Math 1320-9 Notes of 3/2/20 Kepler’s Laws 1. Planets move in elliptical orbits, with the sun at one of its foci. 2. A line from the planet to the sun sweeps out equal areas in equal times. 3. The square of a planet’s orbital period is proportional to the cube of its mean distance from the sun. • Johannes Kepler (1571–1630) found these laws empirically, using data accumulated by his teacher Tycho Brahe (1546- 1601). Isaac Newton (1642-1727) later derived these laws mathematically using his law of motion (F = ma) and his law of gravity (gravity is inversely proportional to the square of distance), using the Calculus that he had in- vented (as had, independently, Gottfried Leibniz, 1646- 1716). • The textbook has a lengthy derivation of Kepler’s first law in section 10.4. It has hints on deriving the second and third laws on page 726. • In these notes, we’ll derive the second law, which is much simpler, using the approach in the Math 2210 textbook. Math 1320-9 Notes of 3/2/20 page 1 hw exam
Transcript
Math 1320-9 Notes of 3/2/20
Kepler’s Laws
1. Planets move in elliptical orbits, with the sun at one of itsfoci.
2. A line from the planet to the sun sweeps out equal areasin equal times.
3. The square of a planet’s orbital period is proportional tothe cube of its mean distance from the sun.
• Johannes Kepler (1571–1630) found these laws empirically,using data accumulated by his teacher Tycho Brahe (1546-1601). Isaac Newton (1642-1727) later derived these lawsmathematically using his law of motion (F = ma) andhis law of gravity (gravity is inversely proportional to thesquare of distance), using the Calculus that he had in-vented (as had, independently, Gottfried Leibniz, 1646-1716).
• The textbook has a lengthy derivation of Kepler’s first lawin section 10.4. It has hints on deriving the second andthird laws on page 726.
• In these notes, we’ll derive the second law, which is muchsimpler, using the approach in the Math 2210 textbook.
Math 1320-9 Notes of 3/2/20 page 1
hw exam
Kepler’s Second Law
2. A line from the planet to the sun sweeps out equal areasin equal times.
Figure 1. Kepler’s Second Law.
from: https://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Kepler’s law 2 en.svg/2000px-Kepler’s law 2 en.svg.png
Math 1320-9 Notes of 3/2/20 page 2
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• Let A(t) denote the area swept out by time t, startingwith A(0) = 0. We will show that A!(t) is constant. Thismeans precisely that the area swept out in a time intervalof given length is constant.
• Place the sun in the origin, let r(t) denote the positionvector of planet at time t, and let r(t+!t) be its position!t time units later. Let !A be the area swept out in thetime interval [t, t+!t].
• !A is approximately half the area of the parallelogramformed by r(t) and r(t+!t). This is the same as the areaof the triangle formed by r(t) and
!r = r(t+!t)! r(t)
which is half the magnitude of r(t)" r(t+!t):
!A #1
2$r(t)"!r$.
• Dividing by !t:
!A
!t#
1
2
!!!!r(t)"
!r
!t
!!!!.
• Taking the limit as !t !% 0 gives
dA
dt=
1
2$r(t)" r!(t)$.
• The only force acting on the planet is the gravitationalattraction of the sun which is given by
F = !GMm
$r$2r
$r$
where M is the mass of the sun, m the mass of the planet,and G the gravity constant.
• By Newton’s Law of motion
Math 1320-9 Notes of 3/2/20 page 3
!GMm
$r$2r
$r$= ma(t) = mr!!(t),
or, after dividing by m and simplifying:
r!!(t) = !GM
$r$3r.
• Now let’s compute
d
dtr(t)" r!(t) = r(t)" r!!(t) + r!(t)" r!(t)
" #$ %
=0
=!GM
2$r(t)$3r(t)" r(t)
= 0.
• This implies that r(t)"r!(t), and hence its norm and A!(t),are constant, as required.
! Note that this argument actually would have worked forany situation where r!! has the same direction as r.
Math 1320-9 Notes of 3/2/20 page 4
Crossing the River
• This is problem 29 on page 725 of our textbook: Watertraveling along a straight portion of a river normally flowsfastest in the middle, and the speed slows to almost zero atthe banks. Consider a long straight stretch of river flowingnorth with parallel banks 40m apart. If the maximumspeed of the river is 3m/s, we can use a quadratic functionas a basic model for the rate of water flow x m from thewest bank:
f(x) =3
400x(40! x).
(a) A boat proceeds at a constant speed of 5m/s from a pointA on the west bank while maintaining a heading perpen-dicular to the bank. How far down the river on the oppo-site bank will the boat touch the shore? Graph the pathof the boat.
(b) Suppose we would like to pilot the boat to land at the pointB on the east bank directly opposite A. If we maintaina constant speed of 5 m/s and a constant heading, findthe angle at which the boat should head. Then graphthe actually path the boat follows. Does the path seemrealistic?
• First let’s draw a picture.
Math 1320-9 Notes of 3/2/20 page 5
• To begin with, let’s use variables instead of specific num-bers. Let w be the width of the river, vB the speed of theboat, vR the speed of the river in the center, and
v0 =4vRw2
the factor corresponding to the factor 3/400 in the originalproblem.
• Next, observer that the boat will take
T =w
vB
seconds to cross the river. Moreover, letting x denote thedistance from the west bank we get
x(t) = tVB .
• We can now compute our displacement D in the northdirection by integration of our vertical velocity:
D =
& T
0
v0x(w ! x)dt
=
& T
0
v0vBt(w ! vBt)dt
= v0vB
& T
0
wt! vbt2dt
= v0vB
'
wt2
2! vB
t3
3
(T
0
= v0vB
)
wT 2
2! vB
T 3
3
*
= v0vB
)
ww2
2v2B!
vBw3
3v2B
*
=v0vB
w3
)1
2!
1
3
*
=v0w
3
6vB
Math 1320-9 Notes of 3/2/20 page 6
• substituting the numbers
w = 40, v0 =3
400, vB = 5
given in the problem gives the answer
D =3" 403
400" 6" 5= 16m.
• Now suppose we keep the boat pointed in a direction 90"+! (relative to flow, i.e., north, direction. We want to choose! so that we arrive at the point B directly opposite thestarting point A.
• Our velocity vector relative to the moving water is vB <cos !, sin ! >.
• We now reach the opposite shore at time
T =w
vB cos !
and our velocity vector (relative to the shore) is
v(t) =<w
vB cos !,vv0wt
vB cos !
)
w !wy
vB cos !
*
! vB sin ! >
• We want ! such that our displacement in the direction ofthe river is zero, i.e.,
& w
vB cos !
0
v0vB(cos !)t(w ! vB cos !t)! vB sin !dt = 0.
• At this stage I became impatient and decided to use maple.The following maple code solves the problem and plots thepaths taken in the two parts of the problem.
Math 1320-9 Notes of 3/2/20 page 7
1 |\^/| Maple 2019 (X86 64 LINUX)2 ._|\| |/|_. Copyright (c) Maplesoft, a divisionof Waterloo Maple Inc. 2019
3 \ MAPLE / All rights reserved. Maple is a trade-mark of
4 <____ ____> Waterloo Maple Inc.5 | Type ? for help.6 > restart:7 > s:=sin(theta):8 > c:=cos(theta):9 > x:=vB*c*t:
• The two paths are shown in Figures 2 and 3. They domatch our expectations.
Math 1320-9 Notes of 3/2/20 page 8
0
2
4
6
8
10
12
14
16
1 2 3 4 5 6 7 8
t
Figure 2. Pointing East.
The Osculating Circle
• We saw that the curvature of a plane or space curve is thereciprocal of radius of the osculating circle.
• Let’s explore this idea for the special case of an ordinary(scalar valued) function of one variable:
y = y(x) = f(x).
• We saw in class that in that case
"(x) =|f !!(x)|
+
1 + (f !(x))2,3/2
. (1)
Math 1320-9 Notes of 3/2/20 page 9
i oik
–1.5
–1
–0.5
0
0.5
1
1.5
2 4 6 8
t
Figure 3. Going Straight.
• In this note, let’s compute the osculating circle directlyand show that the reciprocal of its radius does in factequal the expression in equation (1).
• The osculating circle, at the point (x, f(x)), is the circlepassing through that point that at that point has the samefirst and second derivative as f .
! This is similar to what we did with Taylor Series. Therewe find a polynomial that matches the function value andsome of the derivatives. Here, we find a circle that matchesthe function value and the first two derivatives.
• Suppose we want to compute (the radius of) the osculatingcircle at the point (x0, f(x0)). Let’s define
y0 = f(x0), y!0 = f !(x0), and y!!0 = f !!(x0).
• We want to find the circle
(x! a)2 + (y ! b)2 = r2 (2)
Math 1320-9 Notes of 3/2/20 page 10
that contains the point (x0, y0) and matches first and sec-ond derivatives at that point. Thus we think of the equa-tion (2) as defining y as a function of x that satisfies therequirement
