment chain on one of the answers got me to kick open Geogebra to try to see what the set of all points where the two objects appeared to be the same size was.

Calculating the subtended angle of a sphere of radius $r$ whose center is distance $d$ from point $P$ as:

$$\theta=2 \arcsin\left(\frac{r}{d}\right)$$

GeoGebra Graph of points where Earth and Moon have the Same Angular Size
GeoGebra Graph of points where Earth and Moon have the Same Angular Size
parameter value
Radius of Earth $6\,371\,\mathrm{km}$
Radius of the Moon $1\,737\,\mathrm{km}$
Semi-major axis of the Moon's orbit $384\,000\,\mathrm{km}$

In GeoGebra, I graphed the set of points where the angular size of Earth and the Moon were equal. Based on the GeoGebra graph, it looks very much like the Equal Angular size region is a sphere of radius $113\,000\, \mathrm{km}$ centered on a point about $415\,000\, \mathrm{km}$ from Earth.

My question is: Is the set of all points where two spheres of different sizes subtend the same angle actually a sphere, or just an ovoid that is too close to a sphere for me to tell in GeoGebra?

  • $\begingroup$ If the angle is fixed, the locus is a circle. As the angle varies, this circle morphs somehow. Not sure if that helps. $\endgroup$
    – bubba
    Dec 8, 2021 at 8:13
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    $\begingroup$ I guess it's not exactly a circle (in the plane). In your case, $r<< d$, so $\arcsin (r/d) \approx r/d$. If you want to know when $r_{Moon}/d_{Moon} = r_{Earth}/d_{Earth}$, that's asking when $d_{Earth}/d_{Moon} = r_{Earth}/r_{Moon}$, the solution of which is the en.wikipedia.org/wiki/Circles_of_Apollonius The solution to your real problem seems to be a very close approximation to the problem Appolonius solved. $\endgroup$
    – Ingix
    Dec 8, 2021 at 12:29
  • $\begingroup$ I suggest making more tests with Geogebra, where the distances between the spheres in question are much smaller than here, and on the same order of magnitude as the diameters of the spheres. Then, with the $\arcsin$ approximation no longer valid, I assume it will show a different kind of locus. $\endgroup$
    – Ingix
    Dec 8, 2021 at 12:33
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    $\begingroup$ I've since expanded my answer on Astronomy, which answers your question in the affirmative. I guess I could copy my derivation from there to here, since there's no rule against cross-site duplicates. ;) $\endgroup$
    – PM 2Ring
    Dec 8, 2021 at 15:38
  • $\begingroup$ @PM2Ring I'd welcome that, if you're willing. $\endgroup$
    – notovny
    Dec 9, 2021 at 13:33

1 Answer 1


It's convenient to use coordinates where the Moon and Earth are on the X axis, with the origin being the point where the Moon and Earth have the same angular diameter.

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

$$\sin\left(\frac\theta2\right) = \frac{r_1}{d_1} = \frac{r_2}{d_2} = \frac{r_1+r_2}{d}$$

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

Let $M=(-d_1, 0)$ be the centre of the Moon and $E=(d_2, 0)$ be the centre of the Earth. 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.

Equal angular size anim

Using the previous values of $d_1$ & $d_2$, $q\approx113249.4$ km. 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.

  • $\begingroup$ This answer is derived from my answer on Astronomy.SE $\endgroup$
    – PM 2Ring
    Dec 9, 2021 at 14:13
  • $\begingroup$ Since space is three-dimensional, I suppose the large purple circle is really a sphere. $\endgroup$ Dec 11, 2021 at 17:15
  • $\begingroup$ @JeppeStigNielsen Yes, it is. I mentioned that it's "(actually a sphere)" in my Astronomy answer. $\endgroup$
    – PM 2Ring
    Dec 11, 2021 at 17:30

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