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Transcript
Ch 11: Gravity
For the 1st three sections of this chapter, I’ve interjected most of the notes
covered in 11th grade. You’ll see they are in small (12-pt) font. Read them if
you like to help remind you of the conceptual feel of the material as you work
to put the relevant math behind the theory. 
11-1 Kepler’s Laws
As an intro to Kepler’s Laws and Gravitational PE, let’s take a look at the
notes we saw from 11th grade:
Kepler’s Laws pf Planetary Motion- Notes
The chain
of events…
From Brahe  Kepler  Newton  Einstein*
Brahe – measured positions of planets
Took 20 years
So accurate, still valid today
Entrusted his data to Kepler
Q: What is an ellipse? How are a circle and an ellipse related?
A: The set of all points for which r1 + r2 = constant. A circle is a special case ellipse where the two foci
coincide.
Kepler’s Laws of Planetary Motion
Law 1: Each planet moves in an elliptical orbit with the sun at one of the foci.
The speed varies – faster the closer they are to the sun / slower the farther away they are from the
sun.
Sweeps out equal areas in “spoke-like” fashion
The triangular area swept out when a planet is far from the sun is equal to the area swept out
when it is much closer to the sun.
NOTE: On p 340, there are two sentences that read: “The earth’s orbit is nearly circular, with
the distance to the sun at perihelion (closest point) being 1.48 x 10 11 m and at aphehelion (farthest point)
being 1.52. x 1011 m. The semi-major axis equals the average of these two, which is 1.50 x 1011 m (93
million miles) for the earth’s orbit. The mean earth-sun distance defines the astronomical unit, 1 AU
where 1 AU = 1.50 x 1011 m. You may want to dig up your notes on parsecs and AU’s!
You will need to use the concept of AVERAGING your two distances when you do problem 14. Also,
problem 14 is NOT in the SSM, as denoted!
Law 2: The line from the sun to any planet sweeps out equal areas of space in
equal time intervals.
http://astro.unl.edu/naap/pos/animations/kepler.swf
Kepler was first to coin the term “satellite” though he didn’t understand that satellites were
simply projectiles moving under the force of gravity. A satellite moves slower when working
against gravity, and faster when working with it. Because of this, he spent 10 “needless” years
crankin’ out useless geometrical formulas. Then, he came up with his third law.
Law 3: The squares of the times of revolutions (periods) of the planets are
proportional to the cubes of their average distances from the sun.
T2 ~ R3 or T2 = C R3
where C is the same for all planets.
Table 11-1, p 340, gives the mean orbital radii and orbital periods for all the
planets. (Yep, Pluto is still on the list!)
Also, one last note: the mean sun-earth distance defines the following
astronomical unit, AU where 1 AU = 1.50 x 10 11 m = 93.0 x 106 miles. This
unit, AU, is frequently used in problems dealing with our solar system/
NOTE: When working the problems, be mindful of the following things:
Convert km to m when necessary
Make SURE you square and / or cube, when necessary
Use parentheses (when needed) around your numbers to avoid unnecessary errors!!
See Ex 11-1, p 341
The mean distance from the sun to Jupiter is 5.20 AU. What is the period of
Jupiter’s orbit around the sun? (NOTE: This means that Jupiter is about 5
times farther from the sun than the earth!)
Use T2 = C R3 where TE = 1 year and RE = 1 AU = the period and mean
distance for the Earth. Then, TJ and RJ = 5.20 AU = the period and mean
distance for Jupiter. If you set up a ratio, and knowing that the C’s will
cancel out, you get:
T2Jupiter = R3Jupiter
T2Earth R3Earth
So, TJupiter = TEarth (RJupiter / REarth)3/2 = 1 y(5.20 AU/1 AU)3/2 = 11.9 y
As a last note to this problem, if you look at Fig 11-5, part (a) depicts the
periods versus the mean distance from the sun, while in (b), the squares of
the periods are plotted versus the cubes of their mean distances from the
sun.
Q: What does this do to the data? What would you get if you plotted the
logs of the periods versus the logs of their mean distances from the sun?
A: Linearizes it. You’d also get a straight line. 
Q: What is the slope of the line in graph b?
A: C, of course!
Kepler’s Laws apply not only to planets, but to any satellite. He was familiar with all of Brahe’s
calculations, which initially led him to think the orbits were circular. He was also familiar with
Galileo’s ideas about inertia and accelerated motion, but did not use them in his work. He did not,
however, appreciate the concept of inertia, and really, not until Newton, did inertia figure
prominently in the discussion of planetary motion.
11-2 Newton’s Law of Gravity
You might also remember from 11th grade that Newton said there was a force
of attraction between any pair of particles that was jointly proportional to
their masses, and inversely proportional to the square of the distance between
them.
where G = 6.67 x 10-11 N m2 / kg2.
Q: Is it necessary to memorize the units for G?
A: Nope. Just use the units that would commonly cancel out any mass units
on top, m units on the bottom, and leave you with kg m/s2, which is the same as
a Newton. After all, the SI unit of Force is the Newton.
This opens up some possibilities for calculating and / or manipulating
acceleration, as F ~ ma. See Ex 11-2, p 344.
Ex 11-2, p 344: What is the free-fall acceleration of an object
at the altitude of the space shuttle orbit, about 400 km above
the Earth’s surface?
a = F = GmMEarth/r2 = GMEarth
m
m
r2
To find r, use the radius of the Earth, plus the altitude, so
r = REarth + h = 6370 km + 400 km = 6770 km. Convert this to m and you get
6.77 x 106 m.
Then, a = (6.67 x 10-11 N m2 / kg2)(5.98 x 1024 kg)
(6.77 x 106 m)2
= 8.70 m/s2
One last note: a = F/m = GME / r2 also can be rewritten as gR2Earth / r2. In
other words:
GME = gRE2
TRY THIS!! At what distance above the surface of the earth is the
acceleration of gravity half of its value at sea level? (Don’t forget that “r” is
the TOTAL of RE + rabove the earth)!!!!
Make sure you read through the equation derivations on p 345 – 347
(equations 11-10 – 11-16). Many are just simple algebraic manipulations of the
same equations, but contain useful shortcuts and explanations with regard to
the measurement, manipulation, and accuracy of G.
Derivation of Kepler’s Laws
Newton showed that when an object moves around a force center such as the
sun, the object’s path is a conic section (parabola, hyperbola, or ellipse). The
two former types are not closed paths; only the ellipse is closed, thus, Kepler’s
1st Law is really a consequence of Newton’s Law of Gravity. His second law of
equal areas follows from the understanding that the force exerted on an
orbiting object is centripetal. See Fig 11-8 (a) and (b) below.
(a)
In a given time, dt, the planet moves a distance v dt which is │r x v dt│. So, the
area, dA, swept out by the radius r in time dt is given by
dA = ½ │r x v dt│= 1/2m │r x mv│dt or
NOTE: The ½ is there because you are assuming, for very small values of time
(dt), that your shapes are triangles; hence, the formula A = ½ b  h.
Furthermorr, they are putting m under the 1 (in the fraction ½) so that they
can also out m into the cross product. That way, they can make it look like
angular momentum! Lastly, though the cross product typically has a sin 
component, when dt is very small, sin  is basically sin 900 which = 1, so that
part is not shown.
dA = _1_ L where L = │r x mv│= the magnitude of the orbital angular
dt
2m
momentum of the planet about the sun.
What this really means is that the area swept out in a given time is
proportional to the magnitude of the angular momentum.
Since the force on a planet is along the line from the planet to the sun, there is
no torque about the sun. Thus the magnitude of the angular momentum is
conserved. Another way of saying that is that L is constant!! Further, what
this really means is that the rate at which the area is swept out is the same
for all parts of the orbit. This is Kepler’s 2nd Law.
Now, Check out the rest of the derivations in your text that bring you to:
So,
C
4 2
GM s
where Ms = the mass of the sun.
Try Ex 11-3, p 347-348 now. (It’s an easy one. ) You can either solve it like
they do in the text, or by using the information in Table 11.1.
11-3 Gravitational Potential Energy
Energy Conservation and Satellite Motion
Q: What is KE? PE?
For a planet moving in a perfectly circular orbit, PE and KE would have to be constant. Gravity is
completely centripetal, so there is no component of force acting in the direction of the motion. Thus,
speed, and consequently KE, don’t change. In elliptical orbits, PE is greatest when the satellite is farthest
away (apogee), and least when closest (perigee). Obviously, the opposite is true for KE. So, in the case of
elliptical orbits, KE and PE vary, but their sum is constant. In an elliptical orbit there is a component of
force in the direction of motion, which will change the speed in most every place except at apogee and
perigee.
KE +
PE
KE + PE
KE + PE
KE + PE
Near or at the surface of the earth, g is basically uniform because r = R E + h is
essentially RE, for h <<RE. That’s why when you have a coconut on top of a
tree, and it is subject to gravity, it’s essentially the same amount of g even if
you double the height of the tree. Unless that tree starts at the core of the
earth the change in height is irrelevant, compared to RE.
Q: If an object is far from the earth, by how much does the force of g
decrease?
A: 1 / r2, according to Newton’s Law of Gravity.
In Chapter 6, we learned that PE can be defined as dU = - F∙ ds where F is a
conservative force on a particle, and ds is its general displacement. ( W =  F∙ ds
which is related to KE.)
To find the radial gravitational force, we can substitute dU = - F∙ ds = - Fr∙ ds
= - Fr dr = - (- GMEm ) dr = GMEm dr
r2
r2
Q: What do you get in you integrate both sides of the equation?
A: U = - GMEm + U0 where U0 is the constant of integration
r
Really, all that matters is the change in PE, so, we can choose U0 = 0 most of
the time.
Q: What does U equal if r → ∞?
A: U = 0, of course!
Figure 11-9 below shows a plot of U(r) versus r for the choice U = 0 at r = ∞ for
an object of mass m, and an earth mass of ME. As you can see, the function
begins at the negative value U = - GMEm / RE = - mgRE at the earth’s surface
and increases as r increases. Ultimately, it approaches zero as r → ∞. The
slope of the curve at r = RE is GMEm / R2E = mg. The equation of the tangent
line is shown in blue, and is given by f(h) = U(RE) + mgh, where h = r - RE is
the distance above the earth’s surface. From the Figure, you can see that if h is
very small, U(RE) + mgh ~ U(r).
Figure 11-9
Escape Speed / Velocity
Q: What would happen to a rock launched horizontally at 8 km / s? Why?
Q: What would happen to a rock launched vertically at 8 km / s? Why?
Q: Is there a critical vertical speed / velocity that lets a projectile “outrun” gravity?
DEF: Escape Speed / Velocity = The critical speed / velocity required to outrun gravity = 11.2 km / s.
2GM E
D
Q: What is the KE required for an object to escape gravity?
The formula is v =
A: mgR, of course!!!!
The reason for this value is as you launch something up with some initial KE,
as the object rises, its KE decreases and its PE increases where the maximum
increase in PE is GMEm / RE. So this is also the maximum amount by which
the KE can decrease. If the initial KE is greater than GMEm / RE, then the
total energy will be greater than zero, and the object will still have some KE
even when r is very large. What this really means is that the object will be able
to escape the earth’s gravity.
Further, since the PE at the earth’s surface is - GMEm / RE, the total energy,
E = KE + PE or K + U must be ≥ 0, and the object escapes. See Figure 11-10.
Figure 11-10
The KE of an object at a distance r from the center of the earth is E – U(r).
When the total energy is less than zero (E1 in the figure) the KE is zero at r =
rmax and the object is bound to the earth. When the total energy is greater
than zero (E2 in the figure), the object has enough speed to escape the earth’s
gravity.
This speed is called escape speed, and is given by:
When you substitute the value for g and the radius of the earth, you get
ve = 11.2 km/s.
If superman
flies to the top of a mountain that’s high enough to escape air drag, and hurls a
rock horizontally at 8 km / s, what will happen to the rock?
How much time will go by before he can turn around and catch it?
What if he hurls it faster, but less than 11.2 km / s—what will be the shape of its orbit? Will it take:
shorter – same – longer for the rock to return? Why?
What if he hurls it at a speed greater than 11.2 km / s?
Guess what happens if he hurls it at a speed > 42.5 km / s? (He is, after all, Superman .)
Q: How much work would be required to lift a payload against gravity to a distance far, far away
(like infinity)?
Most of the work done in lifting a payload occurs within the first 10,000 km of the Earth. The change of
PE for an object moved to an infinite distance is 62 MJ, which mathematically corresponds to 11.2 km / s.
If given morr than 62 MJ, it will escape and never return.
Q: What happens to its PE and KE as it continues outward?
Q: Will its velocity ever be zero?
Q: What if a payload is launched such that
11.2 km / s < payload speed < 42.2 km / s ?
Classification of Orbits by Energy
Going back to figure 11-10, a negative total energy means KE < + GMEm / RE
so that KE + PE (or K + U) is never greater than zero. Also, if the total energy
is negative, the total energy line intersects the PE curve at some maximum
separation, rmax, and the system is “bound.” If the total energy is either zero,
or positive, there is no such intersection, and the system is “unbound.”
Hmmm…sounds “calculusy.” 
Further, when E is negative, its absolute value, │E│ is called the binding
energy, which is how much energy has to be added to the system to bring its
total energy up to zero. The PE of an object, such as a planet or comet, having
mass m at a distance r from the sun is given by U(r) = - GMsm / r. Does that
formula sound familiar?
One last bit: KE is defined as ½ mv2. If KE + PE < 0, then the orbit is either
an ellipse or a circle, and the object will be bound to the sun. If KE + PE > 0,
the orbit is a hyperbola. The object will make one swing around the sun, and
vamoose out of our solar system. If KE + PE = 0, the orbit will be a parabola,
and the net effect will be the same as when KE + PE > 0. In essence, when TE
≥ 0, the object escapes. So far, no comet or asteroid has satisfied this criterion,
and thus all are bound to our solar system! Pretty neat!
Try Ex’s 11-4 – 11-6 now. Ex 11-4 will lead you into doing the other two. 