CHAPTER
10
Motion in a
Noninertial Reference Frame
10-1.
The accelerations which we feel at the surface of the Earth are the following:
:
980 cm/sec 2
(1)
Gravitational
(2)
Due to the Earth’s rotation on its own axis:
 2π rad/day 
rω = 6.4 × 10 cm × 

 86400 sec/day 
(
2
)
8
(
) (
= 6.4 × 10 8 × 7.3 × 10 −5
)
2
2
= 3.4 cm/sec 2
(3) Due to the rotation about the sun:
(
rω = 1.5 × 10
2
13


2π rad/year
cm × 

 86400 × 365 sec/day 
)
2
2
 7.3 × 10 −5 
= 1.5 × 1013 × 
= 0.6 cm/sec 2

 365 
(
10-2.
)
The fixed frame is the ground.
y
a
x
θ
The rotating frame has the origin at the center of the tire and is the frame in which the tire is at
rest.
From Eqs. (10.24), (10.25):
+ a + ω × r + ω × ( ω × r ) + 2ω × v
af = R
f
r
r
333
334
CHAPTER 10
Now we have
= − a cos θ i + a sin θ j
R
f
r = r0 i
ω=
vr = ar = 0
V
k
r0
ω =
a
k
r0
Substituting gives
a f = − a cos θ i + a sin θ j + a j −
v2
i
r0
 v2

a f = −i  + a cos θ  + j ( sin θ + 1) a
 r0

2
We want to maximize a f , or alternatively, we maximize a f :
af
2
=
v4
2 av 2
2
2
a
cos
cos θ + a 2 + 2 a 2 sin θ + a 2 sin 2 θ
+
θ
+
2
r0
r0
=
v4
2av 2
2
a
+
+
2
cos θ + a 2 sin 2 θ
r02
r0
d af
dθ
2
=−
2av 2
cos θ + 2a 2 cos θ
r0
= 0 when tan θ =
ar0
v2
(Taking a second derivative shows this point to be a maximum.)
tan θ =
ar0
implies cos θ =
v2
v2
a 2 r02 + v 4
and
sin θ =
ar0
a r + v4
2 2
0
Substituting into (1)
 v2
av 2
a f = −i  +
 r0
a 2 r02 + v 4
 

ar0
 + j
+ 1 a
  a 2 r02 + v 4

This may be written as
a f = a + a 2 + v 4 r02
(1)
335
MOTION IN A NONINERTIAL REFERENCE FRAME
A
θ
This is the maximum acceleration. The point which experiences this acceleration is at A:
where tan θ =
10-3.
ar0
v2
We desire Feff = 0 . From Eq. (10.25) we have
− mω × r − mω × ( ω × r ) − 2mω × v
Feff = F − mR
f
r
r
0
ω
The only forces acting are centrifugal and friction, thus µs mg = mω 2 r , or
r=
µs g
ω2
10-4. Given an initial position of (–0.5R,0) the initial velocity (0,0.5ωR) will make the puck
motionless in the fixed system. In the rotating system, the puck will appear to travel clockwise
in a circle of radius 0.5R. Although a numerical calculation of the trajectory in the rotating
system is a great aid in understanding the problem, we will forgo such a solution here.
10-5.
The effective acceleration in the merry-go-round is given by Equation 10.27:
x = ω 2 x + 2ω y
(1)
y = ω 2 y − 2ω x
(2)
These coupled differential equations must be solved with the initial conditions
x0 ≡ x ( 0 ) = −0.5 m , y0 ≡ y ( 0 ) = 0 m , and x ( 0 ) = y ( 0 ) = v0 2 m ⋅ s −1 , since we are given in the
problem that the initial velocity is at an angle of 45° to the x-axis. We will vary v0 over some
range that we know satisfies the condition that the path cross over ( x0 , y0 ) . We can start by
looking at Figures 10-4e and 10-4f, which indicate that we want v0 > 0.47 m ⋅ s −1 . Trial and error
can find a trajectory that does loop but doesn’t cross its path at all, such as v0 = 0.53 m ⋅ s −1 .
From here, one may continue to solve for different values of v0 until the wanted crossing is
eyeball-suitable. This may be an entirely satisfactory answer, depending on the inclinations of
the instructor. An interpolation over several trajectories would show that an accurate answer to
the problem is v0 = 0.512 m ⋅ s −1 , which exits the merry-go-round at 3.746 s. The figure shows
this solution, which was numerically integrated with 200 steps over the time interval.
336
CHAPTER 10
1
y (m)
0.5
0
0.5
–1
–0.5
–1
0
0.5
1
x (m)
10-6.
z
z = f(r)
m
r
Consider a small mass m on the surface of the water. From Eq. (10.25)
− mω × r − mω × ( ω × r ) − 2mω × v
Feff = F − mR
f
r
In the rotating frame, the mass is at rest; thus, Feff = 0 . The force F will consist of gravity and the
force due to the pressure gradient, which is normal to the surface in equilibrium. Since
= ω = v = 0 , we now have
R
f
r
0 = mg + Fp − mω × ( ω × r )
where Fp is due to the pressure gradient.
Fp
θ
mω2r
θ′
mg
Since Feff = 0 , the sum of the gravitational and centrifugal forces must also be normal to the
surface.
Thus θ ′ = θ.
tan θ ′ = tan θ =
ω 2r
g
337
MOTION IN A NONINERTIAL REFERENCE FRAME
but
tan θ =
dz
dr
Thus
z=
ω2
2g
r 2 + constant
The shape is a circular paraboloid.
10-7. For a spherical Earth, the difference in the gravitational field strength between the poles
and the equator is only the centrifugal term:
g poles − gequator = ω 2 R
For ω = 7.3 × 10 −5 rad ⋅ s −1 and R = 6370 km, this difference is only 34 mm ⋅ s −2 . The disagreement
with the true result can be explained by the fact that the Earth is really an oblate spheroid,
another consequence of rotation. To qualitatively describe this effect, approximate the real Earth
as a somewhat smaller sphere with a massive belt about the equator. It can be shown with more
detailed analysis that the belt pulls inward at the poles more than it does at the equator. The
next level of analysis for the undaunted is the “quadrupole” correction to the gravitational
potential of the Earth, which is beyond the scope of the text.
10-8.
ω
z
y
λ
x
Choose the coordinates x, y, z as in the diagram. Then, the velocity of the particle and the
rotation frequency of the Earth are expressed as
v = ( 0, 0, z )


ω = ( −ω cos λ , 0, ω sin λ ) 
(1)
so that the acceleration due to the Coriolis force is
a = −2ω × r = 2ω ( 0, − z cos λ , 0 )
(2)
338
CHAPTER 10
This acceleration is directed along the y axis. Hence, as the particle moves along the z axis, it
will be accelerated along the y axis:
y = −2ω z cos λ
(3)
Now, the equation of motion for the particle along the z axis is
z = v0 − gt
z = v0 t −
where v0 is the initial velocity and is equal to
h:
(4)
1 2
gt
2
(5)
2gh if the highest point the particle can reach is
v0 = 2 gh
(6)
y = −2ω z cos λ + c
(7)
From (3), we have
but the initial condition y ( z = 0 ) = 0 implies c = 0. Substituting (5) into (7) we find
1


y = −2ω cos λ  v0 t − gt 2 


2
(
= ω cos λ gt 2 − 2v0 t 2
)
(8)
Integrating (8) and using the initial condition y(t = 0) = 0, we find
1

y = ω cos λ  gt 2 − v0 t 2 
3


(9)
From (5), the time the particle strikes the ground (z = 0) is
1 

0 =  v0 − gt t

2 
so that
t=
2v0
g
(10)
Substituting this value into (9), we have
 1 8v 3
4v 2 
y = ω cos λ  g 30 − v0 20 
g 
3 g
v3
4
= − ω cos λ 02
3
g
If we use (6), (11) becomes
(11)
339
MOTION IN A NONINERTIAL REFERENCE FRAME
4
8h3
y = − ω cos λ
3
g
(12)
The negative sign of the displacement shows that the particle is displaced to the west.
10-9. Choosing the same coordinate system as in Example 10.3 (see Fig. 10-9), we see that the
lateral deflection of the projectile is in the x direction and that the acceleration is
ax = x = 2 ω z vy = 2 (ω sin λ )(V0 cos α )
(1)
Integrating this expression twice and using the initial conditions, x ( 0 ) = 0 and x ( 0 ) = 0 , we
obtain
x ( t ) = ω V0 t 2 cos α sin λ
(2)
Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis force. In
this approximation, we have
z ( t ) = V0 t sin α −
1 2
gt
2
(3)
from which the time T of impact is obtained by setting z = 0:
T=
2V0 sin α
g
(4)
Substituting this value for T into (2), we find the lateral deflection at impact to be
x (T ) =
4ωV03
sin λ cos α sin 2 α
2
g
(5)
10-10. In the previous problem we assumed the z motion to be unaffected by the Coriolis
force. Actually, of course, there is an upward acceleration given by −2ω x vy so that
z = 2ω V0 cos α cos λ − g
(1)
from which the time of flight is obtained by integrating twice, using the initial conditions, and
then setting z = 0:
T′ =
2V0 sin α
g − 2ωV0 cos α cos λ
(2)
Now, the acceleration in the y direction is
ay = y = 2ω x vz
= 2 ( −ω cos λ )(V0 sin α − gt )
Integrating twice and using the initial conditions, y ( 0 ) = V0 cos α and y ( 0 ) = 0 , we have
(3)
340
CHAPTER 10
y (t) =
1
ω gt 3 cos λ − ωV0 t 2 cos λ sin α + V0 t cos α
3
(4)
Substituting (2) into (4), the range R′ is
R′ =
4ωV03 sin 3 α cos λ
2V02 cos α cos λ
8 ωV03 g sin 3 α cos λ
−
+
3 ( g − 2ωV0 cos α cos λ ) 3 ( g − 2ωV0 cos α cos λ ) 2 g − 2ωV0 cos α cos λ
(5)
We now expand each of these three terms, retaining quantities up to order ω but neglecting all
quantities proportional to ω 2 and higher powers of ω. In the first two terms, this amounts to
neglecting 2ωV0 cos α cos λ compared to g in the denominator. But in the third term we must
use
2V02 cos α sin α
 2ωV0

cos α cos λ 
g 1 −
g


≅
 2ωV0

2V02
cos α sin α 1 +
cos α cos λ 
g
g


= R0′ +
4ωV03
sin α cos 2 α cos λ
2
g
(6)
where R′0 is the range when Coriolis effects are neglected [see Example 2.7]:
R0′ =
2V02
cos α sin α
g
(7)
The range difference, ∆R ′ = R ′ − R0′ , now becomes
∆R ′ =
4ωV03
1


cos λ  sin α cos 2 α − sin 3 α 
2


g
3
(8)
Substituting for V0 in terms of R′0 from (7), we have, finally,
∆R′ =
2R0′
1


ω cos λ  cot1 2 α − tan 3 2 α 


g
3
(9)
341
MOTION IN A NONINERTIAL REFERENCE FRAME
10-11.
d=
Rθ
R sin θ
θ
This problem is most easily done in the fixed frame, not the rotating frame. Here we take the
Earth to be fixed in space but rotating about its axis. The missile is fired from the North Pole at
some point on the Earth’s surface, a direction that will always be due south. As the missile
travels towards its intended destination, the Earth will rotate underneath it, thus causing it to
miss. This distance is:
∆ = (transverse velocity of Earth at current latitude) × (missile’s time of flight)
= ω R sin θ × T
=
(1)
dω R
 d
sin  
 R
v
(2)
Note that the actual distance d traveled by the missile (that distance measured in the fixed
frame) is less than the flight distance one would measure from the Earth. The error this causes
in ∆ will be small as long as the miss distance is small. Using R = 6370 km, ω = 7.27 × 10 −5
rad ⋅ s −1 , we obtain for the 4800 km, T = 600 s flight a miss distance of 190 km. For a 19300 km
flight the missile misses by only 125 km because there isn’t enough Earth to get around, or
rather there is less of the Earth to miss. For a fixed velocity, the miss distance actually peaks
somewhere around d = 12900 km.
Doing this problem in the rotating frame is tricky because the missile is constrained to be in a
path that lies close to the Earth. Although a perturbative treatment would yield an order of
magnitude estimate on the first part, it is entirely wrong on the second part. Correct treatment
in the rotating frame would at minimum require numerical methods.
10-12.
ε
r0
λ
z
Fs
x
342
CHAPTER 10
Using the formula
Feff = ma f − mω × ( ω × r ) − 2mω × v r
(1)
we try to find the direction of Feff when ma f (which is the true force) is in the direction of the z
axis. Choosing the coordinate system as in the diagram, we can express each of the quantities in
(1) as
vr = 0


ω = (−ω cos λ , 0, ω sin λ ) 


r = (0, 0, R)


ma f = (0, 0, − mg0 )

(2)
ω × r = Rω cos λ e y
(3)
Hence, we have
and (1) becomes
ex
ey
Feff = − mg0 ez − m −ω cos λ
0
ez
0
ω sin λ
Rω cos λ
0
(4)
from which, we have
Feff = − mg0 ez + mRω 2 sin λ cos λ ex + mRω 2 cos 2 λ e z
(5)
Therefore,
( Ff )x = mRω 2 sin λ cos λ



2
2
( Ff )z = − mg 0 + mRω cos λ 
(6)
The angular deviation is given by
tan ε =
( Ff )x
( Ff )z
=
Rω 2 sin λ cos λ
g 0 − Rω 2 cos 2 λ
(7)
Since ε is very small, we can put ε ≅ ε . Then, we have
ε=
Rω 2 sin λ cos λ
g 0 − Rω 2 cos 2 λ
(8)
It is easily shown that ε is a maximum for λ 45° .
Using R = 6.4 × 10 8 cm , ω = 7.3 × 10 −5 sec −1 , g = 980 cm/sec 2 , the maximum deviation is
ε≅
1.7
≅ 0.002 rad
980
(9)
343
MOTION IN A NONINERTIAL REFERENCE FRAME
10-13.
ω
ε
z
z′
Earth
x
x′
λ
The small parameters which govern the approximations that need to be made to find the
southerly deflection of a falling particle are:
height of fall
h
=
R radius of Earth
(1)
centrifugal force
Rω 2
=
g0
purely gravitational force
(2)
δ≡
and
α≡
The purely gravitational component is defined the same as in Problem 10-12. Note that
although both δ and α are small, the product δα = hω 2 g 0 is still of order ω 2 and therefore
expected to contribute to the final answer.
Since the plumb line, which defines our vertical direction, is not in the same direction as the
outward radial from the Earth, we will use two coordinate systems to facilitate our analysis. The
unprimed coordinates for the Northern Hemisphere-centric will have its x-axis towards the
south, its y-axis towards the east, and its z-axis in the direction of the plumb line. The primed
coordinates will share both its origin and its y′-axis with its unprimed counterpart, with the z′and x′-axes rotated to make the z′-axis an outward radial (see figure). The rotation can be
described mathematically by the transformation
x = x ′ cos ε + z ′ sin ε
(3)
y = y′
(4)
z = − x ′ sin ε + z ′ cos ε
(5)
Rω 2
sin λ cos λ
g
(6)
where
ε≡
as found from Problem 10-12.
a) The acceleration due to the Coriolis force is given by
a X ≡ −2ω × v ′
Since the angle between ω and the z′-axis is π – λ, (7) is most appropriately calculated in the
primed coordinates:
(7)
344
CHAPTER 10
x ′ = 2ω y ′ sin λ
(8)
y ′ = −2ω ( z ′ cos λ + x ′ sin λ )
(9)
z ′ = 2ω y ′ cos λ
(10)
In the unprimed coordinates, the interesting component is
x = 2ω y ( sin λ cos ε + cos λ sin ε )
(11)
At our level approximation this becomes
x 2ω y sin λ
(12)
Using the results for y and z , which is correct to order ω (also found from Example 10.3),
x 2ω 2 gt 2 sin λ cos λ
(13)
Integrating twice and using the zeroth order result for the time-of-fall, t = 2 h g , we obtain for
the deflection
dX =
2 h2 2
ω sin λ cos λ
3 g
(14)
b) The centrifugal force gives us an acceleration of
a c ≡ −ω × ( ω × r ′ )
(15)
The component equations are then
x ′ = ω 2 sin λ x ′ sin λ + ( R + z ′ ) cos λ 
(16)
y′ = ω 2 y′
(17)
z ′ = ω 2 cos λ x ′ sin λ + ( R + z ′ ) cos λ  − g0
(18)
where we have included the pure gravitational component of force as well. Now transform to
the unprimed coordinates and approximate
x ω 2 ( R + z ) sin λ cos λ − g0 sin ε
(19)
We can use Problem 10-12 to obtain sin ε to our level of approximation
sin ε ε Rω 2
sin λ cos λ
g0
(20)
The prompts a cancellation in equation (19), which becomes simply
x ω 2 z sin λ cos λ
(21)
Using the zeroth order result for the height, z = h − gt 2 2 , and for the time-of-fall estimates the
deflection due to the centrifugal force
dc 5 h2 2
ω sin λ cos λ
6 g
(22)
345
MOTION IN A NONINERTIAL REFERENCE FRAME
c) Variation in gravity causes the acceleration
ag ≡ −
GM
r + g0 k
r3
(23)
where r = x ′i + y ′j + ( R + z ′ ) k is the vector pointing to the particle from the center of the
spherical Earth. Near the surface
r 2 = x ′ 2 + y ′ 2 + ( z ′ + R) R2 + 2Rz ′
2
(24)
so that (23) becomes, with the help of the binomial theorem,
ag −
g0
( x ′i + y ′j − 2 z ′k )
R
(25)
Transform and get the x component
g0
( − x ′ cos ε + 2z ′ sin ε )
R
(26)
=
g0
 − ( x cos ε − z sin ε ) cos ε + 2 ( x sin ε + z cos ε ) sin ε 
R
(27)
g0
( − x + 3z sin ε )
R
(28)
x Using (20),
x 3ω 2 z sin λ cos λ
(29)
where we have neglected the x R term. This is just thrice the part (b) result,
dg 5 h2 2
ω sin λ cos λ
2 g
(30)
Thus the total deflection, correct to order ω 2 , is
d4
h2 2
ω sin λ cos λ
g
(31)
(The solution to this and the next problem follow a personal communication of Paul Stevenson,
Rice University.)
10-14. The solution to part (c) of the Problem 10-13 is modified when the particle is dropped
down a mineshaft. The force due to the variation of gravity is now
ag ≡ −
g0
r + g0 k
R
(1)
As before, we approximate r for near the surface and (1) becomes
ag −
In the unprimed coordinates,
g0
( x ′i + y ′j + z ′k )
R
(2)
346
CHAPTER 10
x − g0
x
R
(3)
To estimate the order of this term, as we probably should have done in part (c) of Problem
10-13, we can take x ~ h 2ω 2 g , so that
x ~ ω 2 h ×
h
R
(4)
which is reduced by a factor h R from the accelerations obtained previously. We therefore have
no southerly deflection in this order due to the variation of gravity. The Coriolis and centrifugal
forces still deflect the particle, however, so that the total deflection in this approximation is
d
10-15.
3 h2 2
ω sin λ cos λ
2 g
(5)
The Lagrangian in the fixed frame is
( )
1
mv 2f − U rf
2
L=
(1)
where v f and rf are the velocity and the position, respectively, in the fixed frame. Assuming
we have common origins, we have the following relation
v f = v r + ω × rr
(2)
where vr and rr are measured in the rotating frame. The Lagrangian becomes
L=
m 2
2
vr + 2v r ⋅ ( ω × rr ) + ( ω × rr )  − U ( rr )


2
(3)
∂L
= mv r + m ( ω × rr )
∂v r
(4)
The canonical momentum is
pr ≡
The Hamiltonian is then
H ≡ v r ⋅ pr − L =
1
1
2
mvr2 − U ( rr ) − m ( ω × rr )
2
2
(5)
H is a constant of the motion since ∂L ∂t = 0 , but H ≠ E since the coordinate transformation
equations depend on time (see Section 7.9). We can identify
Uc = −
1
2
m ( ω × rr )
2
(6)
as the centrifugal potential energy because we may find, with the use of some vector identities,
−∇U c =
m  2 2
2
∇ ω rr − ( ω ⋅ rr ) 


2
= m ω 2 rr − ( ω ⋅ rr ) ω 
(7)
(8)
347
MOTION IN A NONINERTIAL REFERENCE FRAME
= − mω × ( ω ⋅ rr )
(9)
which is the centrifugal force. Computing the derivatives of (3) required in Lagrange’s
equations
d ∂L
= ma r + mω × v r
dt ∂v r
∂L
= m∇ ( v r × ω ) ⋅ rr  − ∇ (U c + U )
∂rr
= − m ( ω × v r ) − mω × ( ω × rr ) − ∇U
(10)
(11)
(12)
The equation of motion we obtain is then
ma r = −∇U − mω × ( ω × rr ) − 2m ( ω × v r )
(13)
If we identify Feff = ma r and F = −∇U , then we do indeed reproduce the equations of motion
given in Equation 10.25, without the second and third terms.
10-16. The details of the forces involved, save the Coriolis force, and numerical integrations
in the solution of this problem are best explained in the solution to Problem 9-63. The only thing
we do here is add an acceleration caused by the Coriolis force, and re-work every part of the
problem over again. This is conceptually simple but in practice makes the computation three
times more difficult, since we now also must include the transverse coordinates in our
integrations. The acceleration we add is
a c = 2ω  vy sin λi − ( vx sin λ + vz cos λ ) j + vy cos λ k 
(1)
where we have chosen the usual coordinates as shown in Figure 10-9 of the text.
a) Our acceleration is
a = − gk + a C
(2)
As a check, we find that the height reached is 1800 km, in good agreement with the result of
Problem 9-63(a). The deflection at this height is found to be 77 km, to the west.
b) This is mildly tricky. The correct treatment says that the equation of motion with air
resistance is (cf. equation (2) of Problem 9-63 solution)

v
a = − g k + 2
vt


v  + aC

The deflection is calculated to be 8.9 km.
c)
Adding the vaiation due to gravity gives us a deflection of 10 km.
d)
Adding the variation of air density gives us a deflection of 160 km.
(3)
348
CHAPTER 10
Of general note is that the deflection in all cases was essentially westward. The usual small
deflection to the north did not contribute significantly to the total transverse deflection at this
precision. All of the heights obtained agreed well with the answers from Problem 9-63.
Inclusion of the centrifugal force also does not change the deflections to a significant degree at
our precision.
10-17. Due to the centrifugal force, the water surface of the lake is not exactly perpendicular
to the Earth’s radius (see figure).
B
m
g
β
Ta
n
ge
nt
Wa
ter
sur
β
to
fac
rth
A
α
C
e
Ea
su
r
fa
ce
The length BC is (using cosine theorem)
BC =
AC 2 + (mg ) 2 − 2 ACmg cos α
where AC is the centrifugal force AC = mω 2 R cos α with α = 47° and Earth’s radius
R ≅ 6400 km ,
The angle β that the water surface is deviated from the direction tangential to the Earth’s surface
is
BC
AC
=
sin α sin β
⇒ sin β =
AC sin α
= 4.3 × 10−5
BC
So the distance the lake falls at its center is h = r sin β where r = 162 km is the lake’s radius.
So finally we find h = 7 m.
349
MOTION IN A NONINERTIAL REFERENCE FRAME
10-18.
Let us choose the coordinate system Oxyz as shown in the figure.
α
O
νy
β
νx
y
ν
x
The projectile’s velocity is
 vx   v0 cos β 
G   
v = v y = v0 sin β − gt 
  

0
 0 

where β = 37°
The Earth’s angular velocity is
 −ω cos α 
ω =  −ω sin α 


0


G
where α = 50°
So the Coriolis acceleration is
G
G G
ac = 2v × ω = −2v0 ω cos β sin α + 2 ( v0 sin β − gt ) ω cos α e z
(
)
The velocity generated by Coriolis force is
t
vc = ∫ ac dt = 2v0 ω t ( cos β sin α − sin β cos α ) − gt 2ω cos α
0
And the distance of deviation due to the Coriolis force is
gt 3ω cos α
zc = ∫ vc dt = −v0 ω t sin (α − β ) −
3
0
t
2
The flight time of the projectile is t =
2v0 sin β
. If we put this into zc , we find the deviation
2
distance due to Coriolis force to be
zc ~ 260 m
350
CHAPTER 10
10-19.
The Coriolis force acting on the car is
G
G
G G
Fc = 2m v × ω ⇒ Fc = 2mvω sin α
where α = 65°, m = 1300 kg, v = 100 km/hr.
G
So Fc = 4.76 N.
10-20. Given the Earth’s mass, M = 5.976 × 1024 kg , the magnitude of the gravitational field
vector at the poles is
g pole =
GM
= 9.866 m/s 2
2
R pole
The magnitude of the gravitational field vector at the equator is
g eq =
GM
− ω 2 R e q = 9.768 m/s 2
2
Req
where ω is the angular velocity of the Earth about itself.
If one use the book’s formula, we have
g (λ = 90°) = 9.832 m/s 2 at the poles
and
g (λ = 0°) = 9.780 m/s 2 at the equator
10-21.
The Coriolis acceleration acting on flowing water is
G
G G
ac = 2v × ω ⇒
G
ac = 2vω sin α
Due to this force, the water is higher on the west bank. As in problem 10-17, the angle β that the
water surface is deviated from the direction tangential to Earth’s surface is
sin β =
ac
=
g +a
2
2
c
2vω sin α
g + 4v ω sin α ∝
2
2
2
2
= 2.5 × 10−5
The difference in heights of the two banks is
∆h = A sin β = 1.2 × 10−3 m
where A = 47 m is the river’s width.
10-22.
G
G
G
G
The Coriolis acceleration is ac = 2v × ω . This acceleration ac pushes lead bullets
G
eastward with the magnitude ac = 2vω cos α = 2 gt ω cos α , where α = 42°.
The velocity generated by the Coriolis force is
351
MOTION IN A NONINERTIAL REFERENCE FRAME
vc (t ) = ∫ a dt = gt 2ω cos α
and the deviation distance is
∆xc = ∫ vc (t ) dt =
The falling time of the bullet is t =
∆xc =
gt 3
ω cos α
3
2h g . So finally
ω 8h3
3
g
cos α = 2.26 × 10−3 m
352
CHAPTER 10