Answer:
Approximately \(11.1\%\) (approximately \(0.111\)) 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 \({\rm A}\) denote the center of the Earth, and let point \({\rm B}\) denote the space station. The distance between \({\rm A}\) and \({\rm B}\) would be the sum of the radius of the Earth (approximately \(6371\; {\rm km}\)) and the height of the space station above the surface of the planet: approximately \((6\, 371\,000\; {\rm m} + 408\,773\; {\rm m})\).
Let \({\rm C}\) and \({\rm D}\) denote the two points in this imaginary plane where the line-of-sight from the space station \({\rm B}\) intersects the surface of the Earth. These two lines would be tangent to the circle, such that \(\angle {\rm ACB} = 90^{\circ} = \angle {\rm ADB}\).
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 \({\rm C}\) and \({\rm D}\). That ratio would be the same as the ratio between the center angle \(\angle {\rm DAC}\) and the angle of a full circle, \(2\, \pi\) radians (or \(360^{\circ}\).)
To find the measure of angle \(\angle {\rm DAC}\), observe that \(\angle {\rm DAB} = \angle {\rm CAB}\) since the two right triangles \(RT \triangle {\rm ACB}\) and \(RT \triangle {\rm ADB}\) would be congruent by "HL": because \({\rm AC} = {\rm AD}\) while \({\rm AB}\) is shared between the two triangles, \(RT \triangle {\rm ACB} \cong RT \triangle {\rm ADB}\). Hence, \(\angle {\rm DAC}\) would be equal to twice of either of these two angles:
\(\angle {\rm DAC} = \angle {\rm DAB} + \angle {\rm CAB} = 2\, \angle {\rm DAB} = 2\, \angle{\rm CAB}\).
Since \(\angle {\rm DAB}\) is part of right triangle \(RT \triangle {\rm ADB}\), the measure of this angle can be found using trigonometry:
- Length of hypotenuse \({\rm AB}\) is \((6\, 371\,000\; {\rm m} + 408\,773\; {\rm m})\).
- Length of side \({\rm DA}\), which is adjacent to \(\angle {\rm DAB}\), is \(6\, 371\, 000\; {\rm m}\)- same as the radius of the planet.
Hence:
\(\begin{aligned}\angle {\rm DAB} &= \arccos\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) \\ &\approx \arccos\left(\frac{6\, 371\, 000\; {\rm m}}{6\, 371\,000\; {\rm m} + 408\,773\; {\rm m}}\right) \\ &\approx 20.00^{\circ}\end{aligned}\).
Hence, \(\angle {\rm DAC} = 2\, \angle {\rm DAB} \approx 40.0^{\circ}\). The fraction of visible circumference (the shorter arc between \({\rm C}\) and \({\rm D}\)) would be equal to the ratio between the center angle \(\angle {\rm DAB}\) and the angle of a full circle, \(360^{\circ}\):
\(\begin{aligned}\frac{40.00^{\circ}}{360^{\circ}} \approx 11.1\%\end{aligned}\).
