*Michael Fowler 6/7/12*

From the earliest times, gravity meant the tendency of most bodies to fall to earth. In contrast, things that leaped upwards, like flames of fire, were said to have “levity”. Aristotle was the first writer to attempt a quantitative description of falling motion: he wrote that an object fell at a constant speed, attained shortly after being released, and heavier things fell faster in proportion to their mass. Of course this is nonsense, but in his defense, falling motion is pretty fastit’s hard to see the speed variation when you drop something to the ground. Aristotle most likely observed the slower motion of things falling through water, where buoyancy and fluid resistance dominate, and assumed that to be a slowed-down version of falling through airwhich it isn’t.

Galileo was the first to get it right. (True, others had improved on Aristotle, but
Galileo was the first to get the big picture.)
He realized that a falling body *picked up speed at a constant rate*in
other words, it had constant *acceleration*
(as he termed it, the word means “addition of speed” in Italian). He also made the crucial observation that, if
air resistance and buoyancy can be neglected, *all bodies fall with the same
acceleration*, bodies of different weights dropped together reach the ground
at the same time. This was a
revolutionary ideaas
was his assertion that it should be checked by *experiment* rather than by the traditional method of trying to
decipher what ancient authorities might have meant.

Galileo also noted that if a ball rolls without interference on a smooth horizontal surface, and friction and air resistance can be neglected, it will move with constant speed in a fixed directionin modern language, its velocity remains constant.

He considered the motion of an object when not subject to
interference as its “*natural*” motion.

Using his terminology, then,* natural horizontal motion*
is motion at *constant velocity*, and *natural vertical motion* is
falling at *constant acceleration*.

But he didn’t stop therehe
took an important further step, which made him the first in history to derive
useful quantitative results about motion, useful that is to his boss, a duke
with military interests. The crucial
step was the realization that for a cannonball in flight, *the horizontal and
vertical motions can be analyzed independently*. Here’s his picture of the path of a
horizontally fired cannonball:

The vertical drop of the cannonball at the end of successive
seconds, the lengths of the vertical lines *ci*,
*df*, *eh* are the same vertical distances fallen by something dropped from
rest. If you drop a cannonball over a
cliff it will fall 5 meters in the first second, if you fire it exactly
horizontally at 100 meters per second, it will *still* fall 5 meters below a horizontal line in the first
second. Meanwhile, its horizontal motion
will be at a steady speed (again neglecting air resistance), it will go 100
meters in the first second, another 100 meters in the next second, and so
on. Vertically, it falls 5 meters in the
first second, 20 meters total in two seconds, then 45 and so on.

Galileo drew the graph above of the cannonball’s position as
a function of time, and proved the curve was parabolic. He went on to work out
the range for given muzzle velocity and *any*
angle of firing, much to the gratification of his employer.

*really fast*? The cannonball would still fall 5 meters in
the first second (ignoring the minor point that *g* goes down a bit on a
really high mountain), *but* if it’s going fast enough, don’t forget the
curvature of the earth! The surface of
the earth curves away below a horizontal line, so if we choose the right speed,
after one second the cannonball will have reached a point where the earth’s
surface itself has dropped away by 5 meters below the originally horizontal
straight line. In that case, the
cannonball won’t have lost any height at alldefining
“height” as distance above the earth’s surface.

Furthermore, “vertically down” has turned around a bit (it
means perpendicular to the earth’s surface) so the cannonball is still moving “horizontally”,
meaning moving parallel to the earth’s surface directly beneath it. And, since it’s above the earth’s atmosphere,
it won’t have lost any speed, so exactly the same thing happens in the next
second, and the nextit
therefore goes in a circular path.

We can find how fast the cannonball must move to maintain the circular orbit by using Pythagoras’ theorem in the diagram below (which grossly exaggerates the speed so that you can see how to do the proof).

The cannonball fired from point *P* goes *v* meters
horizontally in one second and drops 5 meters vertically, and, if *v* has the right value, the cannonball
will still be the same distance *R*
from the earth’s center it was at the beginning of the second. (Bear in mind that *v* is actually about a thousandth of *R*, so the change in the direction of “down” will be imperceptible,
not like the exaggerated figure here.)

Knowing that the radius of the earth *R* is 6400 km, there is enough information in the above diagram to
fix the value of *v*. Notice that there is a right angled triangle
with sides *R* and *v* and hypoteneuse . Applying Pythagoras’ theorem,

(*R* + 5)^{2} = *R*^{2}
+ *v*^{2},

*R*^{2} + 10*R* + 25 = *R*^{2} + *v*^{2}.

Newton knew (in different units) that *R* = 6400 km, so
the 25 in the above equation can be neglected to give:

*v*^{2 }= 10*R* =
10×6400×1000 so *v* = 8000.

The units for *v* are
of course meters per second, on our diagram we show *v* as a distance,
that traveled in the first second.

So the cannonball must move at 8 km per second, or 5 miles per second if its falling is to match the earth’s curvaturethis is 18,000 mph, once round the earth in a little less than an hour and a half. This is in fact about right for a satellite in low earth orbit.

It occurred to Newton one day (possibly because of a falling apple) that this familiar gravitational force we experience all the time here near the surface of the earth might extend outwards as far as the moon, and in fact might be the reason the moon is in a circular orbit. The radius of the moon’s orbit (384,000 km) and its speed in orbit (about 1 km per second) had long been known (see my notes here if you’re interested in how it was measured), so it was easy to find, using the same Pythagorean arguments as used for the cannonball above, that the moon “falls” 1.37 millimeters below a straight line trajectory in one second.

*That is to say, the ratio
of the moon’s acceleration towards the center of the earth to the cannonball’s
is 1.37/5000, or about 1/3600. *

But the radius of the moon’s orbit is about *60 times greater than the cannonball’s*
(which is just the radius of the earth, approximately). Since 60×60 = 3600, Newton concluded that the gravitational force
decreased with distance as 1/*r*^{2}.

*Every body in the universe
attracts every other body with a gravitational force that decreases with
distance as 1/r ^{2}.*

But actually he knew more about the gravitational force: from
the fact that bodies of different masses near the earth’s surface accelerate
downwards at the same rate, using *F* = *ma* (his Second Law) if two
bodies of different masses have the same acceleration they must be feeling
forces in the same ratio as their masses (so a body twice as massive feels
twice the gravitational force), that is, ** the
gravitational force of attraction a body feels must be proportional to its mass**.

Now suppose we are considering the gravitational attraction
between two bodies (as we always are), one of mass *m*_{1}, one of
mass *m*_{2}. By *m*_{1}
as the earth, the force *m*_{2} feels is proportional to *m*_{2},
as argued aboveso
this must be true whatever *m*_{1} is. And, since the situation is
perfectly symmetrical, *the force must also be proportional to* *m*_{1}.

Putting all this together, the magnitude of the
gravitational force between two bodies of masses *m*_{1} and *m*_{2}
a distance *r* apart

*F* = *Gm*_{1}*m*_{2}/*r*^{2}.

The constant *G* = 6.67×10^{-11} N.m^{2}/kg^{2}.

*It is important to realize that G cannot be measured by
any astronomical observations*. For
example, *g* at the surface of the earth is given by

*g* = *Gm _{E}*/

where *m _{E}* is the mass and

The first measurement of *G* was made in 1798 by
Cavendish, a century after

Cavendish said he was “*weighing the earth*” because
once *G* is measured, he could immediately find the mass of the earth *m _{E}*
from

Newton’s first clue that gravitation between bodies fell as the inverse-square of the distance may have come from comparing a falling apple to the falling moon, but important support for his idea was provided by a detailed description of planetary orbits constructed half a century earlier by Johannes Kepler.

Kepler had inherited from Tycho Brahe a huge set of precise
observations of planetary motions across the sky, spanning decades. Kepler himself spent eight years
mathematically analyzing the observations of the motion of Mars, before
realizing that Mars was moving in an *elliptical*
path.

To appreciate fully how Kepler’s discovery confirmed

A *circle* can be
defined as the set of all points which are the same distance *R* from a given point, so a circle of
radius 1 centered at the origin *O* is
the set of all points distance 1 from *O*.

An *ellipse* can be
defined as the set of all points such that the *sum* of the distances from two fixed points is a constant length
(which must obviously be greater than the distance between the two
points!). This is sometimes called the
gardener’s definition: to set the outline of an elliptic flower bed in a lawn,
a gardener would drive in two stakes, tie a loose rope between them, then pull
the rope tight in all different directions to form the outline.

In the diagram, the stakes are at *F _{1}*,

*CA* is called the
semimajor axis length *a*, *CB* the semiminor axis, length *b*.

*F _{1}*,

Notice first that **the
string has to be of length 2 a**,
because it must stretch along the major axis from

Suppose now we put *P*
at *B*.
Since *F _{1}B* =

We get a useful result by applying Pythagoras’ theorem to
the triangle *F _{1}BC*,

(We shall use this shortly.)

Evidently, for a circle, *F*_{1}*C* = 0.
The ** eccentricity **of the ellipse is defined as the ratio of

so

eccentricity *e* =

The eccentricity of a circle is zero. The eccentricity of a long thin ellipse is just below one.

*F*_{1} and *F*_{2} on the diagram
are called the *foci* of the ellipse
(plural of *focus*) because if a point
source of light is placed at *F*_{1}, and the ellipse is a mirror,
it will reflectand
therefore *focus*all
the light to *F*_{2}. (This
can be proved using the string construction.)

An ellipse is essentially a circle scaled shorter in one
direction: in (*x*, *y*) coordinates it is described by the equation

a circle being given
by *a* = *b*.

In fact, in analyzing planetary motion, it is more natural
to take the origin of coordinates at the center of the Sun rather than the
center of the elliptical orbit. It is
also more convenient to take coordinates instead of (*x*, *y*)
coordinates, because the strength of the gravitational force depends only
on *r*. Therefore, the relevant equation describing a
planetary orbit is the equation with the origin at one focus. For an ellipse of semi major axis *a *and
eccentricity *e* the equation is:

It is not difficult to prove that this is equivalent to the
traditional equation in terms of *x*, *y* presented above.

**Kepler summarized his
findings about the solar system in his three laws:**

**1**. The planets
all move in elliptical orbits with the Sun at one focus.

**2**. As a planet
moves in its orbit, the line from the center of the Sun to the center of the
planet sweeps out equal areas in equal times, so if the area *SAB* (with curved side *AB*) equals the area *SCD*, the planet takes the same time to move from *A* to *B*
as it does from *C* to *D*.

For my Flashlet illustrating this law, click here.

** **

**3**. The time it
takes a planet to make one complete orbit around the sun *T* (one planet
year) is related to its average distance from the sun *R*:

In other words, if a table is made of the length of year *T* for each planet in the solar system,
and its average distance from the sun *R*,
and is computed for each planet, the numbers are
all the same.

These laws of Kepler’s are precise, but they are only *descriptive*Kepler
did not understand why the planets should behave in this way.

Surprisingly, the first of Kepler’s lawsthat
the planetary paths are ellipticalis
the toughest to prove beginning with

The best strategy turns out to be to attack the laws in reverse order.

It is easy to show how Kepler’s Third Law follows from the
inverse square law if we assume the planets move in perfect circles, which they
*almost* do. The acceleration of a
planet moving at speed *v* in a circular orbit of radius *R* is *v*^{2}/*R*
towards the center. (Of course you
already know this, but it is amusing to see how easy it is to prove using the
Pythagoras diagram above: just replace the 5 meters by *at*^{2}, the “horizontal”
distance *v* by *vt*, write down Pythagoras’ theorem and take the
limit of a very small time.)

*F* = *ma* for a planet in orbit becomes:

Now the time for one orbit is so dividing both sides of the equation above
by *R*, we find:

.

This is Kepler’s Third Law: has the same numerical value for all the sun’s planets.

*Exercise*: how are *R*, *T* related if the
gravitational force is proportional to 1/*R*? to 1/*R*^{3}?
To *R*?

*The point of the
exercise is that Kepler’s Third Law, based on observation, forces us to the
conclusion that the Law of Gravity is indeed inverse square.*

* *

In fact, Newton went furtherhe
generalized the proof to elliptic orbits, and established that for the inverse
square law *R* must (for ellipses) be
replaced by *a*, the semimajor axis of
the ellipse, that is to say is the same for all planets. This is in fact exactly what Kepler found to
be the case.

*It follows immediately that all elliptic orbits
with the same major axis length, whatever their eccentricity, have the same orbital time.*

*A planet in its path around the sun sweeps out equal areas in equal
times.*

Suppose at a given instant of time the planet is at point *P* in its orbit, moving with a velocity meters per second in the direction along the
tangent at *P* (see figure). In the next second it will move *v* meters, essentially along this line
(the distance is of course greatly exaggerated in the figure) so the area swept
out in that second is that of the triangle *SPQ*,
where *S* is the center of the sun.

The area
of triangle *SPQ* is just base × height.
The base *PQ* is *v* meters long, the height is the
perpendicular distance from the vertex of the triangle at the sun *S* to the baseline *PQ*, which is just the tangential velocity vector .

Hence

rate of sweeping out of area = .

Comparing this with the angular momentum *L* of the planet as it moves around the
sun,

it becomes apparent that Kepler’s Second Law, the constancy
of the area sweeping rate, is telling us that *the angular momentum of the planet around the sun is constant*.

In fact,

rate of sweeping out of area =

To see what this means, think of applying a force to a wheel on a fixed axle:

If the force is above the axle, as shown, the wheel will begin to turn anticlockwise, if it is below the axle the wheel will turn the other wayassuming no friction, the rate of change of angular momentum is equal to the torque, the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the center of rotation, the axle. If the line of action of the force passes through the middle of the axle, there is no torque, no rotation, no change of angular momentum.

For the planet in orbit, the fact that the angular momentum
about the sun does not change means that the force acting on the planet has no
torque around the sunthe
force is directly towards the sun. This
now seems obvious, but Kepler himself thought the planets were pushed around
their orbits by spokes radiating out from the sun.

Note: I’m including this derivation of the elliptic orbit just so you can see that it’s calculus, not magic, that gives this result. This is an optional section, and will not appear on any exams.

We now back up to Kepler’s *First* Law: proof that the
orbit is in fact an ellipse if the gravitational force is inverse square. As usual, we begin with *F* = *ma*, in
vector form. The force is *GMm*/*r*^{2} in a radial inward
direction. But what is the radial acceleration? Is it just *d*^{2}*r*/*dt*^{2}? Well, no, because if the planet’s moving in a
circular orbit it’s still accelerating inwards at (same as *v*^{2}/*r*) even
though *r* is not changing at all. The total acceleration is the *sum*,
so *ma* = *F* becomes:

This isn’t ready to integrate yet, because varies too. But since the angular momentum *L*
is constant, ,
we can get rid of in the equation to give:

This equation can be integrated, using two very unobvious
tricks, figured out by hindsight. The
first is to change go from the variable *r* to its inverse, *u* = 1/*r*.
The other is to use the constancy of angular momentum to change the variable *t*
to .

Anyway,

so

Therefore

and similarly

Substituting in the equation of motion gives:

This equation is easy to solve! The solution is

where *A* is a constant of integration, determined by
the initial conditions.

This is equivalent to
the standard equation of an ellipse of semi major axis *a*
and eccentricity *e*, with the origin at one focus, which is:

The time it takes a planet to make one complete orbit around
the Sun *T* (one planet year) is related to the semi-major axis *a* of its elliptic orbit by

We have already shown how this can be proved for circular orbits, however, since we have gone to the trouble of deriving the formula for an elliptic orbit, we add here the (optional) proof for that more general case.

(Note that this same result is derived in the next lecture using energy and angular momentum conservationthe proof given here is quicker, but depends on knowing the equation for the ellipse.)

The area of an ellipse is and the rate of sweeping out of area is *L*/2*m*,
so the time *T* for a complete orbit is evidently

* *

Putting the equation

in the standard form

we find

Now, the top point *B* of the semi-minor axis of the
ellipse (see the diagram above) must be exactly *a* from *F*_{1}
(visualize the string *F*_{1}*BF*_{2}), so using
Pythagoras’ theorem for the triangle *F*_{1}*OB* we find

Using the two equations above, the square of the orbital time

**We have established,
then, that the time for one orbit depends only on the semimajor axis of
the orbit: it does not depend on how eccentric the orbit is.**