Central Force Fields

z p

Unitvector

� r OP= rR �

^y in direction of

0 P

_. Figure 1 A central force field

F(P)

₌ f(r)R acts in the direction

-->

of the position vector

0 P

_if

f (r) >

0

and in the direction opposite to the position vector if f (r) <

0.

10.5 Applications of Vector-Valued Functions to Motion 837 It follows that

ar(t) =

d2s dt2 dt

₌

!"!

+ t2 = _{+ t2} t .

Also,

ll

a(t)

ll

2=

ll

-sin(t)i+cos(t)j -k

ll

2=

(

-sin(t)

)

2+

(

cos(t)

)

2+(-1)2=2.

According to Theorem 2,

2 2 t2

aN(t)

=

ll

a(t)

ll

-

(ar(t))

= 2 - 1 + t2 =

In conclusion, equation (10.5.6) becomes

t

�

^+t2

a(t) =

J1+t2

T(t) + ^-1 2N(t).

1+t2 +t (10.5.8) ....

INSIGHT Equation

(10.5.6)

can be used in conjunction with Theorem 2b as an alternative calculation of the principal unit normal vector

N(t).

For instance, in Example

1,

we can use equation

(10.5.8)

to solve for

N(t):

The values for vectors

a(t)

_and

T(t)

can then be substituted into this formula to yield

N(t).

Let

0

be a fixed point in space. We will use it as the origin of our coordinate axes.

For each point

P =I= 0,

let

rp

=

llDPll

and

up= dir(OP).

We say that a force Fis a

central force field

if there is a scalar-valued function

f

of a real variable such that

F(P)

=

f(rp)up

for every

P =I= 0

(see Figure 1).

Iff(r)

> 0, then the force is directed away from

0.

If

f(r)

_< 0, then the force is directed toward

0

and is said to be

attractive.

As an example of an attractive central force field, consider the sun as a point mass that is fixed at a point

0

in space. If a planet is at point

P,

then, according to Newton's Universal Law of Gravitation, the gravitational force exerted by the sun on the planet is equal to

F(P)

= - GMm

Up

(10.5.9)

where G is a universal constant (i.e., the same for all planets), Mis the mass of the sun, m is the mass of the planet,

rp

is the distance of the planet to the sun, and

Up

is the direction of

OP.

By settingf(r) = -GMm/r2, we see that the force of gravity is an attractive central force field.

� E X A M P L E 2 Suppose that

a

> b > 0. A force

F

acts on a particle of mass m in such a way that the particle moves in the xy-plane with position described by r(t) = acos(t)i + bsin(t)j. Show that Fis a central force field.

.A Figure 2

Solution According to Newton's Second Law of Motion, ^{F =}

ma(t)

and therefore, F =

mr"(t)

⁼

m(

^-acos

(t)

^{i - bsin(t)j}

)

⁼

-mr(t)

⁼

-m llr(t) 11

^dir

(r(t)).

This formula for F shows that it is a central force field directed toward the origin. <11111

A key fact about central force fields is that they always give rise to trajectories that lie in a plane.

THEOREM3 If a particle moving in space is subject only to a central force field, then the particle's trajectory lies in a plane.

Proof. Let

r(t)

be the position vector of the particle, and let

m

be its mass. The force on the particle can be written as ^{F =}

f(r)u

where

r

⁼

llr(t) II

and

u

⁼

(1/r)r(t).

Notice that

r(t)

^{X F =}

0

because

r(t)

and

u

have the same direction. Newton's Second Law of Motion, force equals mass times acceleration, allows us to conclude that

r(t)

^X

(mr"(t))

⁼

0,

^or

r(t)

^X

r"(t)

⁼

0.

We can write this as

r(t)

^X

dt r'(t) d

= 0.

From this last equation, we deduce that

d d

dt (r(t)

^X

r'(t))

⁼

r'(t)

^X

r'(t)

⁺

r(t)

^X

dt r'(t)

^{= 0 + 0 = 0.}

It follows that

r(t)

^X

r'(t)

^{= c}

(10.5.10)

for some constant vector c. This tells us that

r(t)

is perpendicular to c whatever the value oft may be. In other words,

r(t)

is in a plane that is normal to c. •

When we study the trajectory of a body under a central force field, the motion is planar, so it is convenient to assume that the motion is in the xy-plane. As the body moves along its trajectory, its position vector sweeps out a region in the plane of motion (see Figure

2).

^Let

A(t)

denote the area of the region swept out by the position vector

r(T)

for

0

:5

T

:5

t .

THEOREM 4 If a moving particle is subject only to a central force field, then the particle's position vector sweeps out a region whose area

A(t)

has a constant rate of change with respect to

t.

Proof. Let

b..t

be an increment of time,

b..r(t)

the corresponding increment of position, and M the increment of area swept out (Figure 3). We see that Mis

10.5 Applications of Vector-Valued Functions to Motion ⁸³⁹

_. Figure ³

approximately equal to half the area of the parallelogram determined by the vectors

r(t)

_and

r(t

⁺

�t)

=

r(t)

⁺

�r(t).

The area of this parallelogram is

11 r(t)

^X

r(t

⁺

�t) 11

=

11 r(t)

^X

(r(t)

⁺

�r(t)) 11

⁼

llr(t)

^X

�r(t) 11

^· Thus M

_�t

^>:::J

! II r(t) ₂

^X

_�t �r(t) II

=

! ₂ ll ^r(t)

^X

^�r(t) _�t II·

Letting

�t

-+ 0 gives

A'(t)

= ^ll.t--+O lim ^{= _}21

ll ^r(t)

^{X ll.t--+O lim}

II= ^{! llr(t) 2}

^X

^{r'(t) II (10�10)}

^{_1 2}

^llcll

^·

We conclude that area

A(t)

is swept out at a constant rate. This is Kepler's second

law of planetary motion. •

Ellipses

For the remainder of our work in this section, we will need an in-depth under­

standing of one particular planar curve, the ellipse. Fix two distinct points F1 and F2 Y in the xy-plane. Let

c

denote half of the distance between the two points. Suppose that

a

is a fixed positive constant that is greater than

c.

The set of points P ⁼ (x, y) with the property that the distance from P to F1 plus the distance from P to F2 equals

2a

is called an

ellipse

(see Figure

4).

Each of the points F1 and F2 is called a

focus

of the ellipse. Together they are called the

foci

(plural of "focus") of the ellipse. The midpoint of line segment F1F2 is called the

center

of the ellipse. The chord of the ellipse passing through the two foci is called the

major axis

of the ellipse. The chord perpendicular to the major _. Figure ⁴

llPF1ll

⁺

llPF2ll =2a

axis and passing through the center is called the

minor axis

of the ellipse. These

features are shown in Figure 5.

a>c>O

Length of major ^axis _. Figure ⁵

Let us derive the formula for an ellipse with major axis along the x-axis and center at the origin. Let c be a positive constant. We place the foci of the ellipse at points F1=(-c,0) and F2 = (c, 0), symmetrically situated on the x-axis. Let

a

be a number that exceeds c. A point

(x, y)

lies on the ellipse provided that

or

( _{(x- (-}

_c

)) +(y-0) ^{2 2} ) ^1/2

₊

(

(x- c)

²

+

(y-

0

) ² ) ^1/2

=2

a

,

distance to F1 distance to F2

( ₍

_x-(-c)

_{) +(y-0)} ^{2 2} ) ^1/2

₌₂

_a- (

_{(x- c)}

² _+(y-0) ² ) ^1/2 _.

We square both sides and simplify to obtain

a2 -

ex=

a(x2 -

2cx

+ c2 + y2 )112.

Squaring both sides again and simplifying gives

Because

a

^> c ^> 0, it follows that each of the coefficients appearing in this equation is positive. Let us simplify matters by setting

(

₁0.5.11

)

We then have

b2x2 + a2y2

=

a2b2

or, equivalently,

x2 y2

-

+

- =1

a2 b2

with

a

^>

b

^> 0. This last equation is the standard form for an ellipse with foci on the x-axis and center at the origin. By definition, the major axis will be a segment in the x-axis. Because the x-intercepts of the ellipse are

(±a,

0) (just set

y

= 0 to find these), we see that the major axis is the segment connecting

(-a,

0) to

(a,

0).

Similarly the minor axis is the segment connecting (0,

-b)

to (0,

b)

(see Figure 5).

If we were to repeat the preceding calculation with foci (0, -c) and (0, c) and

b

=

a2 -

c

2

, then we would obtain the equation

In general, an ellipse with center

(h, k),

major axis length 2a, minor axis length 2b, and axes parallel to the coordinate axes has equation

(x-h)2 (y-k)2

- 1

a2

₊

b2 -

^or

foci at (h±c,k)

(x- h)2 (y-k)2

_-1

b2

₊

a2 - ,

foci at (h,k±c)

(b2

₊

c2

=

a2 ). (

10

.

5

.

12

)

oo�

Eccentricity Eccentricity Eccentricity

0.1 0.75 0.98

.A Figure ⁶

10.5 Applications of Vector-Valued Functions to Motion 841

� EXAM P L E 3

Suppose that a > b > 0. Show that the curve described by r(t)

⁼

acos(t)i

⁺

b sin(t)j, 0 :st :s 27f, is an ellipse. Where are the foci located?

Solution

If x

⁼

acos(t) and y

⁼

bsin(t), then x 2 /a2

⁺

y2 /b2

⁼

cos2(t)

⁺

sin2(t)

⁼ 1.

Therefore r(t) describes an ellipse. Although the half-distance c between the two foci does not explicitly appear in the equation x2 / a2

⁺

y2 / b2

⁼ 1,

we can use the formula b

⁼

(a2 -c2)1/2 to determine that c

⁼

(a2 -b2)1/2. Because a> b, the major axis of the ellipse lies along the x-axis. The foci are therefore located at

and .

^.,..

Let c denote the half-distance between the foci of an ellipse. Let

a denote half the length of the major axis of the ellipse. The quantity e

⁼

c/a is

In document Calculus Multivariable 2nd Edition Blank & Krantz - Vector Calculus PDF (Page 135-139)