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Geostationary Orbit

Pasha

New member
Sep 14, 2019
2
Hi, everyone can you help me with this question, please?
photo_2019-09-14_13-24-12.jpg
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
(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
Sep 14, 2019
2
(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
Jan 29, 2012
1,151
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.