$R = $$ \cdot m = $ Earth radius.
$\Delta r = $$ \cdot m = $ viewing altitude above Earth surface.
$r = R + \Delta r = $ {{ view_alt_in_m + earth_radius_in_m }} = viewing distance from Earth center.



Viewer is at the origin.
Viewing down z axis.
Sphere is distance $z$ away (along the $-z$ axis), and $R$ down beneath you (along the $-y$ axis), such that its north pole is directly in front of you at all times.
How does the profile of its surface change with respect to your distance?

$r_0 = [0,0,0]$ = view origin
$v = [-sin(\phi), tan(\theta), -cos(\phi)]$ = view direction
$c = [0, -R, -z]$ = sphere origin

So sphere's distance from the origin is $r = \sqrt{z^2 + R^2}$
So the altitude above the surface is $\Delta r = r - R = \sqrt{z^2 + R^2} - R$.
So the sphere distance as a function of the view altitude is: $z^2 = (\Delta r + R)^2 - R^2$,
i.e. $z = \sqrt{(\Delta r)^2 + 2 R \Delta r}$.
$z = $ {{ Math.sqrt( view_alt_in_m * ( view_alt_in_m + 2 * earth_radius_in_m )) }}

What's the profile of the sphere in the view?
i.e. what are all rays that intersect the sphere surface perpendicular to its origin?

$(r_0 + s v - c) \cdot v = 0$ (right angle)
$|r_0 + s v - c|^2 = R^2$ (sphere surface)

right angle:
$(r_0 + s v - c) \cdot v = 0$
$r_0 \cdot v + s v \cdot v - c \cdot v = 0$
$s |v|^2 = (c - r_0) \cdot v$
$s (1 + (tan(\theta))^2) = -R tan(\theta) + z cos(\phi)$

sphere surface:
$(s v - c) \cdot (s v - c) = R^2$
$s^2 v \cdot v - 2 s v \cdot c + c \cdot c = R^2$
$s^2 (1 + (tan(\theta))^2) - 2 s (-R tan(\theta) + z cos(\phi)) + R^2 + z^2 = R^2$
...substitute right angle results for s:
$s^2 (1 + (tan(\theta))^2) - 2 s^2 (1 + (tan(\theta))^2) + z^2 = 0$
$s = \pm z / \sqrt{1 + (tan(\theta))^2}$
...let's just use the positive parameter:
$s = z / \sqrt{1 + (tan(\theta))^2}$
...substitute back into ray equation:
$r_0 + s v$
$= z / \sqrt{1 + (tan(\theta))^2} [-sin(\phi), tan(\theta), -cos(\phi)]$
That's the point of intersection of ray with sphere.
What's its declination?
What is $\theta$ as a function of $\phi$ ?
$s (1 + (tan(\theta))^2) = -R tan(\theta) + z cos(\phi)$
$(1 + (tan(\theta))^2) z / \sqrt{1 + (tan(\theta))^2} = -R tan(\theta) + z cos(\phi)$
$\sqrt{1 + (tan(\theta))^2} = -(R / z) tan(\theta) + cos(\phi)$
$1 + (tan(\theta))^2 = (R / z)^2 (tan(\theta))^2 - 2 cos(\phi) (R / z) tan(\theta) + cos(\phi)^2$
$((R / z)^2 - 1) (tan(\theta))^2 - 2 cos(\phi) (R / z) tan(\theta) - sin(\phi)^2 = 0$
$tan(\theta) = \frac{1}{2 ((R / z)^2 - 1)} \left( 2 cos(\phi) (R / z) \pm \sqrt{ (2 cos(\phi) (R / z))^2 + 4 ((R / z)^2 - 1) sin(\phi)^2 } \right)$
$tan(\theta) = \frac{1}{2 ((R / z)^2 - 1)} \left( 2 cos(\phi) (R / z) \pm \sqrt{ 4 cos(\phi)^2 (R / z)^2 + 4 sin(\phi)^2 ((R / z)^2 - 1) } \right)$
$tan(\theta) = \frac{1}{(R / z)^2 - 1} \left( cos(\phi) (R / z) \pm \sqrt{ cos(\phi)^2 (R / z)^2 + sin(\phi)^2 ((R / z)^2 - 1) } \right)$
$tan(\theta) = \frac{1}{(R / z)^2 - 1} \left( cos(\phi) (R / z) \pm \sqrt{ (R / z)^2 - sin(\phi)^2 } \right)$
$tan(\theta) = \frac{R / z}{(R / z)^2 - 1} \left( cos(\phi) \pm \sqrt{ 1 - sin(\phi)^2 / (R / z)^2 } \right)$
$tan(\theta) = \frac{1}{R / z - z / R} \left( cos(\phi) \pm \sqrt{ 1 - (sin(\phi) z / R)^2 } \right)$

$\theta = atan\left( \frac{1}{R / z - z / R} \left( cos(\phi) \pm \sqrt{ 1 - (sin(\phi) z / R)^2 } \right) \right)$


For an altitude of 1.5 m we get ...

...a declination angle of just 0.03 degrees.

For an altitude of 10,000 m we get ...

...a declination angle of just 3 degrees.

For an altitude of 1/3 Earth radii we get ...

...a declination angle of down to 60 degrees.

For an altitude of 3x Earth radii we get ...

... the ability to finally see a distinct circle beneath us.