Determine the direction of the effective value of vector g at a latitude of 60 N

Question

Determine the direction of the effective value of vector g at a latitude of 60 North on the Earth.

Hi, This question really confuses me because while I know the effective value of vector G at latitude 60, I don't understand why it would be different at a latitude of 60 North of Earth. I drew a circle to rep Earth. Above the horizontal is theta = 60 degrees, with a line r that reps. Earth's radius, and the horizontal line reps. Centripetal acceleration. I know Centripetal acceleration is ac = V^2 / r and v = 2(pi)r over T

I'm not really sure where to go from there expect that I was thinking of finding the hypotenuse of that triangle. But I wouldn't exactly understand why.

Explanation / Answer

The centripetal force is just the force of gravity; that always points to the center of mass CM. That is, the force of gravity is always downward perpendicular to the surface of your idealized Earth. Thus, g points downward perpendicular to the surface of Earth. It's value g = 9.81 m/sec^2.

Centrifugal force, on the other hand, points outward perpendicular from the rotation axis. The radius r of the spin axis is r = R cos(lat); where R is Earth's idealized radius and lat = 60 deg. Let a = v^2/r = v^2/(R cos(lat)) be the acceleartion outward due to centrifuga force. [Note...this is where you erred. The rotation axis is not the radius of Earth as you used it. It is the perpendicular to the rotation axis at latitude 60 deg.]

Thus, the net g = G = g + a in vector format. If we set the direction of a to be theta = 0 degrees, then the direction of g is theta = 225 deg. In which case, we find the X and Y components of each acceleration.

gx = g cos(240) and gy = g sin(240); and ax = a cos(0) and ay = a sin(0). Thus the X components of G are Gx = g cos(240) + a cos(0) = g cos(240) + a, and Y are Gy = g sin(240) + a sin(0) = g sin(240).

In which case the angle omega = arctan(Gy/Gx), which will give the angle for the net g = G. The magnitude is found from G^2 = Gx^2 + Gy^2 = (g cos(240) +a)^2 + (g sin(240))^2 g = 9.81 m/sec^2 and you solved for a = v^2/r; so you can solve for G, Gx, and Gy.

Remember, the Earth rotates around an axis through the N and S poles; so the radii of rotation become shorter as we go towards either of the poles. That is r = R cos(lat); where R is the Earth's radius at the equater where lat = 0 deg.

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