5.1. Accelerated Coordinate Systems:

5.1. Accelerated Coordinate Systems:

Recall: Uniformly moving reference frames (e.g. those considered at 'rest' or moving with constant velocity in a straight line) are called inertial reference frames.

Sometimes it is necessary, to employ a coordinate system that is not inertial.

Let us first consider the case of a coordinate system that undergoes pure translation.

Assume Oxyz are the primary fixed coordinate axes, and O'x'y'z' are the moving axes. In the case of pure translation, the respective axes Ox and O'x', and so on, remain parallel.

The position vector of a

particle P is denoted by r in the fixed system and by r' in the moving system. The displacement OO' of the moving origin is denoted by R0. Thus, from the triangle OO'P, we have

r = R

0

+ r'

(115) Taking the first and second time derivatives gives

v = V

₀

+ v'

(116)

a = A

0

+ a'

(117)

in which V0 and A0 are, respectively, the velocity and acceleration of the moving system, and v' and a' are the velocity and acceleration of the particle in the moving system.

If the moving system is not accelerating, i.e. it is also inertial, so that A0 = 0, then

a = a'

In this case we cannot specify a unique coordinate system, because Newton's laws will be the same in both systems.

For example, Newton's second law in fixed system F = ma becomes F' = ma' in the moving system.

On the other hand if the moving system is accelerating, then Newton's second law becomes

F = mA

0

+ ma'

or

F - mA

₀

= F'

(118) where (-mA0) is known as the inertial term or inertial force.

Such "force" is not due to interactions with other bodies;

rather, it happens as a result of the acceleration of the reference system.

5.2. Rotating Coordinate Systems

In this section, we show how velocities, accelerations, and forces transform between an inertial frame of reference and a noninertial one that is rotating.

Assume that the axes of the both coordinate systems have a common origin. Let the rotation of the rotated system takes place about some specific axis of rotation, whose direction is designated by a unit vector, n.

The angular velocity of the rotating system then is;

ω = ωn

The direction of the velocity vector is given by the right- hand rule.

The position of any point P in space can be designated by the vector r in the fixed system and by the vector r' in the rotating system.

Because the coordinate axes of the two systems have the same origin, these vectors are equal, that is,

r = r' (119) or;

ix + jy+ kz = i'x' + j'y'+ k' z'

When we differentiate with respect to time to find the velocity, we must keep in mind the fact that the unit vectors i', j', and k' are not constant. Thus, we can write the velocity vector v in the fixed system as;

dt k z d dt

j y d dt

i

x d ′

+ ′

′ ′

′ + + ′

= v′

v (120)

Where v' is the velocity in the rotating system.

From the definition of the cross product, we can write;

i

dt

d ′ = × ′ i ω

, j dt

d ′ = × ′ j ω

and k dt

d ′ = × ′ k ω

. Hence;

( ) ( ) (

r ω

k ω j

ω i

k ω j

i

′

)

×

=

× ′ + ′

× ′ + ′

× ′

= ′

′ ′

′ + + ′

′ ′ x y z

dt z d dt y d dt x d

This is the velocity of P due to rotation of the coordinate system. Accordingly, Eq. (120 ) can be rewritten as

v = v′+ω×r′ (121) Taking the first time derivatives gives the acceleration in the fixed system in terms of the position, velocity, and acceleration in the rotating system;

^{a =a′+ω}_& ^{×r′+2ω×v′+ω×}

(

^ω×r′

)

(122) If the moved system is undergoing both translation and rotation, the general equations for transforming from a fixed system to a moving and rotating system will be:

v = v′+ω×r′+V0 (123)

And;

^{a =a′+ω}_& ^{×r′+2ω×v′+ω×}

(

^ω×r′

)

^+A₀ (124)

‰ The term 2ω × v' is known as the Coriolis acceleration, which appears whenever a particle moves in a rotating coordinate system except when the velocity v' is parallel to the axis of rotation.

‰ The term ω×(ω× r') is called the centripetal acceleration. which is the result of the

particle being carried around a circular path in the rotating system. It is always directed toward the axis of rotation and is perpendicular to the axis as shown in the figure.

‰ The term is called the transverse acceleration, because it is perpendicular

to the position vector r'. It appears whenever the rotating system has an angular acceleration, i.e. if the angular velocity vector is changing in either magnitude or direction, or both.

r

ω& × ′

5.3. Dynamics of a Particle in a Rotating System:

It is well known that the equation of motion of a particle in an inertial frame of reference is ;

F = ma

where F is the sum of all real, physical forces acting on the particle.

According to Eq.(124) , we can write the equation of motion of a particle in a noninertial frame of reference as ;

^F−mω_& ^{×r′−2mω×v′−mω×}

(

^ω×r′

)

^−mA₀ ^=ma′ ₍₁₂₅₎

All inertial forces have names corresponding to their relevant accelerations. Thus;

‰ The force -mω& ×r′ is called the transverse force, because it is perpendicular to the position vector r'. It is present only if there is an angular acceleration (or deceleration) of the rotating coordinate system.

‰ The force -2mω×v' is the Coriolis force, which appears whenever a particle moves in a rotating coordinate system. Its direction is always perpendicular to v', thus it seems to deflect the moving particle at right angles to its direction of motion.

‰ The force - mω×(ω× r') is the centrifugal force, which is the result of the particle being carried around a circular path in the rotating system. It is directed outward away from the axis of rotation and is perpendicular to that axis.

If r' is perpendicular to ω, the magnitude of the centrifugal force is mr'ω².

A no inertial observer in an accelerated frame of reference must include all, or some, of these inertial forces along with the real forces F to calculate the correct motion of the particle.

In other words, such an observer writes the fundamental equation of motion as;

F' = ma'

in which the sum of the vector forces F' acting on the particle is given by

F' = F

_physical

+ F'

_trans

+ F'

_Cor

+ F'

_cel1trif

– mA

₀

F (or Fphysical) forces are the only forces that a no inertial observer claims are actually acting upon the particle.

5.4. Effects of Earth's Rotation

‰ Consider a coordinate system that is moving with the Earth. Because the angular speed of Earth's rotation is 2π radians per day, the effects of such rotation is relatively small.

‰ Nevertheless, it is the spin of the Earth that makes the equatorial radius is some 13 miles greater than the polar radius, i.e. equatorial bulge.

5.4.1. Static Effects: The Plumb line

Let us describe the motion of the plumb bob in a local frame of reference whose origin is at the position of the bob. Our frame of reference is attached to the surface of the Earth, so it is undergoing

translation as well as rotation.

‰ The translation of the frame takes place along a circle whose radius is ρ = re cosλ, where re is the radius of the Earth and λ is the geocentric latitude of the plumb bob. Hence;

A₀ =ω²ρ =ω² r_e cosλ (126)

‰ Its rate of rotation is ω, the same as that of the Earth about its axis. Let us now examine the terms of Eq. (124):

‰ The force - ma' is zero, because the bob is at rest in the local frame of reference, i.e. a' = 0.

‰ The Coriolis force -2mω × v' is zero, because v'= 0.

‰ The transverse force -mω& ×r′ is zero, because ω is constant.

‰ The centrifugal force - m ω×(ω× r') is zero, because the origin of the local coordinate system is centered on the bob. I.e, r'= 0.

The only surviving terms in Eq. (124) are the real forces F and the inertial term –mA0, which arises because the local frame of reference is accelerating. Thus,

F – mA0 = 0 (127) In other words, the rotation of the Earth causes the acceleration A0 of the local frame. The bob does not hang on a line pointing toward the center of the Earth because the inertial force –mA₀ throws it outward, away from Earth's axis of rotation. The magnitude of this force is;

mω² r_e cosλ

It is a maximum when λ = 0 at the Earth's equator and a minimum at either pole.

The tension T in the string balances out the real gravitational force mg₀ and the inertial force –mA₀, i.e;

(T + mg₀) – mA₀ = 0 (128) Now, when we hang a plumb bob,

we normally think that the tension T balances out the local force of gravity, which we call mg. We can see from the Figure that:

mg = mg₀ – mA₀ or

g = g₀ –A₀ (129)

As can be seen the inertial reaction –mA0, directed away from Earth's axis, causes the direction of the plumb line to deviate by a small angle ε away from the direction toward Earth's center.

We can easily calculate the value of the angle ε. From the Figure we have;

m r_e mg λ λ

ω

ε sin cos

sin

2 =

or, because ε is small

ε ε ω sin2λ sin 2

2

g r_e

=

≈ (130)

Thus, ε vanishes at the equator (λ = 0) and the poles (λ =±90). The maximum deviation of the direction of the plumb line from the center of the Earth occurs at λ = 45°

where; g^{e o} r 0.1 2

2

max = ω ≈

ε

5.4.2. Dynamic Effects:

1- Falling Body

A body that is dropped from a height h above the ground, as it falls, it will drift to the east. The eastward drift is given by:

ω cosλ g

x h ⎟⎟⎠

⎜⎜ ⎞

⎝

= ⎛

′ ₃¹ 8 ³

(131) For a height of 100 m at latitude of 45°, the drift is 1.55 cm 2- Deflection of a Rifle Bullet

If we fire a projectile to east with high initial speed v0 , the projectile will bend to the south. If H is the horizontal range of the projectile, the transverse deflection is then;

ω λ v sin

H

0

≈ 2

Δ (132) This is the same for any direction in which the projectile is initially aimed, provided the trajectory is flat.

6.1. Newton’s Law of Gravitation:

‰

During his study of the motions of the planets and of the moon, Newton discovered the fundamental charter of the gravitational attraction between any two bodies. In 1687, Newton published the law of gravitation which may be stated as follows:

Every particle of matter in the universe attracts every other particle with force (F_g) that is directly proportional to the product of the masses (m1, m2) of the particles and inversely proportional to the square of the distance between them (r).

Or;

2 2 1

r m F_g = Gm

(131) G is a fundamental physical constant called the gravitational constant. The numerical value of G (in SI units ) is

G = 6.67×10^-11 N.m²/kg²

‰

Equation (131) tells us that if the distance ^r is doubled, the force is only one-fourth as great, and so on. Although many of the stars in the night sky are more massive than the sun, they are so far away that their gravitational force on earth is negligible.

‰

Gravitational forces always act along the line joining the two particles, which form an action-reaction pair. Even when the masses of the particles are different, the two interaction forces have equal magnitude.

‰

Gravitational forces combine vectorially. If each of two masses exerts a force on a third, the total force on the third mass is the vector sum of the individual forces of the first two. This property is often called superposition of forces.

‰

The earth's gravitational force on a body of mass m at any point outside the earth is given by ; Fg = GmEm/r², where mE is the mass of the earth and r is the distance of the body from the earth's center. Therefore, we can express the gravitational potential energy (U) in more general form as;

U_g = − r^E

Gm m (132)

6.2. Kepler’s laws and the motion of planets:

‰

One of the great intellectual events of the 16th and 17th centuries was the threefold realization;

1- that the earth is also a planet, 2- that all planets orbit the sun,

3- and that the apparent motions of the planets as seen from the earth can be used to determine the orbits of the planets precisely.

The first and second of these ideas were published by Nicolaus Copernicus in 1543. The determination of planetary orbits was carried out between 1601 and 1619 by the German astronomer and mathematician Johannes Kepler.

‰

By trial and error, Kepler discovered three observed laws that accurately described the motions of the planets.

Two hundred years later, Newton discovered that each of Kepler's laws can be derived using Newton's laws of motion and the law of gravitation.

First law: Law of Ellipses.

Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse.

The following Figure shows the geometry of the ellipse with its main properties;

* The longest dimension 2a is the major axis, with half- length a known as the semi- major axis.

* S and S' are the foci (plural of focus). The sun is at S, and the planet is at P.

* The sum of the distances from S to P and from S' to P is the same for all points on the curve.

* The distance of each focus from the center of the ellipse is ea, where e is a dimensionless number between 0 and 1 called the eccentricity. If e = 0, the ellipse is a circle. The actual orbits of the planets are somewhat circular; their eccentricities range from 0.007 for Venus to 0.248 for Pluto. The earth's orbit has e = 0.017.

* The point in the planet's orbit closest to the sun is the perihelion, and the point most distant from the sun is the aphelion.

Scecond law: Law of Equal Areas

A line from the sun to a given planet sweeps out equal areas in equal times.

In a small time interval dt, the line from the sun S to the planet P turns through an angle dθ. The area swept out is the dA = ½ r² dθ .The rate at which area is swept out, dA/dt, is called the sector velocity :

dt r d dt

dA ₂ θ

21

=

The real meaning of Kepler's second law is that the sector velocity has the same value at all

points in the orbit. When the planet is close to the sun, r is small and dθ/dt is large; when the planet is far from the sun, r is large and dθ/dt is small.

Third law: Harmonic Law

The square of the period of a planet is directly proportional to the cube of the semi-major axis of the plant’s orbit.

This law can be expressed as;

T² = ka³

If the distance measured in astronomical units (1AU=1.5

×10⁸ km) and periods are measured in Earth years then: k=1.

EXAMPLE (6.6.1):

Find the period of a comet whose semi-major axis is 4 AU.

Solution:

With T measured in years and a in astronomical units, we have

T² = a³= (4)³ =64 yrs² T = 8 yrs

However, using Newton’s laws of motion and the inverse- square law of gravity, one can found (see problem 6.5) that the constant k in SI unit is equals to;

Gms

k

4π2

= Where ms is the sun's mass.

Kepler’s 3^rd law can be then rewritten as

Gms

T = 2π a^3/2

(133)

Note:

(

^{6.6721067 10}

)(

^{1.99) 10}

)

^{2.38 10 75.5 years}

2 ⁹

30 11

2 / 3

12 = × =

×

×

= ×

− s

. T π (

The period does not depend on the eccentricity e. I.e.

an asteroid in an elliptical orbit with semi-major axis a will have the same orbital period as a planet in a circular orbit of radius a. The key difference is that the asteroid moves at different speeds at different points in its elliptical orbit, while the planet's speed is constant around its circular orbit.

Example : Comet Halley

Comet Halley moves in an elongated elliptical orbit around the sun. At perihelion, the comet is 8.75 × 10⁷ km from the sun; at aphelion it is 5.26 × 10⁹ km from the sun. Find the semi- major axis, eccentricity, and period of the orbit.

Solution:

- The length of the major axis is;

2a = 8.75 × 10⁷ + 5.26 × 10⁹ So; a = 2.67 × 10⁹ km

- Since the comet-sun distance at perihelion is given by a - ea = a (1- e) = 8.75 × 10⁷

Then; e = 0.967 - From Eq. (133),