Geostationary satellite question using the pythagorean theorem...?

Question

I have calculated the distance from the Equator to the satellite to be 35,848 km. I know that the radius of the Earth is 6,378 km. Using the Pythagorean Theorem, I calculated a hypotenuse of 36,411 km.

But I am asked to determine above and below which latitudes the satellite cannot see, and this is where I'm stuck. When I solve for the angle using trig identities, I get 10 degrees...which obviously isn't correct.

Does anybody have any suggestions? Thank you!!!

Answer 1

Actually, ten degrees is about right. Closer than that to the poles, the satellite would be below the horizon. (Sorry; no Oprah re-runs at South Pole Station.) Incidentally, is your 35,848 km distance measured to the surface of the earth, or to its center? Makes a difference, of course.

Answer 2

The angle is arccos(6378/35848) = 79.75º The distance to the satellite is the hypotenuse, the radius of earth is side adjacent. Equator = 0º latitude, north pole = 90º

Answer 3

If it is 10 degrees from the poles, or 80 degrees latitude, it sounds plausible.

Answer 4

♠ F=G*M*m/r^2 = m*r*w^2, w=2pi/(24*3600);
GM = r^3*52.8852*10^(-10);
GM/R^2 = g = (R+h)^3/R^2 *7.27(2)*10^(-5); (R+h)^3 =g*R^2/*7.27(2)*10^(-5) = 9.8*6378^2*10^(6+10)/52.8852 = 75380813*10^15; R+h =422,429*10^5m = 42243km; h=35865km; Ok with you!
♣ sin(θ) = R/(R+h) =0.151, θ=8.68° all around are not visible.
♦ yes, you are right!

Answer 5

i think its only about 10degrees....the answer isthat it should be near the poles and so it is possible i think.......