Where in space would the Earth and Moon appear to be the same size?

Question

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?

Answer 1

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.

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.

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.

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.

Answer 2

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.