Imagine there are two points on planet Earth and a light is shone from one to the other by reflecting off an object 500km up (think of this as a mirror oriented parallel to the surface right below it). Let us assume the Earth is a perfect sphere. As the distance between the points increases, the angle that the light is received at increases with a limit of 90 degrees which corresponds to a tangent of the sphere. I would like to know how far apart the two points along the sphere are as a function of this angle.

To try to solve it I first notice that the angle of incidence equals the angle of reflection. So we can draw an isosceles triangle with the reflector at the top and the two points as the other two vertices. The height of the triangle is a function of how far apart the two points are. But now I am stuck.

The radius of the Earth is 6371km.

  • 2
    $\begingroup$ Not one hundred percent sure I understand the problem.. But a figure may help $\endgroup$ Jun 14, 2020 at 7:53
  • $\begingroup$ Can you just think of the sphere as being of radius $6371+r$ to simplify things a little? $\endgroup$
    – user35671
    Jun 14, 2020 at 7:55
  • 1
    $\begingroup$ @Anush: That's kilometers, probably; you need to multiply by $1000$, perhaps? $\endgroup$
    – Brian Tung
    Jun 14, 2020 at 7:58
  • $\begingroup$ The reverse problem is easier: given the distance $D$ between the two 'observers', half distance angle is $\alpha=\frac{D}{2R}$. Then half of the angle at the reflector is given by $\tan \delta = \frac{R\sin\alpha}{R(1-\cos\alpha)+r}$. $R$ is Earth radius, while $r$ is the altitude of the reflector. $\endgroup$
    – N74
    Jun 16, 2020 at 19:36
  • $\begingroup$ This object that's 500km up, is it directly above one of the surface objects? Above the point on the surface halfway between them? Somewhere else? $\endgroup$
    – user307169
    Jun 17, 2020 at 16:21

5 Answers 5


I interpreted the problem as asking for a function that, for general $R$ and $h$, gives the distance between the two points for any reflection angle (i.e., not only for the case in which the beams are tangent), and allows to calculate the maximal reflection angle and the maximal distance. This function can then be used to calculate the maximal angle and the maximal distance in the specific scenario provided in the OP, with $R=6371$ Km and $h=500$ Km.

Let us consider the Earth circumference as represented by the circle $x^2+y^2=R^2$, with the center in the origin and where $R=6371$. We can place our object in $A(0,R+h)$ on the $y$-axis, representing a point that is $h$ Km up the Earth surface.

Now let us draw two lines passing through $A$, symmetric with respect to the $y$-axis and intersecting the circumference. For each line, let us consider the intersection point that is nearer to the $y$-axis. Let us call the two new points $B$ (in the first quadrant) and $C$ (in the second quadrant). These represent the two points on Earth surface.

enter image description here

Due to the symmetry of the construction, we can continue by analyzing only one of these two points, e.g. $C$. The equation of the line containing $AC$ can be written as $y=sx+R+h$, where $s$ is its positive slope. To determine where this line crosses the circumference, we can set


whose solutions are

$$x=\frac{-sR-sh \pm \sqrt{\Delta}}{s^2 + 1}$$

where $\Delta=s^2 R^2 - 2h R - h^2$.

As stated above, we are interested in the less negative solution for $x$, as it is that nearer to the $y$-axis. So we get that the $x$-coordinate of $C$ is

$$X_C=\frac{-sR-sh + \sqrt{\Delta}}{s^2 + 1}$$

and the $y$-coordinate is

$$Y_C=\frac{s(-sR-sh + \sqrt{\Delta})}{s^2 + 1}+R+h$$

As a result, the equation $y=tx$ of the line $OC$ has slope


Now setting $\angle{BAC}=\alpha$ and $\angle{BOC}=\beta$, we have $s=\cot(\alpha/2)$ and $t=-\cot(\beta/2)$.

Thus, we get

$$ \cot(\beta/2) =\frac{(\cot^2(\alpha/2)+1)(R+h)}{\cot(\alpha/2)(R+h)-\sqrt{\Delta}} - \cot(\alpha/2)$$

and then

$$ \beta =2\cot^{-1}\left[\frac{(\cot^2(\alpha/2)+1)(R+h)}{\cot(\alpha/2)(R+h)-\sqrt{\Delta}} - \cot(\alpha/2)\right] $$

So the length of the arc $D$ corresponding to $ \beta$, which is the distance along the spherical surface asked in the OP, is

$$ D =2R\cot^{-1}\left[\frac{(\cot^2(\alpha/2)+1)(R+h)}{\cot(\alpha/2)(R+h)-\sqrt{\Delta}} - \cot(\alpha/2)\right] $$

where $\Delta=\cot^2(\alpha/2)R^2 - 2h R - h^2$.

The last equation can be simplified as

$$ D =2R\cot^{-1}\left[\frac{(R+h)+\sqrt{\Delta}}{\cot(\alpha/2)(R+h)-\sqrt{\Delta}} \right] $$

For example, for $\alpha=\pi/2$ and $h= (\sqrt{2}-1)R$, as expected we have $\Delta=0$ (this is the situation where $\alpha$ is a right angle and the light beams are tangent to the surface). In this case, $\beta$ is also a right angle and $D=\pi/2\,R$. Accordingly the formula above gives this result, as shown by WA here.

For any value of $R$ and $h$, the maximal angle $\alpha_{max}$ and the maximal distance $D_{max}$ (i.e., those obtained with the beams tangent to the surface) can be determined by considering the case in which $\Delta=0$. This case occurs when $\cot^2(\alpha/2)R^2 - 2h R - h^2=0$. Solving for $\alpha$ in the range $0 \leq \alpha \leq \pi$ we get

$$\alpha_{max} = 2 \cot^{-1}\left(\frac{\sqrt{h(h+2R)}}{R}\right)$$

Interestingly, when $\Delta=0$, the formula for the distance is considerably simplified, and by few calculations reduces to

$$ D_{max} =2R\cot^{-1}\left[\tan(\alpha/2)\right] $$

As shown here, in the specific scenario described in the OP, substituting $R=6371$ and $h=500$, we get

$$\alpha= 2 \cot^{-1}\left(\frac{10 \sqrt{66210}}{6371}\right) \approx 2.3739 \,\,\text{radians}$$

which corresponds to about $136$ degrees. Here is the plot of the distance $D$ (in Km) as a function of $\alpha$ (in radians) for $R=6371$ and $h=500$, as obtained by WA. The plot confirms the maximal real value of $\alpha$, concordant with the predicted value of $2.3739$. The blue and red lines indicate the real and imaginary part, respectively.

enter image description here

Lastly, from the simplified formula for the maximal distance, taking $R=6371$ and $\alpha=2.3739$, we get

$$D_{max}\approx 4891 \, \text{Km}$$

  • $\begingroup$ Please note that the angle $\angle$ given in this answer is not the one specified in the question. $\endgroup$
    – user35671
    Jun 24, 2020 at 5:14

We could make a triangle with the center of the sphere, one of the observer and the mirror. From this triangle, we know

  • the angle at the observer is a right angle since the radius is perpendicular to the tangent.

  • the side from the center to the observer mesure $6371$ km (radius).

  • the side from the center to the mirror mesure $6871$ km (radius $+$ height of the mirror).

We are able to find the angle at the center of the sphere. $$\cos \theta=\frac{6371}{6871}$$ $$\theta=0.383848\ \text{rad}$$ The same triangle could be built with the second observer. So the angular distance between the two observers is $$2\theta=0.767696\ \text{rad}$$ We multiply this value with the radius to get the distance between the two observers. $$\text{distance}=2\theta\times6371=4891\ \text{km}$$

More generaly, if an observer is on a sphere of radius $r$ uses a mirror placed at a height $h$ above the surface. The further distance that he could reach is given by $$\text{distance}=2\times r\times\arccos\left(\frac{r}{r+h}\right)$$


I understood the problem as follows (see the picture).

enter image description here

We have $OA=OA'=R$ and $OM=R+h$, where $R=6371$ km is the radius of Earth and $H=500$ km its the height of the reflector over the Earth’s surface. The point $P$ on the tangent is chosen to provide $AP||OM$. Given the reflection angle $\angle PMA=\alpha$, find the distance $d$ between the points $A$ and $A’$ along the sphere. But $d=2R\angle AOM=2R\beta$. We have $PM=AM\cos\alpha=OA\sin\beta=R\sin\beta$. By theorem of cosinuses

$$AM^2=OA^2+OM^2-2OA\cdot OM\cos\beta=R^2+(R+H)^2-2R(R+H) \cos\beta.$$


$$(R^2+(R+H)^2-2R(R+H) \cos\beta) \cos^2\alpha= R^2\sin^2\beta.$$

Putting $h=H/R$, we have

$$(1+(1+h)^2-2(1+h) \cos\beta)\cos^2\alpha=\sin^2\beta=1-\cos^2\beta.$$

This is a quadratic equation for $\cos\beta$, whose solution gives



Consider the triangle formed by one of the points, the mirror and the centre of the Earth. If $\alpha$ is "the angle the light is received" then $\alpha-\beta$ is the angle at the mirror in the triangle, where $\beta$ is the angle at the centre of the Earth in the same triangle.

If $R$ is the radius of the Earth and $d$ is the distance of the mirror from the surface of the Earth $(500~\text{km})$, by the sine law we have: $$ R:\sin(\alpha-\beta)=(R+d):\sin\alpha. $$ We can expand $\sin(\alpha-\beta)$ and solve for $\sin\beta$: $$ \sin\beta={R\sin\alpha\over R+d} \left(-\cos\alpha+\sqrt{{(R+d)^2\over R^2}-\sin^2\alpha}\right). $$ The distance between the two points is then $2R\sin\beta$ (straight line), or $2R\beta$ along the surface of the Earth ($\beta$ measured in radians).

enter image description here


Analytical geometry calculation. You want intersection between ( meridians of ) sphere and cone of semi-vertical angle $\theta$ from a celestial mirror at $O$.

$$ z^2+r^2=R^2 \tag1$$

and a ray ( generator of cone ) from the mirror

$$ r \cot \theta-z = R+h\tag2$$

Eliminate $z$ between (1),(2) $$ r^2( 1+\cot^2 \theta) -2 a r \frac{\cos \theta}{\sin \theta} + (a^2-R^2) \tag3$$

$$ r^2-2 a r \frac{\cos \theta}{\sin \theta}+ h (2R+h)\sin^2 \theta $$

The quadratic has two roots

$$ \frac{r}{\sin \theta}= a \cos \theta \pm \sqrt{ a^2 \cos^2 \theta -h(2R+h)} \tag4$$

$-ve $ sign for required spherical cap nearby and $+ve,$ if the ray is produced to pierce second half hemisphere.

By geometry of right triangles we can find when the ray is tangential to the sphere:

$$ (z_m,r_m)= \dfrac{h(2R+h)}{(R+h)},\dfrac{R\sqrt{h(2R+h)}}{(R+h)}\tag5$$ is the required relation, graphed.

If $OM$ is along polar axis connecting north/south poles, the the latitude, co-latitude of required circle circle of horizon which has

$$ \cos^{-1}\frac{r_m}{R}, \sin^{-1}\frac{y_m}{R}\tag6$$

enter image description here


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