Lecture 20 The Effects of the Earth s Rotation

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1 Lecture 20 The Effects of the Earth s Rotation 20.1 Static Effects The Earth is spinning/rotating about an axis with (aside from slight variations) uniform angular velocity ω. Let us examine the effects of this rotation on a stationary body. For example, consider hanging a plumb line (a mass/bob suspended from a cord/rope) from the top of the tower of Pisa (h = 50meters). We would have originally expected that the gravitational attraction would pull the bob towards the center of the Earth. But, referring back to our general equation 19.1, it is clear that the centrifugal force will cause the bob to be swung outwards through a small angle? What is the magnitude of this force and deflection? We define a coordinate system where the origin is at the center of the Earth and the z axis points in the radial direction. Take the x,y axes of our coordinate system to lie on the surface of the Earth, as shown in the following figure. The coordinates of the tower are (0, 0, R), where R is the radius of the Earth. From the above figure, it is clear that, ω = ω y ĵ + ω zˆk or [ ] ω = ω cos αĵ + sin αˆk (20.1) Since r = Rˆk, ω r = ωr cos αî Therefore, the centrifugal force becomes, [ ] F cent = m ω ( ω r) = ω 2 R cos α sin αĵ cos 2 αˆk (20.2) We thus see that,

2 N (y) W E (x) ω S α r α F Latitude (a) The centrifugal force weakens the gravitational force by subtracting off ω 2 R cos 2 α. (b) There is a deflection in the y direction or southern direction by ω 2 R cos α sin α. The angle of deflection between the true g and the effective g following figure. is seen in the ω F g g* β This angle is calculated as the ratio of the horizontal and vertical components of the centrifugal force: tan β = F horizontal F vertical = mω2 R cos α sin α mg + mω 2 R cos 2 α = ω2 R cos α sin α g ω 2 R cos 2 α (20.3)

3 Note that R = 6370km, ω = s 1, therefore ω 2 R 34mm/s 2 << g. Hence, the deflection angle is very small and can be approximated by: tan β β ω2 R cos α sin α g (20.4) Note that the maximum value occurs at α = 45 o, which leads to a deflection of the plumb line of β radians. Therefore, if buildings were built according to this plumb line they would be tilted by 1.7milliradians! As a side note, if you want to appear lighter then move to the equator where the reduction in g is maximal g = g ω 2 R. Therefore, the weight reduction is about 0.34%. The actual measured difference is a bit larger: g = g pole g eq = 52mm/s 2 This occurs because the Earth is not a perfect sphere, but is flattened at the poles. Therefore, g is larger at the pole then the equator independent of the centrifugal contribution Dynamic Effects of Rotations on Moving Objects We return now to the tower of Pisa, but this time we will drop a marble and see what happens. The centrifugal force will contribute by causing a deflection of the marble, as we saw for our static plumb line. However, the Coriolis force makes a much larger contribution to the resultant motion of the marble. Lets examine this motion by successive approximations. If the marble is moving initially with velocity, v (0) = gtˆk, where (0) denotes a zeroeth order approximation, then the Coriolis force becomes: where we have employed expression F (0) Cor = 2m ω v = 2mωgt cos αî (20.5) 1 actually, the flattening is due to rotations, so they are not really independent in a sense

4 What is the change of momentum due to this force? p (1) = t 0 F (0) Cordt = mωgt 2 cos αî where (1) denotes a first-order approximation or correction. Thus, the change in velocity due to this force is just, v (1) = ωgt 2 cos αî which means that, to first-order, the velocity becomes, v (1) = v (0) + v (1) (20.6) Since for the zeroeth-order approximation there is no Coriolis force, the zeroeth-order approximation to the position of the particle is 0, r (0) = 0. Hence, r (1) = r (1) = ωgt3 3 cos αî (20.7) where we have simply integrated Thus, as the marble drops it falls to the east. From standard kinematics, z(t) = h 1/2gt 2, where h is the height above the ground. Thus, the time for the marble to hit the ground is z = 0, t 2 = 2h/g and the deflection is: r (1) = 1 ( ) 8h 3 1/2 3 ω cos αî (20.8) g For a height of 50m at a latitude of 45 o (very approximately), the drift is: 1 3 ( rad/s) ( 8 (50m) 3 9.8m/s 2 ) 1/2 1 2 î = 5.47mm Let us now calculate the second-order correction/approximation for the deflection: v (1) = v (0) + v (1) = gtˆk + ωgt 2 cos αî which gives rise to a first-order Coriolis correction of, F (1) Cor = F (0) Cor + F (1) Cor = 2m ω v (1) (20.9) = 2mωgt cos αî 2mω 2 gt 2 sin α cos αĵ In turn, this gives rise to a second-order correction in the velocity, (20.10) v (2) = 2 3 ω2 gt 3 sin α cos αĵ (20.11)

5 If we integrate the above equation, we obtain the second-order correction to the position, r (2) = ω2 gt 4 sin α cos αĵ 6 = ω2 h 2 sin(2α) (20.12) 3g This gives us an additional deviation to the south of 0.45 microns!