Originally posted by SuzianneI haven't done the calculations yet but I don't think it would be quite be inversely proportional. More like:
It would have to be inversely proportional to the latitude.
As a function of the latitude, tho, I'm not at all sure about the actual angle of the tilt.
EDIT: (And yes, I did take trig in school, but I forget most of it, hehe)
F = ((v^2)/r) cos (l)
where l is latitude.
Will try and verify this and work out the angle of tilt relative to gravity if the earth stops moving. 🙂
Ok, have confirmed that it:
Starting from F = (v^2)/r I decided to represent the equation in terms of F, r and t to make things a little easier.
d = 2*pi*r
therefore F = ((2*pi*r/t)^2)/r = (4 * pi^2 * r) / (t^2)
r would be proportional to cos(l) therefore
F = ((4 * pi ^ 2 * r) / (t^2)) cos (l)
Which effectively does confirm that F = ((v^2)/r) cos (l)
Agree that it is frustrating not being able to represent equations properly on here.
Still working on the tilt.
Diagram: http://www.lunacy.force9.co.uk/images/trig.jpg
My vector addition is a little rusty but I think I got it (and referring to my very badly done drawing). After resolving:
latitute = l = theta
a = angle of tilt = alpha
g = force of gravity
z and p are just variables I used while deriving.
h = F sin (l)
p = F cos (l)
z = F cos (l) + g
tan (a) = (F sin (l)) / (F cos (l) + g)
a = arctan ((F sin (l)) / (F cos (l) + g))
I haven't checked this through but I believe it is correct. I am also sure it can be simplified further but haven't bothered to go that far.
Edit: Forgot to substitute in F as in my previous post.
F = ((4 * pi ^ 2 * r) / (t^2)) cos (l)
a = arctan (((4 * pi ^ 2 * r) / (t^2)) cos (l) sin (l) / (((4 * pi ^ 2 * r) / (t^2)) cos^2 (l) + g))
I will simplify it tomorrow, getting late. 😛