Answer:
Approximately [tex]11.1\%[/tex] (approximately [tex]0.111[/tex]) of the entire circumference.
Step-by-step explanation:
Refer to the diagram attached (not to scale.) Assume that the Earth is a sphere. The diagram is a cross-section view along a plane that contains both the space station and the center of the Earth.
Let point [tex]{\rm A}[/tex] denote the center of the Earth, and let point [tex]{\rm B}[/tex] denote the space station. The distance between [tex]{\rm A}[/tex] and [tex]{\rm B}[/tex] would be the sum of the radius of the Earth (approximately [tex]6371\; {\rm km}[/tex]) and the height of the space station above the surface of the planet: approximately [tex](6\, 371\,000\; {\rm m} + 408\,773\; {\rm m})[/tex].
Let [tex]{\rm C}[/tex] and [tex]{\rm D}[/tex] denote the two points in this imaginary plane where the line-of-sight from the space station [tex]{\rm B}[/tex] intersects the surface of the Earth. These two lines would be tangent to the circle, such that [tex]\angle {\rm ACB} = 90^{\circ} = \angle {\rm ADB}[/tex].
The question is asking for the fraction of the circumference visible from the space station. In this diagram, this part of the circumference is equivalent to the shorter arc between [tex]{\rm C}[/tex] and [tex]{\rm D}[/tex]. That ratio would be the same as the ratio between the center angle [tex]\angle {\rm DAC}[/tex] and the angle of a full circle, [tex]2\, \pi[/tex] radians (or [tex]360^{\circ}[/tex].)
To find the measure of angle [tex]\angle {\rm DAC}[/tex], observe that [tex]\angle {\rm DAB} = \angle {\rm CAB}[/tex] since the two right triangles [tex]RT \triangle {\rm ACB}[/tex] and [tex]RT \triangle {\rm ADB}[/tex] would be congruent by "HL": because [tex]{\rm AC} = {\rm AD}[/tex] while [tex]{\rm AB}[/tex] is shared between the two triangles, [tex]RT \triangle {\rm ACB} \cong RT \triangle {\rm ADB}[/tex]. Hence, [tex]\angle {\rm DAC}[/tex] would be equal to twice of either of these two angles:
[tex]\angle {\rm DAC} = \angle {\rm DAB} + \angle {\rm CAB} = 2\, \angle {\rm DAB} = 2\, \angle{\rm CAB}[/tex].
Since [tex]\angle {\rm DAB}[/tex] is part of right triangle [tex]RT \triangle {\rm ADB}[/tex], the measure of this angle can be found using trigonometry:
Hence:
[tex]\begin{aligned}\angle {\rm DAB} &= \arccos\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) \\ &\approx \arccos\left(\frac{6\, 371\, 000\; {\rm m}}{6\, 371\,000\; {\rm m} + 408\,773\; {\rm m}}\right) \\ &\approx 20.00^{\circ}\end{aligned}[/tex].
Hence, [tex]\angle {\rm DAC} = 2\, \angle {\rm DAB} \approx 40.0^{\circ}[/tex]. The fraction of visible circumference (the shorter arc between [tex]{\rm C}[/tex] and [tex]{\rm D}[/tex]) would be equal to the ratio between the center angle [tex]\angle {\rm DAB}[/tex] and the angle of a full circle, [tex]360^{\circ}[/tex]:
[tex]\begin{aligned}\frac{40.00^{\circ}}{360^{\circ}} \approx 11.1\%\end{aligned}[/tex].
