You, the observer are on point \(O\) on the earth, represented by the blue circle (actually a sphere) of radius \(R = 4000\) miles. The semi circle FSE represents the hemispherical dome of stars in your night sky centered on you the observer. The astronomical distances of stars makes it possible to assume the stars sit on dome centered on the observer. The arc FPE is the spherical cap where the planes are flying—at a height \(h\) of 6 miles ( distance exaggerated for clarity)  above the surface of the earth. This spherical cap is centered on the earths’s center \(C\).

Since the stars are equally likely to be anywhere in the night sky, the number per unit area or (probability) density is:

\[
\frac{N}{2\pi S^2}
\]

where S is the assumed imaginary radius of the star dome. We will see that it does not matter. (we could alternatively just work with solid angles)

Similarly that of the planes is:

\[
\frac{N}{2\pi (R+h)h}
\]

The denominator above uses the formula for the area of a spherical cap in terms of radius \(R + h\) and cap height \(h\).

Let \(\theta\) represent the elevation angle where there is equal likelihood that a point of light observed is a star or a plane.

Consider the infinitesimal strips formed between elevations \(\theta\) and \(d \theta\).

For the Stars: this is the circular strip IH whose area is

\[
2\pi S \cos\theta \cdot S \ d\theta = 2 \pi S^2 \cos\theta \ d\theta
\]

The number of stars expected in this strip is the area multiplied by the number per unit area (from above):  

\begin{align*}
&= 2 \pi S^2 \cos\theta \ d\theta \ \cdot \ \frac{N}{2\pi S^2} \\
&= \boldsymbol{ N\cos\theta d\theta }  \tag{1}\label{eq1} 
\end{align*}

which is independent of S as we would expect.

For the Planes:  this strip is on a sphere centered on C (earth centre) where the angle subtended is \(\phi\) and \(d\phi\). So it has area:

\[
2 \pi (R+h) \sin\phi \cdot (R+h) \ d\phi = 2 \pi (R+h)^2 \sin\phi \ d\phi
\]

The number of planes expected on this strip is as before, the strip area multiplied by the density:

\begin{align*}
&= 2 \pi (R+h)^2 \sin\phi \ d\phi \ \cdot \ \frac{N}{2\pi (R+h)h} \\
&= \boldsymbol{ \frac{R+h}{h}\sin\phi\ Nd\phi}  \tag{2}\label{eq2} 
\end{align*}

To derive the relation between \(\theta\) and \(\phi\), we need to solve \(\triangle COM\) where we know sides \(CO = R\) and \(CM = R+h\) and the non included \(\angle COM = \frac{\pi}{2} + \theta \) . The standard triangle solution approach gives us:

\begin{align*}
\phi =  \frac{\pi}{2} – \theta – \sin^{-1}(k\cos\theta) \tag{3}\label{eq3} \\
\end{align*}

where \(k = \displaystyle{\frac{R}{R+h}} \)

Differentiating gives the relation between \(d\phi\) and \(d\theta\):

\begin{align*}
\frac{d\phi}{d\theta} = -1 + \frac{k\sin\theta}{\sqrt{1 – k^2\cos^2\theta}}  
\end{align*} 

This will be negative in the way we have defined \(\phi\), since \(\phi\) decreases as \(\theta\) increases. We need the absolute value of \(d\phi\) in our density calculation. Consequently, we use:

\begin{align*}
d\phi = 1 – \frac{k\sin\theta}{\sqrt{1 – k^2\cos^2\theta}}d\theta \tag{4}\label{eq4}
\end{align*}

We can now use these relations between \(\phi\) and \(\theta\) and \(d\phi\) and \(d\theta\) from \(\eqref{eq3}\) and \(\eqref{eq4}\) in the planes expression \(\eqref{eq2}\) to get the number of planes at the elevation \(\theta\) as:

\[
\frac{R+h}{h}\sin \left(\frac{\pi}{2} – \theta – \sin^{-1}(k\cos\theta) \right) \left (1 – \frac{k\sin\theta}{\sqrt{1 – k^2\cos^2\theta}} \right ) N d\theta
\]

The question states that the probability (hence the number) is the same for the Plane strip and the Stars strip. So all we need to do is equate the above with \(\eqref{eq1}\) and solve for \(\theta\):

\begin{align*}
\frac{R+h}{h}\sin \left(\frac{\pi}{2} – \theta – \sin^{-1}(k\cos\theta) \right) \left (1 – \frac{k\sin\theta}{\sqrt{1 – k^2\cos^2\theta}} \right ) N d\theta &= N\cos\theta d\theta \\
\Rightarrow \qquad \frac{R+h}{h}\sin \left(\frac{\pi}{2} – \theta – \sin^{-1}(k\cos\theta) \right) \left (1 – \frac{k\sin\theta}{\sqrt{1 – k^2\cos^2\theta}} \right ) &= \cos\theta \\
\end{align*}

\(\theta\) is the only unknown in the equation above. Solving for \(\theta\) numerically gives us \( \mathbf{{\color{Red} {\theta \approx 0.106}}}\) radians or \(\mathbf{{\color{Red} {\approx 6.07 ^\text{o}}}} \) which is the required answer.

Note that \(\theta = 90^\text{o} \) is also a (trivial) solution to the equation but a meaningless one since the probability of finding a plane/star at zenith is effectively zero.