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<!DOCTYPE html>
<html lang="en" xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>The Geometry of Planetary Orbits</title>
<meta name="description"
content="A derivation of Kepler's laws from Newton's inverse square law
using as little algebra as possible.">
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<div class="textcolumn">
<h1 style="text-align: center;">
<a href="index.html">The Geometry of Planetary Orbits</a></h1>
<h2>Introduction</h2>
<p>In 1687 Isaac Newton published "Philosophiae Naturalis Principia
Mathematica," arguably the most influential science book ever
written. In addition to providing the framework still used for
all of mechanics, the Principia introduces Newton's theory of
universal gravitation, which posits an attractive force between
all masses varying inversely as the square of their distance. A
centerpiece of Principia is Newton's demonstration that Kepler's
three laws of planetary motion are mathematically equivalent to
his radically simpler inverse square central force law. This is
an exceptionally difficult math problem, even for students who can
bring all the developments of calculus in the intervening 350
years to bear on it. But every math problem has many solutions,
and along one narrow path through the logic of this problem, each
individual step makes no demands beyond the most elementary plane
geometry and algebra. Here we go.</p>
<p>Kepler's three laws are: First, each planet orbits the sun in an
ellipse with the sun at one focus. Second, the planet moves
around its ellipse so that the radius drawn from the sun to the
planet sweeps out equal areas in equal times. Third, the ratio of
the cube of the semi-major axis of the orbital ellipse to the
square of the time for a complete orbit is the same for every
planet. We aim to show that if a planet begins from any point in
the solar system and moving with any velocity, and thereafter is
attracted to the sun by a force varying inversely as the square of
its distance from the sun, it will orbit forever according to
Kepler's laws. (We assume our planet is subject to no other
forces.)</p>
<p>Our elementary path leads through three waypoints, which you may
visit in any order: The first is to master the geometry of the
focus points and tangent lines of an ellipse, in order to
comprehend what Kepler is talking about. The second preliminary
topic is circular motion, introducing the concept of velocity
space. We must extend this study to the velocity space analog of
circular motion, which is motion at a constant speed. The third
preliminary is to demonstrate that all central force laws, whether
an inverse square law or any other, cause motions obeying Kepler's
equal area law. Note that our third preliminary step actually
achieves a part of our final goal. With these three preliminaries
fresh in our minds, the fact that an inverse square law orbit will
be a Keplerian ellipse follows fairly easily. Kepler's third law,
involving cubes and squares, requires several final algebraic
steps to complete the demonstration.</p>
<h2>Ellipse Focus and Tangent Geometry</h2>
<p>Kepler's first law says that a planet P moves around the sun S in
an ellipse with the sun at one focus; this figure explains what
that means. A circle is the set of all points a given distance
from a center. An ellipse is the set of all points P with a given
sum of distances from two focus points S and O. You can imagine
connecting pins at S and O by a string of length SP+PO, which you
stretch taught by a pencil at P. By keeping the string taught,
you can move the pencil along the ellipse. Instead of a single
center, an ellipse has two focus points, S and O.</p>
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viewBox="-10.5 -10.5 21 21" onload="fig1_interactions(evt)"
stroke-width="0.1">
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d="M -6.0,0.0 L 0.0,-3.2 L 0.0,-10.0" />
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d="M -4.3933,-5.8360 L 2.8051,-1.5170" />
<circle fill="none" stroke="#c0c0c0" cx="0" cy="0" r="10.0" />
<ellipse id="ell1" fill="none" stroke="black" stroke-width="0.13"
cx="-3.0" cy="0" rx="5.0" ry="4.0" />
<circle fill="black" cx="0" cy="0" r="0.2" />
<circle id="pntp1" fill="black" cx="0" cy="-3.2" r="0.2" />
<circle id="pntm1" fill="black" cx="-3" cy="-5" r="0.2" />
<path id="snq1" fill="none" stroke="transparent"
d="M 0.0,0.0 L -3.0,-5.0 L 0.0,-10.0 M -6.0,0.0 L -3.0 -5.0" />
<circle id="pntn1" class="clickable" fill="transparent" stroke="transparent"
stroke-width="0.3" cx="-3" cy="-5" r="0.2" />
<circle id="foco1" class="clickable" fill="black" stroke="transparent"
stroke-width="0.3" cx="-6.0" cy="0" r="0.2" />
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stroke-width="0.3" cx="0" cy="-10.0" r="0.2" />
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<text class="svgtxt" id="txq1" x="0.2" y="-9.2">Q</text>
<text class="svgtxt" id="txm1" x="-3.4" y="-5.4">M</text>
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<p>If we extend the sun-planet line SP to point Q such that the blue
segments OP and PQ in the figure are equal, something interesting
happens: Since SP+OP is the same for any point P on the ellipse,
SP+PQ is also fixed, independent of P. Therefore, the point Q
traces a circle with center S as P moves around the ellipse; the
radius of this circle is the length of the string SP+PO.</p>
<p>Next, consider the midpoint M of the segment OQ. Since by our
construction of Q, P is also equidistant from O and Q, the line PM
is the perpendicular bisector of OQ, the set of all points
equidistant from O and Q. The line PM appears to be tangent to
the ellipse at P, that is PM just touches the ellipse at P without
crossing inside. To prove this tangency, note that the sum of the
distances from S and O is less than SP+PO inside the ellipse, and
greater than SP+PO outside the ellipse. For every point N on PM,
the sum SN+NO is equal to the sum SN+NQ, which is always greater
than SP+PQ for any N except P, because SPQ is a straight line -
the shortest distance between S and Q.</p>
<p>The figure is interactive. You can drag the point O to see how the
shape of the ellipse depends on the distance from O to S (at fixed
string length SQ). Notice that when O coincides with S, the ellipse
becomes a circle. You can drag the point Q to watch what happens as
P moves around the ellipse. Finally, you can drag the point M along
PM to create a point N, to help you visualize the proof that SN+NQ
is always greater than SPQ, so that line PM is entirely outside the
ellipse, touching it only at point P.</p>
<h2>Circular and Constant Speed Motion</h2>
<p>Consider an object moving at a constant speed around a circle, as
depicted in the next figure. We introduce the concept of vectors
- quantities with both magnitude and direction which we represent
as arrows in diagrams. The most basic vector is position measured
from a given point. The black vector on the left side of the
figure goes from the center of the circle to the orbiting object;
its magnitude $r$ is the radius of the circular orbit. You can
press and hold the button at the bottom of the figure to animate
this motion.</p>
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d="M 5.0,0.0 L 0.6,0.0" />
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vτ</text>
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ω</text>
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<text class="svgtxt" id="vel2g" fill="#ff8000" x="1.0" y="1.4">
gτ</text>
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<p>At any instant of time, the object is moving tangent to the
circle, at right angles to the radius vector $r$, indicated by the
blue vector $v\tau$. Strictly speaking, any vector we draw in the
position plane is a displacement vector, with its tail at the
starting point of the displacement and its head at the final
point. Hence, we label the blue vector by the product $v\tau$,
the product of the speed $v$ and some short time interval $\tau$,
so that $v\tau$ is the displacement the object would have made if
it had continued in a straight line at constant speed for that
short time.</p>
<p>We can treat the velocity, which is the rate of displacement, as
a vector in its own right, as opposed to a displacement dependent
on an arbitrary short time interval $\tau$. Therefore, we
introduce a second plane, the velocity plane indicated on the
right side of the figure. In this plane, we draw the tail of the
velocity vector at a fixed point representing zero velocity rather
than as a displacement from the moving point at the tip of the
object. Furthermore, the length of vectors in the velocity plane
represents the speed an object is moving (rather than its
position). Notice that tip of the velocity vector orbits a circle
in the velocity plane, staying exactly ninety degrees ahead of the
orbit in the position plane.</p>
<p>To a large degree, Newtonian mechanics takes place in this
velocity plane, not in the ordinary position plane. This is
because $F=ma$, Newton's second law, associates force with
acceleration, which is the rate of change of velocity. Just as
velocity times a short time interval $v\tau$ is a displacement
vector in position space, acceleration times time $g\tau$ is a
displacement in the velocity plane - the change in velocity in
some short time $\tau$, drawn orange in the figure. Because the
acceleration vector stays ninety degrees ahead of the velocity
vector, which stays ninety degrees ahead of the position vector,
the acceleration always points directly toward the center of the
orbital circle. Here the acceleration will be due to a
gravitational attraction of the orbiting object toward the center
of the circle in the position plane, so we use the symbol $g$
(rather than $a$) for acceleration.</p>
<p>Let $T$ be the time it takes to complete one orbit. Since the
perimeter of the circle is $2\pi r,$ the speed $v$ is $2\pi r/T.$
Similarly, the perimeter of the velocity orbit in the velocity
plane is $2\pi v,$ so the acceleration $g$ is $2\pi v/T.$ Now the
angle between any fixed direction and the radius or velocity
vector changes at a constant rate, which we call $\omega.$ If we
use radian units to measure angle, then $\omega = 2\pi/T,$ because
there are $2\pi$ radians in one complete circle. Thus, our
formulas relating orbital radius, speed, and acceleration are
simply $v=\omega r$ and $g=\omega v.$ In fact, the reason people
use radians to measure angle instead of degrees is to make these
important formulas as simple as possible: In radians, velocity is
just radius times the rate of change of direction (heading angle),
and acceleration is velocity times the rate of change of
direction, at least for uniform circular motion.</p>
<p>Since planets do not move in circular orbits, we need to broaden
our study to more complicated kinds of motion. The true shape of
planetary orbits had been sought by ancient Greek and medieval
Islamic astronomers before culminating in Kepler's eillipses, but
it turns out that the position plane is the wrong place to look
for orbital shapes. The genius of Newton was to recognize that
when you look at planetary motion in the velocity plane, it is
circular after all. Therefore, we continue this preliminary
exercise by studying motion with a constant speed, that is, motion
constrained to a circle in velocity space, rather than constrained
to a circle in position space.</p>
<p>As a first step away from uniform circular motion, we turn to an
object moving at constant speed but with a time varying sideways
acceleration $g.$ The following figure demonstrates this general
case of constant speed motion.</p>
<div class="halfwide">
<svg xmlns="http://www.w3.org/2000/svg"
viewBox="-10.5 -6.0 21 11" onload="fig3_interactions(evt)"
stroke-width="0.1">
<path fill="none" stroke="#f0f0f0" stroke-width="0.3"
d="M 0.0,-4.85 L 0.0,3.7" />
<path id="posr3" fill="none" stroke="#c0c0c0" d="M -2.5,0.0 L -3.0,0.0" />
<circle fill="none" stroke="#a0d0ff" cx="5.0" cy="0" r="4.0" />
<path id="posv3" fill="none" stroke="#0080ff" marker-end="url(#arrowb)"
d="M -3.0,0.0 l -1.2,0.0" />
<circle id="dotr3" fill="black" cx="-3.0" cy="0.0" r="0.2" />
<path id="velc3" fill="none" stroke="#c0c0c0" display="none"
marker-end="url(#arrowg)" d="M 5.0,0.0 L 1.0,0.0" />
<path id="velu3" fill="none" stroke="#c0c0c0" display="none"
marker-end="url(#arrowg)" d="M 5.0,0.0 L 5.0,0.0" />
<path id="velv3" fill="none" stroke="#0080ff" marker-end="url(#arrowb)"
d="M 5.0,0.0 L 1.0,0.0" />
<path id="velg3" fill="none" stroke="#ff8000"
marker-end="url(#arrowo)" d="M 1.0,0.0 l 0.0,0.0001" />
<path id="velw3" fill="none" stroke="#0080ff" marker-end="url(#arrowb)"
d="M 5.0,-1.5 a 1.5,1.5 0 0,0 -1.5,1.5" />
<circle fill="#c0c0c0" cx="5.0" cy="0.0" r="0.2" />
<circle id="dotc3" class="clickable" fill="#0080ff" stroke="transparent"
stroke-width="0.3" cx="5.0" cy="0.0" r="0.2" />
<circle id="dotv3" fill="#0080ff" cx="1.0" cy="0.0" r="0.2" />
<path fill="none" stroke="#ff8000"
d="M -4,4.95 L 4,4.95 M -4,4.95 l 0,-0.2 m 2,0.2 l 0,-0.2 m 2,0.3
l 0,-0.4 m 2,0.3 l 0,-0.2 m 2,0.2 l 0,-0.2"/>
<path id="press3" class="clickable" fill="#b0b0b0" stroke="#606060"
stroke-width="0.2" transform="translate(0.0,0)"
d="M -0.4,4.5 l 0.4,0.4 l 0.4,-0.4 a 0.4,0.4 0 0,0 -0.8,0.0" />
<text class="svgtxt" x="-6.6" y="-5.0">position</text>
<text class="svgtxt" fill="#0080ff" x="3.5" y="-5.0">velocity</text>
<text class="svgtxt" id="vel3v" fill="#0080ff" x="1.6" y="-0.3">v</text>
<text class="svgtxt" id="vel3g" fill="#ff8000" x="1.4" y="1.4">
gτ</text>
<text class="svgtxt" id="vel3w" fill="#0080ff" x="3.2" y="-1.1">
ω</text>
</svg>
</div>
<p>Press the button at the bottom to watch the point move at
constant speed. Slide the button left or right to deflect its
course to the left or right - more positive or more negative rate
of change of direction $\omega.$ Think of the slider as the tiller
of a boat or the steering wheel of a car. (In the figure, the
point will jump to the opposite side if you drive off the visible
part of the position plane.) The sideways acceleration $g$ varies
in proportion to $\omega,$ with $g=\omega v$ as for uniform
circular motion. The orange acceleration vector always remains
perpendicular to the blue velocity vector, which is why the speed
of the point remains constant while you steer it around. Hold the
button at a fixed position and watch the point move in a circle -
the uniform circular motion we studied before. Unlike that
uniform case, a varying angular speed $\omega$ only refers to the
angular rate of change of the velocity vector, because there is no
longer a center in the position plane.</p>
<p>Our second step is to study motion constrained to a velocity
space circle centered on a non-zero velocity. In the previous
figure, you can drag the point at the center of the velocity plane
away from the center of the circle to any other point you please.
That new point becomes the origin where velocity is zero. The
displaced blue arrow remains the velocity of the point in position
space, which is now the sum of the two gray velocity vectors. The
first is the now non-zero center velocity, running from the origin
(tail of the blue arrow) to the center of the velocity space
circle. The second is a radius vector, with its tail at the
center and its head at the head of the blue arrow. The length $v$
of the gray radius vector remains fixed, and both the length and
direction of the center vector remain fixed, as the point moves.
Again, you can animate the motion and control the magnitude of the
acceleration by holding and sliding the button at the bottom. The
center velocity acts like a steady current, making the point
harder to steer.</p>
<p>The orange acceleration vector must remain normal to the gray
radius vector - not to the blue velocity vector - in order to keep
the blue vector on its circular path. The magnitude of the
acceleration is still $g=\omega v$, but now $v$ is the radius of
the velocity circle, the second gray vector. The speed of the
point in position space, which is the length of the blue vector,
changes depending on which way it is moving relative to the center
velocity.</p>
<h2>Equal Areas in Equal Times</h2>
<p>Gravity is an attraction between any two masses, according to
Newton. The strength of attraction is proportional to mass, and
in the case of the sun and the planets, the mass of the sun is so
much larger than the mass of any planet, that to a good
approximation you can ignore the attractions of the planets to one
another, and consider only their attraction toward the sun. (It
turns out that this is also the accuracy of Kepler's Laws, which
themselves are only approximate.) A force which is always
directed toward a fixed point - in this case the sun - is called a
central force. For any central force, Newton's first two laws of
motion turn out to produce Kepler's equal area law, as the next
figure indicates.</p>
<div class="halfwide">
<svg xmlns="http://www.w3.org/2000/svg"
viewBox="-10.5 -5.5 21 11" onload="fig4_interactions(evt)"
stroke-width="0.1">
<path id="area4" fill="#d0f0ff" stroke="#d0f0ff"
d="M 0.0,4.0 L 2.0,-1.0 L -0.5,-2.0 Z"/>
<path id="velw4" fill="none" stroke="#0080ff" marker-end="url(#arrowb)"
d="M 1.06067,5.06067 a 1.5,1.5 0 0,0 0.0,-2.12132" />
<path id="path4" fill="none" stroke="#c0c0c0"
d="M 4.5,0.0 L 2.0,-1.0" />
<path id="rect4" fill="none" stroke="#c0d8ff"
d="M 0.0,4.0 L -2.5,3.0 L -0.5,-2.0" />
<circle id="dotw4" fill="#c0d8ff" cx="-2.5" cy="3.0" r="0.2" />
<path id="velv4" fill="none" stroke="#0080ff" marker-end="url(#arrowb)"
d="M 2.0,-1.0 L -0.5,-2.0" />
<circle id="dotv4" class="clickable" fill="#0080ff" stroke="transparent"
stroke-width="0.3" cx="-0.5" cy="-2.0" r="0.2" />
<path id="posp4" fill="none" stroke="black"
marker-end="url(#arrowk)" d="M 0.0,4.0 L 2.0,-1.0" />
<circle id="dotp4" class="clickable" fill="black" stroke="transparent"
stroke-width="0.3" cx="2.0" cy="-1.0" r="0.2" />
<circle id="dots4" fill="black" cx="0.0" cy="4.0" r="0.2" />
<text class="svgtxt" id="txt4o" fill="#0080ff" x="1.7" y="4.2">
ω</text>
<text class="svgtxt" id="txt4s" fill="black" x="-0.2" y="4.9">S</text>
<text class="svgtxt" id="txt4p" fill="black" x="1.973" y="-1.307">P</text>
<text class="svgtxt" id="txt4v" fill="black" x="-1.357" y="-1.973">V</text>
<text class="svgtxt" id="txt4w" fill="black" x="-3.543" y="2.953">W</text>
<text class="svgtxt" id="txt4r" fill="black" x="1.892" y="0.314">r</text>
<text class="svgtxt" id="txt4vt" fill="#0080ff" x="0.244" y="-1.948">
vτ</text>
</svg>
</div>
<p>Here S is the sun, the center of attraction, and P is the planet
at distance $r$ from the sun, moving with velocity $v.$ Again, we
can draw the blue velocity vector in this position plane by
plotting the point V the planet would reach in a time $\tau$ if it
continued in a straight line at its current speed. The shaded
blue triangle is the area which the radius vector would sweep out
were it to follow that course. This area is half the parallelogram
SPVW; we have added point W such that SW is parallel to PV and the
same length.</p>
<p>Now according to Newton's first law of motion, if there were no
force acting on the planet P, it would continue along PV at a constant
speed $v.$ You can drag the point P along this no-force trajectory.
Notice that no matter where the point lies on the line PV, the area
of the parallelogram (hence of triangle SPV) remains unchanged, since
the parallelogram always has base SW and lies between the parallel
lines SW and PV.</p>
<p>On the other hand, according to Newton's second law of motion,
any force on the planet P directed towards (or away from) S causes
its velocity to change in the direction PS, which is parallel to
VW. You can drag the point V along the line VW to see all the
different velocities a central force toward (or away from) S could
produce. Once again, the area of SPVW (hence of SPV) remains the
same, no matter how strong the deflection in the direction of S.</p>
<p>The shaded area is what would be swept over by the radius if P
were to continue along its trajectory for a time $\tau.$ In other
words, the rate of change of area swept out by the radius SP is
the shaded area divided by $\tau.$ In order to calculate how the
point P actually moves, we can break the motion into a sequence of
very short time steps (much smaller than the $\tau$ in the
figure). In each time step, the velocity changes only slightly,
so we can use the current velocity to compute a new position.
Similarly, the position changes only slightly, so we can use the
acceleration due to the central force at the current position to
compute the change in velocity. We take the next step from the
new position with the new velocity. What the figure shows is that
both parts of each of these time steps - the calculation of the
change in position and the calculation of the change in velocity
- leave the rate of change of swept area unchanged. That is, the
rate of change of swept area remains constant as the planet P
moves along its path as long as its acceleration is centrally
directed, no matter how that force changes along its trajectory.</p>
<p>Instead of the area SPV swept out by the radius vector, we will
write a formula for twice that area, SPVW. The rate of change of
parallelogram area is called the angular momentum of the planet P
relative to the center S. (Actually, it is the angular momnetum
per unit planetary mass.) What we have shown is that for a
central force directed toward S, the angular momentum remains
constant throughout the orbit; we will denote this rate of change
of parallelogram area by $L.$ And this conservation of angular
momentum is equivalent to Kepler's law that the planet's radius
vector sweeps out equal areas in equal times.</p>
<p>The area of SPVW is the length of its base SP, which is the
sun-planet distance $r,$ times the component of the velocity
normal to SP times $\tau.$ This normal component of the velocity
is $\omega r,$ where $\omega$ is the angular speed of P around S,
measured in radians per unit time. That is, the area SPVW is
$(r)(\omega r)(\tau).$ The rate of change of this area is the
angular momentum $L=\omega r^2.$ This formula relates the
unchanging angular momentum to the changing distance $r$ and
angular speed $\omega.$</p>
<h2>An Ellipse with the Sun at One Focus</h2>
<p>Our three preliminary studies - the construction of an ellipse
and its tangent lines from its two focus points, motion with
velocity constrained to a circle, and the fact that central force
laws produce orbits which conserve angular momentum - fit together
like puzzle pieces. We now assemble the puzzle, showing that an
inverse square law for gravity is equivalent to elliptical
planetary orbits with the sun at one focus.</p>
<p>We begin with the formula we wrote for angular momentum, which we
rewrite as an equation for the angular speed $\omega$ of the
planet around its center of attraction, the sun: $\omega=L/r^2,$
where the angular momentum $L$ (twice the rate of change of area
swept out by the radius vector) remains constant as the planet
orbits. In other words, for any central force law, the angular
speed of the planet around the sun is inversely proportional to
the square of its distance from the sun - the closer it gets, the
faster it goes around. If the gravitational acceleration of the
planet toward the sun <em>also</em> varies inversely as the square
of the distance - $g=M/r^2,$ where the constant $M,$ Newton
argues, is proportional to the mass of the sun - that means its
acceleration will be proportional to its angular speed. In fact,
$g=\omega M/L,$ where $M$ is the solar mass constant, and $L$ is
the constant angular momentum of the planet as it orbits.</p>
<p>From our study of circular motion in velocity space, we found
that acceleration toward the center in velocity space is
proportional to angular speed around that center: $g=\omega v,$
where $v$ is the radius of the circle in velocity space. Thus, if
we could arrange for the center of the circle in velocity space to
somehow coincide with the sun in position space, we would know the
planetary orbit is a circle in velocity space with radius
$v=M/L.$</p>
<div class="halfwide">
<svg xmlns="http://www.w3.org/2000/svg"
viewBox="-10.5 -10.5 21 21" onload="fig5_interactions(evt)"
stroke-width="0.1">
<path id="os5" fill="none" stroke="#c0c0c0"
marker-end="url(#arrowg)" d="M -6.0,0.0 L 0.0,0.0" />
<path id="sq5" fill="none" stroke="#c0c0c0"
marker-end="url(#arrowg)" d="M 0.0,0.0 L 0.0,-10.0" />
<path id="oq5" fill="none" stroke="#0080ff"
marker-end="url(#arrowb)" d="M -6.0,0.0 L 0.0,-10.0" />
<circle fill="none" stroke="#c0c0c0" cx="0" cy="0" r="10.0" />
<ellipse id="ell5" fill="none" stroke="black" stroke-width="0.13"
cx="-3.0" cy="0" rx="5.0" ry="4.0" />
<circle fill="black" cx="0" cy="0" r="0.2" />
<circle id="foco5" class="clickable" fill="black" stroke="transparent"
stroke-width="0.3" cx="-6.0" cy="0" r="0.2" />
<circle id="pntq5" class="clickable" fill="black" stroke="transparent"
stroke-width="0.3" cx="0" cy="-10.0" r="0.2" />
<path id="inset5" fill="white" stroke="#f0f0f0" stroke-width="0.4"
transform="rotate(0.0)"
d="M 5.5,2.0 L 10.4,2.0 L 10.4,10.4 L 2.0,10.4 L 2.0,5.5 Z"/>
<path id="posu5" fill="none" stroke="#c0c0c0"
marker-end="url(#arrowg)" d="M 0.0,-3.2 L 0.0,-5.18" />
<path id="posr5" fill="none" stroke="#c0c0c0"
marker-end="url(#arrowg)" d="M 0.0,-5.18 L -3.2,-5.18" />
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<circle fill="none" stroke="#a0d0ff" cx="0" cy="0" r="3.2"/>
<path id="velu5" fill="none" stroke="#c0c0c0"
marker-end="url(#arrowg)" d="M 0.0,1.92 L 0.0,0.0" />
<path id="velr5" fill="none" stroke="#c0c0c0"
marker-end="url(#arrowg)" d="M 0.0,0.0 L -3.2,0.0" />
<path id="velg5" fill="none" stroke="#ff8000"
marker-end="url(#arrowo)" d="M -3.2,0.0 L -3.2,0.9375" />
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marker-end="url(#arrowb)" d="M 0.0,1.92 L -3.2,0.0" />
<text class="svgtxt" fill="#0080ff" x="0.0" y="-3.6">velocity</text>
</g>
<text class="svgtxt" x="-9.0" y="-9.0">position</text>
<text class="svgtxt" id="txs5" x="-0.3" y="0.9">S</text>
<text class="svgtxt" id="txo5" x="-6.3" y="0.9">O</text>
<text class="svgtxt" id="txp5" x="0.29" y="-3.188">P</text>
<text class="svgtxt" id="txq5" x="0.2" y="-9.2">Q</text>
</svg>
</div>
<p>Our diagram of ellipse geometry shows us how to arrange for the
velocity space circle to connect to a Keplerian ellipse in exactly
this way: The three vectors OS, SQ, and OQ resemble the
decompostion of the velocity into a drift component (OS) and a
radial component (SQ) to make a total velocity (OQ), which is
constrained to lie on a circle in velocity space. The only
difference is that we must rotate the velocity diagram ninety
degrees clockwise relative to the OSQ position space triangle.
With that rotation, the blue velocity vector indeed becomes
parallel to the ellipse tangent, since that tangent is
perpendicular to OQ. Furthermore, the drift component velocity is
now perpendicular to OS, in other words vertical, and the radial
velocity component is perpendicular to the sun-planet direction SQ
or SP. Since the radial velocity component is perpendicular to
PS, the orange acceleration vector is parallel to PS. That is,
the acceleration of the planet P is directly toward the sun S.
The puzzle pieces indeed fit perfectly.</p>
<p>You can drag the points O and Q around as before to show that
our construction works for any ellipse shape and for any position
of the planet P on its orbital ellipse.</p>
<p>A planet which obeys $F=ma$ with the force directed toward the
sun and varying inversely as the square of its distance will
indeed orbit in an ellipse with the sun at one focus. We already
found that for any central force law the planet's radius vector
will sweep over equal areas in equal times, so we have proven
the first two of Kepler's laws are equivalent to Newton's
inverse square law. We have learned the additional fact, which
Kepler did not notice, that the velocity vector of the planet
orbits in a circle in the velocity plane.</p>
<h2>The Cube-Square Law</h2>
<p>Since Kepler's first two laws merely describe the motion of a
single planet, he wanted a third law to relate the orbits of the
different planets. More than a decade elapsed between his
publication of his ellipse and equal area laws and his discovery
of the third law, that the cubes of the semi-major axes are
proportional to the squares of the orbital periods. In contrast,
Newton explains Kepler's first two laws as consequences of an
inverse square force law, $g=M/r^2,$ in which the constant of
proportionality $M$ is a property of the sun - in fact, a measure
of the mass of the sun. Therefore, in the Newtonian system,
the same force law already applies to all the planets, so he does
not need any additional laws relating the orbits of different
planets.</p>
<p>Our proof that the inverse square law produces elliptical orbits
completely sidestepped the question of when the planet reaches any
point on its orbit. Since we know that the rate of change of area
swept out by the radius is constant, $L/2$ in fact, we can use the
area between any fixed radius, say the point of closest approach
to the sun, and any point P on the ellipse to compute the time
when the planet will reach P. We will not derive this equation of
time in detail (it requires some trigonometry). However, we do
need a formula for the orbital period $T$ - the length of the
planet's year.</p>
<p>The area of an ellipse with semi-major axis $a$ and semi-minor
axis $b$ is $\pi ab,$ a generalization of the celebrated formula
$\pi r^2$ for a circle. Since the constant angular momentum $L$
is twice the rate of change of the area, the time for one complete
orbit must be $T=2\pi ab/L.$</p>
<div class="thirdwide">
<svg xmlns="http://www.w3.org/2000/svg" viewBox="-9.0 -7.5 12 12"
stroke-width="0.1">
<path fill="none" stroke="#c0c0c0"
d="M -3.0,-4.0 L 2.0,-4.0 L 2.0,0.0 L 0.0,0.0"/>
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<path fill="none" stroke="#c0c0c0" d="M 0.0,0.0 L -3.0,-4.0 L -3.0,0.0"/>
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<ellipse fill="none" stroke="black" stroke-width="0.13"
cx="-3.0" cy="0" rx="5.0" ry="4.0"/>
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marker-end="url(#arrowb)" d="M -3.0,-4.0 L -7.0,-4.0"/>
<circle fill="black" cx="0" cy="0" r="0.2"/>
<circle fill="black" cx="-6.0" cy="0" r="0.2"/>
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<circle fill="black" cx="-3.0" cy="-4.0" r="0.2"/>
<text class="svgtxt" x="-0.3" y="0.9">S</text>
<text class="svgtxt" x="-6.3" y="0.9">O</text>
<text class="svgtxt" x="-2.85" y="-4.25">P</text>
<text class="svgtxt" x="-3.65" y="-5.2">u</text>
<text class="svgtxt" x="-5.6" y="-5.5">v</text>
<text class="svgtxt" x="-0.5" y="-4.15">a</text>
<text class="svgtxt" x="-3.65" y="-1.6">b</text>
<text class="svgtxt" x="-2.0" y="-0.15">c</text>
<text class="svgtxt" x="-1.4" y="-2.0">a</text>
<!-- <text class="svgtxt" x="-5.5" y="-7.8">Q</text> -->
</svg>
</div>
<p>Now SQ, the string length for constructing the ellipse, is twice
its semi-major axis, or $2a.$ We define $c$ to be the distance
from the center of the ellipse to either focus, so that OS is
$2c.$ In the velocity plane, we have already named the radius of
the circle $v=M/L.$ We now give the name $u$ to the drift
component of the velocity. Since the velocity plane triangle is
similar to OSQ in the position plane, the ratio OS/SQ equals
$c/a=u/v.$</p>
<p>To derive Kepler's cube-square law, consider the moment when the
planet P is equidistant from the foci O and S. At that instant,
the total velocity is horizontal with magnitude $\sqrt{v^2-u^2}$,
and the point P is a distance $b$ above OS. Since $u/v=c/a,$
$\sqrt{v^2-u^2}=vb/a.$ The angular momentum is therefore
$L=vb^2/a.$ Since $v=M/L,$ we have $M=v^2b^2/a.$ We can eliminate
the velocity $v$ to find an expression relating $M,$ $L,$ $a,$ and
$b,$ namely $M=L^2a/b^2.$ The final step is to square our
expression for the period $T^2=4\pi^2 a^2b^2/L^2.$ Rearranging
this, $4\pi^2 a^3/T^2 = L^2a/b^2,$ which is $a^3/T^2=M/(4\pi^2).$
Thus, we identify Kepler's third law constant as Newton's solar
mass constant $M$ divided by $4\pi^2.$ Since the mass $M$ is a
property of the sun, the constant is the same for every planetary
orbit, completing our demonstration that the Newtonian law of
gravity produces Keplerian orbits.</p>
<h2>Conclusion</h2>
<p>Kepler's three laws describe planetary motion without giving any
insight into what causes the orbital shapes or speeds. Newton's
law of motion $F=ma,$ on the other hand, applies to the motion of
everyday objects on earth. While there isn't any more cause for
Newton's inverse square law of gravitational force $g=M/r^2$ than
for Kepler's laws, it is far simpler and the fact that planets
respond to this simple force law by moving in Keplerian orbits
shows us that planets move according to the same rules as rocks on
earth. A planet is just a large rock, moving as rocks move.
Despite its apparently simpler form, Newtonian gravity is really
dramatically more complicated than any prior description of
planetary motion because of its universal character. Newtonian
gravity acts between every pair of objects, between every atom of
the sun or planet or rock. Before Principia, no one had any idea
that the force holding the planets in their orbits around the sun
is identical to the force that makes a rock fall to earth.</p>
<p>Newtonian gravity gives us a framework for comparing the
acceleration of the moon in orbit around the earth to the downward
acceleration of a rock: A sidereal month is 27.3 days or about
2.36 million seconds and the moon is about 60 earth radii away, so
$\omega^2 r$ for the moon is $(2\pi/2.36\times 10^6)^2\times
60=4.3\times 10^{-10}$ earth radii per second per second.
According to the inverse square law, the acceleration of gravity
at the surface of the earth, 60 times closer, should be
$60^2=3600$ times greater. Since the radius of the earth is about
$6.4\times 10^6$ meters, the gravity holding the moon in its orbit
would cause an acceleration of $4.3\times 10^{-10} \times
3600\times 6.4\times 10^6=9.9$ meters per second per second at the
surface of the earth according to the inverse square law. When
you drop a rock, it falls to earth with an acceleration of 9.8
meters per second per second, "which answers pretty nearly," as
Newton put it; the inverse square law describes how both the moon
and the rock move. No one before Newton connected the two.</p>
<p>Because of all of the gravitational attractions to other planets
as well as to the sun, the planetary orbits should not quite be
ellipses with the sun at one focus sweeping out equal areas in
equal times - their mutual attractions as they move should cause
them to wobble about in slightly irregular orbits. The moon is an
extreme case, because its attraction to the sun is an appreciable
fraction of its attraction to the earth, which causes the orbit of
the moon to be particularly irregular, deviating obviously from a
Keplerian ellipse. Newton himself spent years computing these
orbital perturbations, particularly for the moon. In every case,
the observed deviations from perfect Keplerian ellipses and equal
area sweep rates match what you expect from all those additional
inverse square attractions. Unlike any previous theory, Newtonian
gravity predicts the complexity we observe in the real world,
assigning to it the same cause as the approximate simple
elliptical motion.</p>
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<p>You can enjoy the simple, approximate planetary motion according
to Kepler's laws by pressing and holding the button at the bottom
of our final drawing. The figure shows both the elliptical orbit
with the sun at one focus in the position plane, and the circular
orbit with offset center in the velocity plane. You can drag the
gray focal point of the ellipse in the position plane to change
the shape of the ellipse. The eight planets of the solar system
all describe nearly circular orbits, with the separation between
their focal points only a few percent of the diameter of their
orbit. Comets have highly eccentric orbits, spending a long time
far from the sun, then dramatically plunging close to the sun.</p>
<p>Using no math beyond high school geometry and algebra, we have
demonstrated that planetary motion obeying Newton's laws of motion
and of universal gravitation will also obey Kepler's laws.
However, the length and depth of our reasoning far exceeds what
most people ever experience in high school, or even in college.
After all, this is a famously hard math problem, so it has cost
you some time to follow the reasoning all the way through. It was
time well spent because, apart from the pleasure of solving any
puzzle, this particular problem launched the science of physics.
A lot has changed in the past three and a half centuries, but the
force of Newton's style of reasoning remains irresistible.</p>
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