# Geostationary Orbit

#### skeeter

##### Well-known member
MHB Math Helper
(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

##### New member
(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

##### Well-known member
MHB Math Helper
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.