Free Fall on the Rotating Earth
2
On the earth, the previously derived form of the basic equation of mechanics holds if
we neglect the rotation about the sun and therefore consider a coordinate system at the
earth center as an inertial system.
mr̈ |M = F − mR̈|L − mω̇ × r |M − 2mω × ṙ |M − mω × (ω × r ).
(2.1)
The rotational velocity ω of the earth about its axis can be considered constant in time;
therefore, mω̇ × r = 0.
The motion of the point R, i.e., the motion of the coordinate origin of the system
(x , y , z ), still has to be recalculated in the moving system. According to (2.1), we
have
R̈|L = R̈|M + ω̇|M × R + 2ω × Ṙ|M + ω × (ω × R).
Since R as seen from the moving system is a time-independent quantity and since
ω is constant, this equation finally reads
R̈|L = ω × (ω × R).
This is the centripetal acceleration due to the earth’s rotation that acts on a body
moving on the earth’s surface. For the force equation (2.1) one gets
mr̈ = F − mω × (ω × R) − 2mω × ṙ − mω × (ω × r ).
Hence, in free fall on the earth—contrary to the inertial system—there appear virtual forces that deflect the body in the x - and y -directions.
If only gravity acts, the force F in the inertial system is F = −γ Mmr/r 3. By insertion we obtain
mr̈ = −γ
Mm
r − mω × (ω × R) − 2mω × ṙ − mω × (ω × r ).
r3
We now introduce the experimentally determined value for the gravitational acceleration g:
g = −γ
M
R − ω × (ω × R).
R3
Here we have inserted in the gravitational acceleration −γ Mr/r 3 the radius
r = R + r and kept the approximation r ≈ R, which is reasonable near the earth’s
surface. The second term is the centripetal acceleration due to the earth’s rotation,
W. Greiner, Classical Mechanics,
DOI 10.1007/978-3-642-03434-3_2, © Springer-Verlag Berlin Heidelberg 2010
9
10
2
Free Fall on the Rotating Earth
Fig. 2.1. Octant of the globe:
Position of the various coordinate systems
which leads to a decrease of the gravitational acceleration (as a function of the geographical latitude). The reduction is included in the experimental value for g. We thus
obtain
mr̈ = mg − 2mω × ṙ − mω × (ω × r ).
ω2
In the vicinity of the earth’s surface (r R) the last term can be neglected, since
enters and |ω| is small compared to 1/s. Thus the equation simplifies to
r̈ = g − 2(ω × ṙ )
or
r̈ = −ge3 − 2(ω × ṙ ).
(2.2)
The vector equation is solved by decomposing it into its components. First one suitably evaluates the vector product. From Fig. 2.1 one obtains, with e1 , e2 , e3 the unit
vectors of the inertial system and e1 , e2 , e3 the unit vectors of the moving system, the
following relation:
e3 = (e3 · e1 )e1 + (e3 · e2 )e2 + (e3 · e3 )e3
= (− sin λ)e1 + 0 e2 + (cos λ)e3 .
Because ω = ωe3 , one gets the component representation of ω in the moving system:
ω = −ω sin λe1 + ω cos λe3 .
Then for the vector product we get
ω × ṙ = (−ωẏ cos λ)e1 + (ż ω sin λ + ẋ ω cos λ)e2 − (ωẏ sin λ)e3 .
The vector equation (2.2) can now be decomposed into the following three component equations:
ẍ = 2ẏ ω cos λ,
ÿ = −2ω(ż sin λ + ẋ cos λ),
(2.3)
z̈ = −g + 2ωẏ sin λ.
This is a system of three coupled differential equations with ω as the coupling parameter. For ω = 0, we get the free fall in an inertial system. The solution of such a system
2.1
Perturbation Calculation
can also be obtained in an analytical way. It is, however, useful to learn various approximation methods from this example. We will first outline these methods and then
work out the exact analytical solution and compare it with the approximations.
In the present case, the perturbation calculation and the method of successive approximation offer themselves as approximations. Both of these methods will be presented here. The primes on the coordinates will be omitted below.
2.1 Perturbation Calculation
Here one starts from a system that is mathematically more tractable, and then one
accounts for the forces due to the perturbation which are small compared to the remaining forces of the system.
We first integrate (2.3):
ẋ = 2ωy cos λ + c1 ,
ẏ = −2ω(x cos λ + z sin λ) + c2 ,
(2.4)
ż = −gt + 2ωy sin λ + c3 .
In free fall on the earth the body is released from the height h at time t = 0; i.e., for
our problem, the initial conditions are
z(0) = h,
ż(0) = 0,
y(0) = 0,
ẏ(0) = 0,
x(0) = 0,
ẋ(0) = 0.
From this we get the integration constants
c1 = 0,
c2 = 2ωh sin λ,
c3 = 0,
and obtain
ẋ = 2ωy cos λ,
ẏ = −2ω(x cos λ + (z − h) sin λ),
(2.5)
ż = −gt + 2ωy sin λ.
The terms proportional to ω are small compared to the term gt. They represent the
perturbation. The deviation y from the origin of the moving system is a function of
ω and t; i.e., in the first approximation the term y1 (ω, t) ∼ ω appears. Inserting this
into the first differential equation, we find an expression involving ω2 . Because of the
consistency in ω we can neglect all terms with ω2 , i.e., we obtain to first order in ω
ẋ(t) = 0,
ż(t) = −gt,
and after integration with the initial conditions we get
x(t) = 0,
g
z(t) = − t 2 + h.
2
11
12
2
Free Fall on the Rotating Earth
Because x(t) = 0, in this approximation the term 2ωx cos λ drops out from the
second differential equation (2.5); there remains
ẏ = −2ω(z − h) sin λ.
Inserting z leads to
1 2
ẏ = −2ω h − gt − h sin λ
2
= ωgt 2 sin λ.
Integration with the initial condition yields
y=
ωg sin λ 3
t .
3
The solutions of the system of differential equations in the approximation ωn = 0
with n ≥ 2 (i.e., consistent up to linear terms in ω) thus read
x(t) = 0,
ωg sin λ 3
t ,
3
g
z(t) = h − t 2 .
2
y(t) =
The fall time T is obtained from z(t = T ) = 0:
2h
.
T2 =
g
From this one finds the eastward deflection (e2 points east) as a function of the fall
height:
ωg sin λ2h 2h
y(t = T ) = y(h) =
3g
g
2ωh sin λ 2h
=
.
3
g
2.2 Method of Successive Approximation
If one starts from the known system (2.5) of coupled differential equations, these equations can be transformed by integration to integral equations:
t
x(t) = 2ω cos λ y(u) du + c1 ,
0
t
y(t) = 2ωht sin λ − 2ω cos λ
t
x(u) du − 2ω sin λ
0
1
z(t) = − gt 2 + 2ω sin λ
2
t
y(u) du + c3 .
0
z(u) du + c2 ,
0
2.2
Method of Successive Approximation
Taking into account the initial conditions
x(0) = 0,
ẋ(0) = 0,
y(0) = 0,
ẏ(0) = 0,
z(0) = h,
ż(0) = 0,
the integration constants are
c1 = 0,
c2 = 0,
c3 = h.
The iteration method is based on replacing the functions x(u), y(u), z(u) under
the integral sign by appropriate initial functions. In the first approximation, one determines the functions x(t), y(t), z(t) and then inserts them as x(u), y(u), z(u) on the
right-hand side to get the second approximation. In general there results a successive
approximation to the exact solution if ω · t = 2πt/T (T = 24 hours) is sufficiently
small.
By setting x(u), y(u), z(u) to zero in the above example in the zero-order approximation, one obtains in the first approximation
x (1) (t) = 0,
y (1) (t) = 2ωht sin λ,
g
z(1) (t) = h − t 2 .
2
To check the consistency of these solutions up to terms linear in ω, we have to check
only the second approximation. If there is consistency, there must not appear terms
that involve ω linearly:
t
x
(2)
(t) = 2ω cos λ
t
y
(1)
(u) du = 2ω cos λ
0
= 4ω2 h cos λ sin λ
2ωh(sin λ)u du
0
t2
2
= f (ω2 ) ≈ 0.
Like x (1) (t), z(1) (t) is consistent to first order in ω:
z
(2)
1
(t) = h − gt 2 + 2ω sin λ
2
t
y (1) (u) du
0
g
= h − t 2 + 2ω sin λ
2
t
2ωh(sin λ)u du
0
g
= h − t 2 + i(ω2 ).
2
13
14
2
Free Fall on the Rotating Earth
On the contrary, y (1) (t) is not consistent in ω, since
t
y
(2)
(t) = 2ωht sin λ − 2ω cos λ
t
x
(1)
(u) du − 2ω sin λ
0
= 2ωh sin λ · t − 2ωh sin λ · t + gω(sin λ)
= gω(sin λ)
t3
3
z(1) (u) du
0
t3
3
= 2ωh(sin λ)t + k(ω2 ).
We see that in this second step the terms linear in ω once again changed greatly. The
term 2ωht sin λ obtained in the first iteration step cancels completely and is finally
replaced by gω(sin λ)t 3 /3. A check of y (3) (t) shows that y (2) (t) is consistent up to
first order in ω.
Just as in the perturbation method discussed above, we get up to first order in ω the
solution
x(t) = 0,
gω sin λ 3
t ,
3
g
z(t) = h − t 2 .
2
y(t) =
We have of course noted long ago that the method of successive approximation (iteration) is equivalent to the perturbation calculation and basically represents its conceptually clean formulation.
2.3 Exact Solution
The equations of motion (2.3) can also be solved exactly. For that purpose, we start
again from
ẍ = 2ω cos λẏ,
(2.3a)
ÿ = −2ω(sin λż + cos λẋ),
(2.3b)
z̈ = −g + 2ω sin λẏ.
(2.3c)
By integrating (2.3a) to (2.3c) with the above initial conditions, one gets
ẋ = 2ω cos λy,
(2.5a)
ẏ = −2ω(sin λz + cos λx) + 2ω sin λh,
(2.5b)
ż = −gt + 2ω sin λy.
(2.5c)
Insertion of (2.5a) and (2.5c) into (2.3b) yields
ÿ + 4ω2 y = 2ωg sin λt ≡ ct.
(2.6)
The general solution of (2.6) is the general solution of the homogeneous equation and
one particular solution of the inhomogeneous equation, i.e.,
y=
c
t + A sin 2ωt + B cos 2ωt.
4ω2
2.3
Exact Solution
The initial conditions at the time t = 0 are x = y = 0, z = h, and ẋ = ẏ = ż = 0. It
follows that B = 0 and 2ωA = −c/4ω2 , i.e., A = −c/8ω3 and therefore
sin 2ωt
c
c
c
t
−
,
t
−
sin
2ωt
=
y=
2ω
4ω2
8ω3
4ω2
i.e.,
y=
sin 2ωt
g sin λ
t−
.
2ω
2ω
(2.7)
Insertion of (2.7) into (2.5a) yields
sin 2ωt
ẋ = g sin λ cos λ t −
.
2ω
From the initial conditions, it follows that
2
t
1 − cos 2ωt
x = g sin λ cos λ
.
−
2
4ω2
(2.8)
Equation (2.7) inserted into (2.5c) yields
g sin λ
sin 2ωt
ż = −gt + 2ω sin λ
t−
,
2ω
2ω
sin 2ωt
,
ż = −gt + g sin2 λ t −
2ω
and integration with the initial conditions yields
2
g
1 − cos 2ωt
t
z = − t 2 + g sin2 λ
+ h.
−
2
2
4ω2
(2.9)
Summarizing, one finally has
t 2 1 − cos 2ωt
x = g sin λ cos λ
,
−
2
4ω2
g sin λ
sin 2ωt
y=
t−
,
2ω
2ω
2
1 − cos 2ωt
g 2
t
2
.
−
z = h − t + g sin λ
2
2
4ω2
(2.10)
Since ωt = 2πfall time/1 day, i.e., very small (ωt 1), one can expand (2.10):
x=
gt 2
sin λ cos λ(ωt)2 ,
6
gt 2
sin λ(ωt),
3
gt 2
sin2 λ 2
z=h−
1−
ωt .
2
3
y=
(2.11)
15
16
2
Free Fall on the Rotating Earth
If one considers only terms of first order in ωt, then (ωt)2 ≈ 0, and (2.11) becomes
x(t) = 0,
gωt 3 sin λ
,
3
g
z(t) = h − t 2 .
2
y(t) =
(2.12)
This is identical with the results obtained by means of perturbation theory. However,
(2.10) is exact!
The eastward deflection of a falling mass seems at first paradoxical, since the earth
rotates toward the east too. However, it becomes transparent if one considers that the
mass in the height h at the time t = 0 in the inertial system has a larger velocity
component toward the east (due to the earth rotation) than an observer on the earth’s
surface. It is just this “excessive” velocity toward the east which for an observer on the
earth lets the stone fall toward the east, but not ⊥ downward. For the throw upward
the situation is the opposite (see Exercise 2.2).
Fig. 2.2. Cut through the earth
in the equatorial plane viewed
from the North Pole: M is the
earth center, and ω the angular
velocity
EXAMPLE
2.1 Eastward Deflection of a Falling Body
As an example, we calculate the eastward deflection of a body that falls at the equator
from a height of 400 m.
The eastward deflection of a body falling from the height h is given by
2ω sin λh 2h
.
y(h) =
3
g
The height h = 400 m, the angular velocity of the earth ω = 7.27 · 10−5 rad s−1 ,
and the gravitational acceleration is known.
Inserting the values in y(h) yields
2 · 7.27 · 400 rad m 2 · 400 s2
y(h) =
,
9.81
3 · 105 s
2.3
Exact Solution
where rad is a dimensionless quantity. The result is
y(h) = 17.6 cm.
Thus, the body will be deflected toward the east by 17.6 cm.
EXERCISE
2.2 Eastward Deflection of a Thrown Body
Problem. An object will be thrown upward with the initial velocity v0 . Find the
eastward deflection.
Solution. If we put the coordinate system at the starting point of the motion, the
initial conditions read
z(t = 0) = 0,
ż(t = 0) = v0 ,
y(t = 0) = 0,
ẏ(t = 0) = 0,
x(t = 0) = 0,
ẋ(t = 0) = 0.
The deflection to the east is given by y, the deflection to the south by x; z = 0 denotes
the height h above the earth’s surface.
For the motion in y-direction we have, as has been shown (see (2.4)),
dy
= −2ω(x cos λ + z sin λ) + C2 .
dt
The motion of the body in x-direction can be neglected; x ≈ 0. If one further neglects the influence of the eastward deflection on z, one immediately arrives at the
equation
g
z = − t 2 + v0 t,
2
which is already known from the treatment of the free fall without accounting for the
earth’s rotation. Insertion into the above differential equation yields
g 2
dy
= 2ω t − v0 t sin λ,
dt
2
g 3 v0 2
y(t) = 2ω t − t sin λ.
6
2
At the turning point (after the time of ascent T = v0 /g), the deflection is
v3
2
y(T ) = − ω sin λ 02 .
3
g
It points toward the west, as expected.
17
Example 2.1
18
2
Free Fall on the Rotating Earth
EXERCISE
2.3 Superelevation of a River Bank
Problem. A river of width D flows on the northern hemisphere at the geographical
latitude ϕ toward the north with a flow velocity v0 . By which amount is the right bank
higher than the left one?
Evaluate the numerical example D = 2 km, v0 = 5 km/h, and ϕ = 45◦ .
Solution. For the earth, we have
m
d 2r
= −mge3 − 2mω × v with ω = −ω sin λe1 + ω cos λe3 .
dt 2
The flow velocity is v = −v0 e1 , and hence,
ω × v = −ωv0 sin ϕe2 .
Then the force is
mr̈ = F = −mge3 + 2mωv0 sin ϕe2 = F3 e3 + F2 e2 .
F must be perpendicular to the water surface (see Fig. 2.3). With the magnitude of
the force
F = 4m2 ω2 v02 sin2 ϕ + m2 g 2
one can, from Fig 2.3, determine H = D sin α and sin α = F2 /F . For the desired
height H one obtains
2Dωv0 sin ϕ
2ωv0 sin ϕ
.
≈
H =D
g
4ω2 v02 sin2 ϕ + g 2
Fig. 2.3.
For the numerical example one gets a bank superelevation of H ≈ 2.9 cm.
EXERCISE
2.4 Difference of Sea Depth at the Pole and Equator
Problem. Let a uniform spherical earth be covered by water. The sea surface takes
the shape of an oblate spheroid if the earth rotates with the angular velocity ω.
2.3
Exact Solution
19
Find an expression that approximately describes the difference of the sea depth at
the pole and equator, respectively. Assume that the sea surface is a surface of constant potential energy. Neglect the corrections to the gravitational potential due to the
deformation.
Solution.
Fig. 2.4.
γ mM
Feff (r) = − 2 er + mω2 r sin ϑex ,
r
r2
V |rr21 = − Feff (r) · dr
r = r · sin ϑ,
r1
r2
=−
γ mM
2
− 2 er + mω r sin ϑex dr · er
r
r1
γ mM r2 mω2 r 2 sin2 ϑ r2
=−
−
.
r r1
2
r1
We therefore define
Veff (r) = −
γ mM mω2 r 2 2
−
sin ϑ.
r
2
(2.13)
Let
r = R + r(ϑ);
r(ϑ) R.
The potential at the surface of the rotating sphere is constant by definition:
V (r) = −
γ mM
+ V0 .
R
According to the formulation of the problem, the earth’s surface is an equipotential surface. From this it follows that the attractive force acts normal to this surface.
Because of the constancy of the potential along the surface, no tangential force can
arise.
γ mM
r
r
m
V (r) = −
1−
− ω2 R 2 1 + 2
sin2 ϑ
R
R
2
R
!
=−
γ mM
+ V0 .
R
From this it follows that
V0 =
γ mM
m
r − ω2 R 2 sin2 ϑ − mω2 Rr sin2 ϑ.
2
2
R
As can be seen by inserting the given values, the last term can be neglected:
γ mM
mω2 R sin2 ϑ.
R2
20
2
Free Fall on the Rotating Earth
Exercise 2.4
From this it follows that
γ mM
m
r = V0 + ω2 R 2 sin2 ϑ,
2
R2
or explicitly for the difference r(ϑ),
r(ϑ) =
R2
m
V0 + ω2 R 2 sin2 ϑ .
γ mM
2
(2.14)
The second requirement for the evaluation of the deformation is the volume conservation. Since one can assume r R, we can write this requirement as a simple surface
integral
π/2 2π
da · r(ϑ) = 0,
(2.15)
ϑ=0 ϕ=0
and hence, because of the rotational symmetry in ϕ,
π/2
mω2 R 2 sin2 ϑ
V0 +
2πR · R sin ϑdϑ = 0,
2
0
from which follows
π/2
mω2 R 2 sin3 ϑ
V0 sin ϑ +
dϑ = 0.
2
0
With
π/2
sin ϑdϑ = 1 and
π/2
2
sin3 ϑdϑ = ,
3
0
0
one gets
m 2 2
ω R = 0,
3
m
V0 = − ω2 R 2 .
3
V0 +
By inserting this result into (2.14), one obtains
r(ϑ) =
2
R 4 ω2
sin2 ϑ −
.
γM 2
3
In the last step γ M/R 2 is to be replaced by g; thus we have found an approximate
expression for the difference of the sea depth:
r(ϑ) =
2
ω2 R 2
sin2 ϑ −
.
2g
3
(2.16)
2.3
Exact Solution
Exercise 2.4
By inserting the given values
R = 6370 km,
g = 9.81
m
,
s2
ω=
1
2π
= 7.2722 · 10−5 ,
T
s
we get
π
d = r
− r(0) ≈ 10.94 km.
2
If one wants to include the influence of the deformation on the gravitational potential, one needs the so-called spherical surface harmonics. They will be outlined in
detail in the lectures on classical electrodynamics.1
1
21
See W. Greiner: Classical Electrodynamics, 1st ed., Springer, Berlin (1998).
http://www.springer.com/978-3-642-03433-6