Chap12_Sec4 [Compatibility Mode]

7/23/2019 Chap12_Sec4 [Compatibility Mode]

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VECTOR FUNCTIONS


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12.4

Motion in Space:

Velocity and Acceleration

In this section, we will learn about:

The motion of an object

using tangent and normal vectors.

VECTOR FUNCTIONS


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Here, we show how the ideas of tangent

and normal vectors and curvature can be

used in physics to study:

The motion of an object, including its velocity

and acceleration, along a space curve.

MOTION IN SPACE: VELOCITY AND ACCELERATION


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In particular, we follow in the footsteps of

Newton by using these methods to derive

Keplers First Law of planetary motion.

VELOCITY AND ACCELERATION


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Suppose a particle moves through

space so that its position vector at

time tis r(t).

VELOCITY


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Notice from the figure that, for small values

of h, the vector

approximates

the direction of the

particle moving along

the curve r(t).

VELOCITY

( ) ( )t h t

h

+ r r

Vector 1


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Its magnitude measures the size

of the displacement vector per unittime.

VELOCITY


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The vector 1 gives the average

velocity over a time interval oflength h.

VELOCITY


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Its limit is the velocity vector v(t)

at time t :

VELOCITY VECTOR

0

( ) ( )

( ) lim'( )

h

t h t

t h

t

+ =

=

r r

v

r

Equation 2


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Thus, the velocity vector is also

the tangent vector and points in

the direction of the tangent line.

VELOCITY VECTOR


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The speed of the particle at time t

is the magnitude of the velocity vector,

that is, |v(t)|.

SPEED


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This is appropriate because, from Equation 2

and from Equation 7 in Section 12.3,

we have:

= rate of change

of distance with

respect to time

SPEED

| ( ) | | '( ) | dst tdt

= =v r


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As in the case of one-dimensional motion,

the acceleration of the particle is defined as

the derivative of the velocity:

a(t) = v(t) = r (t)

ACCELERATION


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The position vector of an object moving

in a plane is given by:

r(t) = t3 i + t2j

Find its velocity, speed, and acceleration

when t= 1 and illustrate geometrically.

VELOCITY & ACCELERATION Example 1


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The velocity and acceleration at time t

are:

v(t) = r(t) = 3t2 i + 2tj

a(t) = r (t) = 6t I + 2j



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The speed at tis:


2 2 2

4 2

| ( ) | (3 ) (2 )

9 4

t t t

t t

= +

= +

v


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When t = 1, we have:

v(1) = 3 i + 2j

a(1) = 6 i + 2j

|v(1)| =


13


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These velocity and acceleration vectors

are shown here.



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Find the velocity, acceleration, and

speed of a particle with position vector

r(t) = t2, et, tet



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2 2 2 2

( ) '( ) 2 , , (1 )

( ) '( ) 2, , (2 )

| ( ) | 4 (1 )

t t

t t

t t

t t t e t e

t t e t e

t t e t e

= = +

= = +

= + + +

v r

a v

v


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The figure shows the path of the particle in

Example 2 with the velocity and acceleration

vectors when t = 1.

VELOCITY & ACCELERATION


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The vector integrals that were introduced in

Section 12.2 can be used to find position

vectors when velocity or acceleration vectors

are known, as in the next example.



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A moving particle starts at an initial position

r(0) = 1, 0, 0

with initial velocity

v(0) = ij + k

Its acceleration is

a(t) = 4t i + 6tj + k

Find its velocity and position at time t.



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Since a(t) = v(t), we have:

v(t) = a(t) dt

= (4t i + 6tj + k) dt

=2t2 i + 3t2j + t k + C



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To determine the value of the constant

vector C, we use the fact that

v(0) = i j + k

The preceding equation gives v(0) = C.

So,

C = i j + k



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It follows:

v(t) = 2t2 i + 3t2j + t k + ij + k

= (2t

2

+ 1) i + (3t

2

1) j + (t + 1) k



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Since v(t) = r(t), we have:

r(t) = v(t) dt

= [(2t2 + 1) i + (3t2 1)j + (t + 1) k] dt

= (t3 + t) i + (t3 t)j + (t2 + t) k + D



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Putting t = 0, we find that D = r(0) = i.

So, the position at time tis given by:

r(t) = (t3

+ t + 1) i + (t3

t)j + (t2

+ t) k



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The expression for r(t) that we obtained

in Example 3 was used to plot the path

of the particle here for 0 t 3.



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In general, vector integrals allow us

to recover:

Velocity, when acceleration is known

Position, when velocity is known


0 0( ) ( ) ( )

t

tt t u du= + v v a

00( ) ( ) ( )

t

tt t u du= + r r v


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If the force that acts on a particle is known,

then the acceleration can be found from

Newtons Second Law of Motion.



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The vector version of this law states that if,

at any time t, a force F(t) acts on an object

of mass m producing an acceleration a(t),

then

F(t) = ma(t)



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An object with mass m that moves in

a circular path with constant angular speed

has position vector

r(t) = a cos t i + a sin tj

Find the force acting on the object and

show that it is directed toward the origin.



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To find the force, we first need to know

the acceleration:

v(t) = r(t) = a sin t i + a cos tj

a(t) = v(t) = a2 cos t i a2 sin tj



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Therefore, Newtons Second Law gives

the force as:

F(t) = ma(t)

= m2

(a cos t i + a sin tj)



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Notice that:

F(t) = m2r(t)

This shows that the force acts in the directionopposite to the radius vector r(t).



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Therefore, it points toward

the origin.



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Such a force is called a centripetal

(center-seeking) force.

CENTRIPETAL FORCE Example 4


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A projectile is fired with:

Angle of elevation

Initial velocity v0



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Assuming that air resistance is negligible

and the only external force is due to gravity,

find the position function r(t) of the projectile.



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What value of maximizes the range

(the horizontal distance traveled)?



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We set up the axes so that the projectile

starts at the origin.



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As the force due to gravity acts downward,

we have:

F = ma = mgj

where g= |a| 9.8 m/s2.

Therefore, a = gj



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Since v(t) = a, we have:

v(t) = gtj + C

where C = v(0) = v0.

Therefore,

r(t) = v(t) = gtj + v0



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Integrating again, we obtain:

r(t) = gt2j + t v0

+ D

However,

D = r(0) = 0



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So, the position vector of the projectile

is given by:

r(t) = gt2j + t v0

VELOCITY & ACCELERATION E. g. 5Equation 3


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If we write |v0| = v0 (the initial speed

of the projectile), then

v0 = v0 cos i + v0 sin j

Equation 3 becomes:

r(t) = (v0 cos

)t i + [(v0 sin

)t gt2

]j



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Therefore, the parametric equations

of the trajectory are:

x = (v0 cos )t

y = (v0 sin )t gt2

VELOCITY & ACCELERATION E. g. 5Equations 4


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If you eliminate tfrom Equations 4,

you will see that yis a quadratic function

ofx.



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So, the path of the projectile is part

of a parabola.



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The horizontal distance dis the value

ofx when y= 0.

Setting y= 0, we obtain:

t= 0 or t = (2v0 sin )/g



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That second value of tthen gives:

Clearly, dhas its maximum value when

sin 2= 1, that is, = /4.


0

0

2 20 0

2 sin

( cos )

(2sin cos ) sin 2

v

d x v g

v v

g g

= =

= =

Example 5


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A projectile is fired with muzzle speed 150 m/s

and angle of elevation 45 from a position

10 m above ground level.

Where does the projectile hit the ground?

With what speed does it do so?


VELOCITY & ACCELERATION E l 6


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If we place the origin at ground level,

the initial position of the projectile is (0, 10).

So, we need to adjust Equations 4

by adding 10 to the expression for y.




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With v0 = 150 m/s, = 45, and g= 9.8 m/s2,

we have:


212

2

150 cos( / 4) 75 2

10 150 sin( / 4) (9.8)

10 75 2 4.9

x t t

y t t

t t

= =

= +

= +

Example 6



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Impact occurs when y= 0, that is,

4.9t2 75 t 10 = 0

Solving this quadratic equation (and usingonly the positive value of t), we get:


75 2 11, 250 196

21.749.8t+ +

=

Example 6

2



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Then,

x 75 (21.74)

2306

So, the projectile hits the ground

about 2,306 m away.


2



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The velocity of the projectile is:


( ) '( )

75 2 (75 2 9.8 )

t t

t

=

= +

v r

i j

Example 6



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So, its speed at impact is:


2 2| (21.74) | (75 2) (75 2 9.8 21.74)

151m/s

= +

v

Example 6

ACCELERATIONCOMPONENTS


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When we study the motion of a particle,

it is often useful to resolve the acceleration

into two components:

Tangential (in the direction of the tangent)

Normal (in the direction of the normal)




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If we write v = |v| for the speed of the particle,

then

Thus,

v = vT


'( ) ( )( )| '( ) | | ( ) |

t ttt t v

= = =r v vTr v

ACCELERATIONCOMPONENTS Equation 5


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If we differentiate both sides of that

equation with respect to t, we get:

ACCELERATION COMPONENTS

' ' 'v v= = +a v T T

Equation 5



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If we use the expression for the curvature

given by Equation 9 in Section 12.3,

we have:

ACCELERATION COMPONENTS Equation 6

| ' | | ' | so | ' || ' |

vv

= = =T T Tr



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The unit normal vector was defined

in Section 12.4 as N = T/ |T|

So, Equation 6 gives:

ACCELERATION COMPONENTS

' | ' | v= =T T N N

ACCELERATIONCOMPONENTS Formula/Equation 7


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Then, Equation 5 becomes:

2'v v= +a T N

q

ACCELERATIONCOMPONENTS Equations 8


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Writing aTand aN for the tangential and

normal components of acceleration,

we have

a = aTT + aNN

where

aT= v and aN= Kv2

q



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This resolution is illustrated

here.



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Lets look at what Formula 7

says.

2'v v= +a T N



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The first thing to notice is that

the binormal vector B is absent.

No matter how an object moves through space,

its acceleration always lies in the plane of T and N(the osculating plane).

Recall that T gives the direction of motion andN points in the direction the curve is turning.



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Next, we notice that:

The tangential component of acceleration is v,

the rate of change of speed.

The normal component of acceleration is v2,

the curvature times the square of the speed.



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This makes sense if we think of

a passenger in a car.

A sharp turn in a road means a large value

of the curvature .

So, the component of the acceleration perpendicular

to the motion is large and the passenger is thrown

against a car door.



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High speed around the turn has

the same effect.

In fact, if you double your speed,

aN is increased by a factor of 4.



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We have expressions for the tangential

and normal components of acceleration in

Equations 8.

However, its desirable to have expressions

that depend only on r, r, and r .



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Thus, we take the dot product of v = vT

with a as given by Equation 7:

v a = vT (v T + v2N)

= vv T T + v3T N

= vv (Since T T = 1 and T N = 0)



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Therefore,

'

'( ) "( )| '( ) |

Ta v v

t tt

= =

=

v a

r rr



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Using the formula for curvature given by

Theorem 10 in Section 12.3, we have:

2 2

3

| '( ) "( ) || '( ) |

| '( ) || '( ) "( ) |

| '( ) |

N

t ta v t

t

t t

t

= =

=

r rr

r

r r

r

ACCELERATIONCOMPONENTS Example 7


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A particle moves with position function

r(t) = t2, t2, t3

Find the tangential and normal

components of acceleration.



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2 2 3

2

2 4

( )

'( ) 2 2 3

"( ) 2 2 6

| ( ) | 8 9

t t t t

t t t t

t t

t t t= +

r = i + j + k

r = i + j + k

r = i + j + k

r '



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Therefore, Equation 9 gives the tangential

component as:

3

2 4

'( ) "( )

| '( ) |

8 18

8 9

T

t ta

t

t t

t t

=

+=

+

r r

r



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2

2 2

'( ) "( ) 2 2 3

2 2 6

6 6

t t t t t

t

t t

=

=

ji k

r r

i j



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Hence, Equation 10 gives the normal

component as:

2

2 4

'( ) "( )

| '( ) |

6 2

8 9

N

t ta

t

t

t t

=

=

+

r r

r

KEPLERS LAWS OF PLANETARY MOTION


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We now describe one of the great

accomplishments of calculus by showing how

the material of this chapter can be used toprove Keplers laws of planetary motion.

KEPLERS LAWS OF PLANETARY MOTION


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After 20 years of studying the astronomical

observations of the Danish astronomer

Tycho Brahe, the German mathematician andastronomer Johannes Kepler (15711630)

formulated the following three laws.

KEPLERS FIRST LAW


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A planet revolves around the sun

in an elliptical orbit with the sun at

one focus.

KEPLERS SECOND LAW


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The line joining the sun to

a planet sweeps out equal areas

in equal times.

KEPLERS THIRD LAW


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The square of the period of revolution

of a planet is proportional to the cube

of the length of the major axis of its orbit.

I hi b k P i i i M th ti f 1687

KEPLERS LAWS


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In his book Principia Mathematica of 1687,

Sir Isaac Newton was able to show that

these three laws are consequences oftwo of his own laws:

Second Law of Motion

Law of Universal Gravitation

I h t f ll K l

KEPLERS FIRST LAW


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In what follows, we prove Keplers

First Law.

The remaining laws are proved

as exercises (with hints).

Th it ti l f f th l t

KEPLERS FIRST LAWPROOF


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The gravitational force of the sun on a planet

is so much larger than the forces exerted by

other celestial bodies.

Thus, we can safely ignore all bodies inthe universe except the sun and one planet

revolving about it.

W di t t ith th



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We use a coordinate system with the sun

at the origin.

We let r = r(t) be the position vector

of the planet.

E ll ll ld b th iti



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Equally well, rcould be the position

vector of any of:

The moon

A satellite moving around the earth

A comet moving around a star

The elocit ector is



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The velocity vector is:

v = r

The acceleration vector is:

a = r

We use the following laws of Newton



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We use the following laws of Newton.

Second Law of Motion: F = ma

Law of Gravitation:3

2

GMm

r

GMmr

=

=

F r

u

In the two laws



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In the two laws,

F is the gravitational force on the planet

m and Mare the masses of the planet and the sun

G is the gravitational constant

r= |r|

u = (1/r)r is the unit vector in the direction of r

First we show that



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First, we show that

the planet moves in

one plane.

By equating the expressions for F in



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By equating the expressions for F in

Newtons two laws, we find that:

So, a is parallel to r.

It follows that r x a = 0.

3GMar= r

We use Formula 5 in Theorem 3 in



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We use Formula 5 in Theorem 3 in

Section 12.2 to write:

( ) ' 'd

dt = +

= +

= +

=

r v r v r v

v v r a

0 0

0

Therefore



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Therefore,

rx v = h

where h is a constant vector.

We may assume that h 0;

that is, rand v are not parallel.

This means that the vector r = r(t) is



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This means that the vector r = r(t) is

perpendicular to h for all values of t.

So, the planet always lies in the plane through

the origin perpendicular to h.

Thus the orbit of the planet is



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Thus, the orbit of the planet is

a plane curve.

To prove Keplers First Law we rewrite



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To prove Kepler s First Law, we rewrite

the vector h as follows:

2

2

'

( ) '

( ' ' )

( ') '( )

( ')

r r

r r r

r rr

r

= =

=

= +

= +

=

h r v r r

u u

u u u

u u u u

u u

Then



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Then,

(Property 6,

Th. 8, Sec. 11.4)

[ ]

2

2

( ')

( ')

( ') ( ) '

GMr

r

GM

GM

=

=

=

a h u u u

u u u

u u u u u u

However u u = |u|2 = 1



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However, u u |u| 1

Also, |u(t)| = 1

It follows from Example 4 in Section 12.2

that:

u u = 0

Therefore,



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Therefore,

Thus,

'GM =a h u

( ) ' ' 'GM = = =v h v h a h u

Integrating both sides of that equation,

KEPLERS FIRST LAWPROOF Equation 11


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g g q ,we get:

where c is a constant vector.

GM = +v h u c

At this point, it is convenient to choose the



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p ,coordinate axes so that the standard basis

vector k points in the direction of the vector h.

Then, the planet moves in thexy-plane.

As both v x h and u are perpendicular



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p p

to h, Equation 11 shows that c lies in

thexy-plane.

This means that we can choose



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thex- and y-axes so that the vector i

lies in the direction of c.

If is the angle between c and r,



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g ,

then (r, ) are polar coordinates of

the planet.

From Equation 11 we have:



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where c= |c|.

( ) ( )

| || | cos

cos

GM

GM

GMr

GMr rc

= +

= +

= +

= +

r v h r u c

r u r c

u u r c

Then,


( )h


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where e = c/(GM).

( )

cos1 ( )

1 cos

r

GM c

GM e

=

+ =

+

r v h

r v h

However,



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where h = |h|.

2

2

( ) ( )

| |

h

=

= =

=

r v h r v h

h hh

Thus,



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,2

2

/( )

1 cos

/

1 cos

h GM

r e

eh c

e

= +

= +

Writing d = h2/c, we obtain:

KEPLERS FIRST LAWPROOF Equation 12


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g

1 cos

edr

e

=+

Comparing with Theorem 6 in Section 10.6,th t E ti 12 i th l ti



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we see that Equation 12 is the polar equation

of a conic section with:

Focus at the origin

Eccentricity e

We know that the orbit of a planet is



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a closed curve.

Hence, the conic must be an ellipse.

This completes the derivation



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p

of Keplers First Law.

The proofs of the three laws show thatthe methods of this chapter provide

KEPLERS LAWS


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the methods of this chapter provide

a powerful tool for describing someof the laws of nature.

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Chap12_Sec4 [Compatibility Mode]