Download Lecture 20 The Effects of the Earth`s Rotation

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
~ = ωy ĵ + ωz k̂
or
h
i
ω
~ = ω cos αĵ + sin αk̂
Since ~r = Rk̂, ω
~ × ~r = ωR cos αî Therefore, the centrifugal force becomes,
h
i
2
2
~
F cent = −m~ω × (~ω × ~r) = −ω R cos α sin αĵ − cos αk̂
We thus see that,
(20.1)
(20.2)
N (y)
ω
W
E (x)
F
α
r
S
Latitude
α
(a) The centrifugal force weakens the gravitational force by subtracting off ω 2 R cos2 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~∗ is seen in the
following figure.
ω
F
g
g*
β
• This angle is calculated as the ratio of the horizontal and vertical components of the
centrifugal force:
Fhorizontal
Fvertical
−mω 2 R cos α sin α
=
−mg + mω 2 R cos2 α
ω 2 R cos α sin α
=
g − ω 2 R cos2 α
tan β =
(20.3)
Note that R = 6370km, ω = 7.292 × 10−5 s−1 , therefore ω 2 R ' 34mm/s2 << g. Hence,
the deflection angle is very small and can be approximated by:
tan β ' β '
ω 2 R cos α sin α
g
(20.4)
Note that the maximum value occurs at α = 45o , which leads to a deflection of the
plumb line of β ' 1.7 × 10−3 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 ∗ = gpole
− geq
= 52mm/s2
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
contribution1 .
20.2
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) = −gtk̂,
where (0) denotes a zeroeth order approximation, then the Coriolis force becomes:
(0)
F~ Cor = −2m~ω × ~v = 2mωgt cos αî
where we have employed expression 20.1.
1
actually, the flattening is due to rotations, so they are not really independent in a sense
(20.5)
• What is the change of momentum due to this force?
Z t
(0)
(1)
∆~p =
F~ Cor dt0 = mωgt2 cos αî
0
where (1) denotes a first-order approximation or correction. Thus, the change in velocity due to this force is just,
∆~v (1) = ωgt2 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
cos αî
3
(20.7)
where we have simply integrated 20.6. Thus, as the marble drops it falls to the east.
• From standard kinematics, z(t) = h − 1/2gt2 , where h is the height above the ground.
Thus, the time for the marble to hit the ground is z = 0, t2 = 2h/g and the deflection
is:
µ 3 ¶1/2
1
8h
(1)
∆~r = ω
cos αî
(20.8)
3
g
For a height of 50m at a latitude of 45o (very approximately), the drift is:
1
(7.2722 × 10−5 rad/s)
3
µ
8 × (50m)3
9.8m/s2
¶1/2
1
√ î = 5.47mm
2
• Let us now calculate the second-order correction/approximation for the deflection:
~v (1) = ~v (0) + ∆~v (1) = −gtk̂ + ωgt2 cos αî
which gives rise to a first-order Coriolis correction of,
(1)
F~ Cor =
=
=
(0)
(1)
F~ Cor + ∆F~ Cor
−2m~ω × ~v (1)
2mωgt cos αî − 2mω 2 gt2 sin α cos αĵ
(20.9)
(20.10)
In turn, this gives rise to a second-order correction in the velocity,
2
∆~v (2) = − ω 2 gt3 sin α cos αĵ
3
(20.11)
If we integrate the above equation, we obtain the second-order correction to the position,
ω 2 gt4
sin α cos αĵ
6
ω 2 h2
= −
sin(2α)
3g
∆~r(2) =
This gives us an additional deviation to the south of 0.45 microns!
¨
(20.12)