MHB Geostationary Orbits: Benefits & Challenges

[Pasha]

Hi, everyone can you help me with this question, please?
View attachment 9268

 

[skeeter]

(i) This calculation is straightforward ... what do you get for the length of AC?

(ii) recall $C = \pi \cdot d$, where $d$ is the length of the orbital diameter AC.

(iii) $BC = \sqrt{|AC|^2-|AB|^2}$

(iv) note ... $\cos(\angle{BAC}) = \dfrac{|AB|}{|AC|}$. Use inverse cosine on your calculator to determine the angle measure.

 

[Pasha]

(i) This calculation is straightforward ... what do you get for the length of AC?

(ii) recall $C = \pi \cdot d$, where $d$ is the length of the orbital diameter AC.

(iii) $BC = \sqrt{|AC|^2-|AB|^2}$

(iv) note ... $\cos(\angle{BAC}) = \dfrac{|AB|}{|AC|}$. Use inverse cosine on your calculator to determine the angle measure.

(i) for this one i got 84.300 is it right ?

 

[HallsofIvy]

Yes. Since the orbit is 35,800 km above the Earth surface, the distance from the Earth's surfacr to C, although it is not shown in the picture is also 35,800 km so the distance from A to C is 35,800+ 12,700+ 35,800= 84,300 km.