# How much of the earth can see the moon? [closed]

I framed this into 2d. If you draw two circles, get the common direct tangents, then you need to find the angle between the two intersection points for the two lines and the bigger circle. Except I never figured it out.

Maybe someone could show me a different approach? Something I missed?

Thanks

• 1. The solution / proportion of what you can see depends on where yuo are on the moon. Aug 7 at 12:09
• 2. The proportion of how much you can see is a function of the distance to earth, and it does not depend on the angle under which you see Earth (assuming Earth is a perfect sphere). Aug 7 at 12:10
• 3. Standing on the moon might complicate things, though, because parts of Earth might be hidden behind Moon's horizont. Aug 7 at 12:13
• 4. If you want to include allpoints on Moon: Take a point $P$ beyond Moon that lies on the line with the centers of Moon and Earth. Chose $P$ such that a tangent from $P$ to Earth is also a tangent from $P$ to Moon (and vice versa). Then the proportion of Moon you see from $P$ equals the proportion of Earth you see from $P$ equals the proportion of Earth you see from Moon. Aug 7 at 12:28
• @emacsdrivesmenuts He's not looking at the Earth, he's looking at the moon. Aug 7 at 13:05

Let

1. $$R=6371km$$ be the radius of the earth
2. $$r= 1737km$$ be the radius of the moon
3. $$d= 384467km$$ be the distance between the earth and moon.

Consider the figure below: $$A$$ is the centre of the earth.

($$B$$ is only an artifact of the construction and won't be used.)

$$C$$ is the centre of the moon.

$$D$$ is the intersection between the tangent and the line formed by the centers of the earth and moon.

$$E,F$$ are the intersection points for the common tangent.

By construction it immediately follows that $$\angle DEA=\angle DFC=90^{\circ}$$. Also, $$DEA\sim DFC$$. Our task is to find $$\angle DAE$$, which is the maximum angle wherefrom one can observe the moon. We can also find the latitudes from which the moon is visible if we tack on the precession of the equinoxes. Anyhow,

$$\frac{AE}{FC}=\frac{R}{r}=\frac{AD}{CD}=\frac{d+CD}{CD}$$

$$\therefore CD=\frac{d}{R/r-1}$$

$$\angle DAE = \textrm{arccos}\left(\frac{AE}{AD}\right)=\textrm{arccos}\left(\frac{R}{d+CD}\right)$$

$$=\textrm{arccos}\left(\frac{R}{d+d/(R/r-1)}\right)$$

$$\approx 89.3^{\circ}$$

Attached pic for referenceShort and approximate:

The distance between earth and moon is about 384,400 km, whereas the radius of the earth is near 6,400 km which is very small compared to the previous value. even if you take exact values and solve for the area of earth that can see the moon, it will be very close to half of the surface area of earth.

Long and more accurate:

Given:

$$OF = 384,400 km$$

$$OA=OB=OY=6,400 km$$

$$FC=FD=1,700 km$$

Required measurements for area of spherical cap:

$$r = AX ; h = XY$$

in $$\Delta OAE$$ and $$\Delta FCE$$

$$\angle OAB = \angle FCE = \pi / 2 \ or\ 90$$ degrees (tangents to a circle are always perpendicular to radius)

$$\angle AEO = \angle CEO\$$ (common\ angle)

Therefore $$\Delta OAE \sim \Delta FCE\$$ By AA similarity criterion

So we get:

$$\frac{OA}{FC}=\frac{OE}{FE}$$ $$\implies \frac{6400}{1700}=\frac{OF+FE}{FE}$$ $$\implies 3.764706= \frac{384400+FE}{FE}$$ $$\implies FE= 13889.36111$$ $$\implies OF+FE=OF+13889.36111$$ $$\implies OE=384400+13889.36111=398289.3611$$

Therefore in $$\Delta OAE$$ we have $$\cos \theta = \frac{OA}{OE}=\frac{6400}{398289.3611}=0.016068719$$

Also in $$\Delta OAX$$ $$\cos \theta =\frac{OX}{OA}$$ $$\implies \frac{OX}{6400}=0.016068719$$ $$\implies OX= 102.8398016$$ $$Now \ h=OY-OX=6400-102.8398016$$ $$\implies h=6297.160198$$ Back to cos theta $$\cos \theta = 0.016068719$$ $$\therefore \sin \theta = \sqrt{1-(\cos \theta)^2}=\sqrt{1-(0.016068719)^2}$$ $$=0.9998705$$ so in $$\Delta OAX$$ $$\sin \theta =\frac{AX}{OA}=\frac{r}{6400}=0.9998705$$ $$\implies r=6399.1712$$

Now we have all the required elements to find area of spherical cap.

Area of earth visible to moon = $$\pi(r^2+h^2)$$ $$= 3.14159 \times (6399.1712^2+6297.160198^2)$$

$$=253223522.2$$ square km

total area of earth= $$4\pi R^2= 4\times 3.14159\times 6400^2$$ $$=514718105.6$$

fraction of area that can see the moon:

$$\frac{253223522.2}{514718105.6}=0.491965\approx 49.2\%$$

• Welcome to Math.SE! <> There appears to be an extra decimal point in the calculation of $\cos\theta$; using the stated numerical values gives $\cos\theta \approx 0.004\dots$. Aug 8 at 1:56
• Thankyou for pointing the error out sir, will edit it right away @AndrewD.Hwang Aug 8 at 2:02