If the distance to the center of the earth at 90° is 6357km and at 0° it is 6378km wat would it be at 39.0747°?
im having to prove gravity is 9.8meters per second squared, so im doing the newton equation to figure out if that is how you prove it and i need the distance from where i stand to the center of the earth because as you know it requires the distance from center to center. if anyone can help please help me out.
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Lets assume the earth a perfect oblate spheroid, the circumference could be described as [assuming 0° to be the equator)
x = sin (a) * 6378
y = cos (a) * 6357
(try plotting the x and the y out on a sheet of paper and you'll end up with a slightly oblate sphere)
The distance to the centre of the earth would be
(x^2+y^2)^,5 (Pythagorus)
lets plug in 39.0747° into both formulas
x = 4020.26 km
y = 4935.09 km
The distance is (x^2+y^2)^0,5 = 6365.35 km roughly in the middle of the distances at 0° and at 90°. Just as you'd expect it to be.