4.

**:***Application to Newtonian Gravitational Theory*
A
very important application of algebra is

__Newton__

__'s Universal law of gravitation__. In one form it relates the force of attraction betweeb the mass of the Earth (M_{E}), say, and a smaller mass m, on its surface, at a distance equal to its radius r:
F
= G M

_{E}m/ r^{2}
If
we set the weight (w =mg) equal to the force of gravitational attraction, F, we
obtain:

mg
= G M

^{E}m/ r^{2}
Or:
g = GM

_{E}/r^{2}
In
other words, g is independent of the mass m on the Earth's surface. Now, what
about objects actually orbiting the Earth, say like artificial satellites? In
this case we understand that what keeps the objects orbiting is the centripetal
(or center-directed) force, which is defined as:

F

_{c }= mv^{2}/r
Then
to achieve an orbit (say circular of radius R = r + h where h is the altitude
above the surface)) we need this centripetal force F

_{c}to equal the force of gravitational attraction, F. Or:
mv

^{2}/R = GM_{E}m/ R^{2}
whence:
v

^{2}/R = GM_{E}/R^{2}
But
this can be simplified even further, using the result for g above, and also
using the angular velocity w = 2p/T = v/R, so:

GM

_{E}= gr^{2}and
mRw

^{2}= gr^{2}m/R^{2}, so that:
w

^{2}= gr^{2}/R^{3},
Example
Problem:

*the altitude*) if the period of the satellite is known to be one day or 86,400 secs.

Solution:

T = 86,400s and, solving for R:

R
= [g r

^{2}/w^{2}]^{1/3}
R
= [(10 m/s

^{2})(6.4 x 10^{6}m)^{2}(86400s)^{2})/ 4p^{2}]^{1/3}
R
= 4.24 x 10

^{7}m = 42 400 km
But
we know r = 6400 km so h = R - r

And
h = 42 400 km - 6400 km = 36 000 km

Or
h » 22 500 miles above the
Earth.

We
call such an orbit

*geosynchronous*or "geo-stationary" because the orbiting body retains an essentially fixed position above a point on the Earth and its motion (velocity) in orbit matches the rate of Earth's rotation.
To
find v we have v = wr
= (2p/T) r

=
{ 2p x 42.4 x 10

^{6}m} / 86400 s.
v
= 3100 m/s

The
Newtonian gravitational law of attraction can also be extended to the Sun and
any planet – say of mass m- in the solar system. In
this case, we may write:

GMm/R

^{2}= mv^{2}/R
where R is the distance between centers. Then:

GM/ R

^{2}= v^{2}/R

**Let: v = 2π/P, where P is the period**

GM/R

^{2}= (2π/P)

^{2}1/R

or, in terms of P

or, in terms of P

^{2}:**P**

^{2}= (4π^{2}/ GM) R^{3}**which is just**

**the Newtonian statement of**.

__Kepler’s Harmonic law__
5

**. Electron beam deflection**:
Another
intriguing application of algebra is to

*the deflection of an electron beam*, say in a cathode ray tube such as shown. While it is true cathode ray tubes are seldom if ever used anymore, the example is still important in terms of how particles move in applied electric fields – which one may encounter, say in plasma physics. (Say with a charged particle encountering the Earth's magnetosphere.)
The
illustration below is useful.

Deflection relates to the behavior of a beam of
electrons when fired from an electron "gun" and through a defined
field. The diagram accompanying shows the path of a beam through an electric
field, E, set up inside a cathode ray tube.

Since
the E-field is vertical (+ to -) as shown in the diagram, no horizontal force
acts on the electron entering a region between the charged plates. Thus, the
horizontal velocity component remains unaffected.

The
displacement y in the vertical direction can be obtained from:

y = ½ at

Then, byNewton ’s
second law of motion (resultant force F = mass times acceleration):

m

where m

So the acceleration:

a = eE / m

y = ½ at

^{2}Then, by

m

_{e}a = Eewhere m

_{e}is the mass of the electron, E is the electric field intensity in V/m and e is the unit of electronic charge (e = 1.6 x 10^{-19}C).So the acceleration:

a = eE / m

_{e}
Therefore,
the vertical displacement can be written:

y = ½ {eE/ m

y = ½ {eE/ m

_{e}} t^{2}which the student ought to see is in the same form as the kinematic equation:^{2}

so
it is clear the acceleration in this case is: a
= eE/ m

_{e}Meanwhile, horizontally, the distance displaced is:

x = vt so t = x/v

y = ½ (eE/ m

_{e}) x

^{2}/v

^{2}= (eE/ 2 m

_{e}v

^{2}) x

^{2}

Which
the student should easily see is of the
form:

y = kx

y = kx

^{2}(parabola)
A
special condition obtains when the electron just passes the plates (at distance
x = D) so the value of y there is:

y = eE D

y = eE D

^{2}/ 2 m_{e}v^{2}Then the time for transit between the plates, t is:

t = D/v

and the horizontal component of the velocity is:

v

_{y}= a

_{y}t = (eE/ m

_{e}) D/v

6.

*Quadratic models:*

If
there’s one important equation form in Algebra II it’s

*the quadratic equation*, of the form:
ax

^{2}+ bx + c = 0
which are then solved, either by factoring and solving for
x, or - very often - by using the quadratic formula:

x = [-b

x = [-b

__+__{b^{2 }- 4ac}^{1/2}]/ 2a
Many applied mathematical models can assume the form of
quadratics, so if one can solve for them one can decipher the model.

Example:

Consider the quadratic:
0.3x

^{2}+ 4x + 5 = f(x)
Where
f(x) is an empirical formula found to approximate the probability of a flare
when a sunspot group is on or near the solar central meridian. Here: a = the
proportion of largest sunspot area in the complex sunspot group or groups in
relation to the total area of the complex groups., b = the number of delta
class (complex) sunspot groups, and c = the total number of sunspots of large
area (> 1000 msh or millionths of a solar hemisphere) in the groups used for the formula.

Then
how close to the central meridian are the groups?

Using
the quadratic formula one gets two solutions: x1 = -1.396 and x2 = -11.937

Where
the minus sign is take to mean the group has already passed the central
meridian, and is now west of it in our line of sight. In this case the value of
(-1.396) can be taken as days past central meridian, which means geo-effective
flaring (i.e. causing short wave radio blackouts) is still possible.

7.

**:***Coordinate Equation Transformations*
Galilean
relativity transformations are the prelude to Einsteinian special relativity.
Manipulating the key quantities using algebra to get different transformational
identities, is therefore a useful exercise to build algebra skills that will be
needed later.

Basically,
the interest is in how one transforms from one coordinate system, call it x, y,
z to another, call it x’, y’, z’. In
the diagram below, for example, if systems S and S' are moving relative to each
other, consider a meter stick of length
L (= 1 m) pointed in the +x direction and moving in that direction with
velocity v. (We can also think of it as being at rest in the S' system with one
end at x' = 0 and another end at x' = L', initially. )

We look for a transformation similar to the
Galilean transformation, but which will allow c (velocity of light) to be the same in both S and S'. Since the y
and z coordinates of the position are not affected by the motion in the
x-direction we can say y' = y and z' = z. For the x-coordinate, we try a
transformation of the form: x = a(x' + vt') and x = a(x - vt), where a is an
invariant (unchanged quantity)

We
expect

**to depend on the velocity v in such a way that it becomes equal to 1 when v becomes very small compared with the speed of light. When this happens, the x and x' transformations become the same as the ordinary Galilean transformations. We begin by using x = a(x' + vt') and solve for t' and obtain:***a*
t' = 1/v (x/a - x')

For
t' above, we now insert the value for x' (e.g. x' = a(x- vt)):

t'
= 1/v(x/a - ax - avt) = at - x

^{2}(a^{2}-1)/ va
Similarly,
we find for t:

t
= -at' + x'(a

^{2}- 1)/ va__Problems__:

1.A beam of electrons moving with v = 1.0 x 10

^{7}m/s enters midway between two horizontal plates in a direction parallel to the plates which are 5 cm long and 2 cm apart, and have a potential difference V between them. Find V, if the beam is deflected so that it just grazes the bottom plate. (Take the electron charge to mass ratio: e/ m

_{e}= 1.8 x 10

^{11}C/kg).

2.A
heated filament emits electrons which are accelerated to the anode by a p.d. of
500 V. Find the kinetic energy and velocity of the electron as it strikes the
anode.

3.
Given that x' = 1/a (x - vt) and t' = 1/a (t - vx/c

^{2}), derive similar equations for x and t in terms of x' and t'. (Let: 1/a = (1 - v^{2}/c^{2})^{½})
If
we now substitute x' = a(x - vt) and the equation for t’ into the right hand side of:

r’

^{2}= x’^{2}+ y’^{2}+ z’^{2}- c^{2}t’^{2}what do we get?

4.
Let a quadratic model for central meridian transit of a sunspot group –
indicating days past the central meridian (where the closer the value is to 0,
the greater the incidence of likely flaring) is:

f(x)
= 0.1x

^{2}+ 8x + 12
Is
this complex group more or less likely to see powerful flares than the one
given in the example?

5.
An athlete standing close to the edge on the top of a 160 ft. high building
throws a baseball vertically upward. The quadratic function:

s(t)
= - 16 t

^{2}+ 64t + 160
models
the ball’s height above the ground, with s(t) in feet, t seconds after it’s
thrown:

a)
After
how many seconds does the ball reach its maximum height? What is the maximum
height?

b)
How
many seconds does it take until the ball finally hits the ground? (Round to the
nearest tenth of a second)

6.
Another form of Newton ’s
law of gravitation which incorporates Kepler’s third law is:

P

^{2}= 4(p)^{2}a^{3}/[G(M + M_{p})].
where
P is the planet's period, a the distance (i.e. semi-major axis, we want to
solve for) and G the gravitational constant, M the mass of the Sun, and M

_{p}the planet's mass.
Write
an equation for the planet’s mass, i.e. which might be used to find it if all
the parameters were known, and the units.

__Partial Fraction Decomposition Problem__:
Partial
fraction decomposition uses common denominators to write a sum or difference as
a single rational expression:

Add
the following algebraic expressions and arrive at such a single rational
expression:

3/
(x – 4) + 2/ (x + 2)

*NOTE: Solutions to all the algebra II applied problems will appear over the next five days!*
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