Today, I will talk to you about elliptical orbits
and Kepler's famous laws. I first want to review
with you briefly what we know about
circular orbits, so I wrote on the blackboard
everything we know about circular orbits. There's an object mass little m going in a circle
around capital M. This could be the Sun;
it could be the Earth. It has radius R, circular. We know there in equation one how to derive the time
that it takes to go around. The way we found that was by setting the centripetal
force onto little m the same as
the gravitational force. Also, the velocity in orbit-- maybe I should say
speed in orbit-- also follows through
the same kind of reasoning. Then we have the conservation
of mechanical energy-- the sum of kinetic energy
and potential energy-- is a constant;
it's not changing. You see there first the
component of the kinetic energy, which is
the one-half mv-squared, and then you see the term
which is the potential energy. We have defined potential energy
to be zero at infinity, and that is why all bound orbits
have negative total energy. If the total energy is positive,
the orbit is not bound. And when you add these two up, you have an amazing coincidence
that we have discussed before. We get here
a very simple answer. The escape velocity you find by
setting this E total to be zero, so this part of the equation
is zero. Out pops that speed
with which you can escape the gravitational pull
of capital M, which is the square root
of two times larger than this V. And I want to remind you
that for near Earth orbits, the period to go around
the Earth is about 90 minutes, and the speed--
this velocity, then, that you see in equation two-- is about eight kilometers
per second, and the escape velocity
from that orbit would be about
11.2 kilometers per second. And for the Earth
going around the Sun, the period would be
about 365 days, and the speed
of the Earth in orbit is about 30 kilometers
per second, just to refresh your memory. Now, circular orbits
are special. In general, bound orbits
are ellipses, even though I must add to it that most orbits of our planets
in our solar system-- very close to circular,
but not precisely circular. But the general solutions call
for a elliptical orbit. And I first want
to discuss with you the three famous laws by Kepler
from the early 17th century. These were brilliant statements
that he made. The interesting thing is that before he made these
brilliant statements, he published more nonsense
than anyone else. But finally he arrived
at two... three golden eggs. And the first golden egg then is
that the orbits are ellipses-- he talked always about planets--
and the Sun is at one focus. That's Kepler's law number one. These are from
around 1618 or so. The second...
Kepler's second law is-- quite bizarre how he found that
out, an amazing accomplishment. If you take an ellipse, and you
put the Sun here at a focus-- this is highly exaggerated because I told you that most
orbits look sort of circular-- and the planet goes
from here to here in a certain amount of time, and you compare that with the
planet going from here to here in a certain amount of time, then Kepler found out
that if this area here is the same as that area here, that the time to go
from here to here is the same as to go
from there to there. An amazing accomplishment
to come up with that idea. And this is called
"equal areas, equal times." Somehow, it has the smell of some conservation
of angular momentum. And then his third law was that if you take the orbital period
of an ellipse, that is proportional
to the third power of the mean distance to the Sun. And he was so pleased
with that result that he wrote jubilantly
about it. I will show you here the data that Kepler had available
in 1618, largely from the work done
by, of course, astronomers, observers like Tycho Brahe
and others. You see here the six planets
that were known at the time, and the mean distance
to the Sun. For the Earth, it is one because
we work in astronomical units. Everything is referenced
to the distance of the Earth. This is 150 million kilometers. And it takes the Earth 365 days
to go around the Sun; Jupiter, about 12 years;
and Saturn, about 30 years. And then when he takes this
number to the power of three and this number squared,
and he divides the two, then he gets numbers
which are amazingly constant. And that is his third law. The third law leads immediately to the inverse square dependence
of gravity, which he was not aware of, but Newton later
put that all together. But he very jubilantly writes: And he wrote that in 1619. So, orbits in general
are ellipses. And now I want
to review with you what I have there on
the blackboard about ellipses. You see an ellipse there? Capital M-- could be the Earth,
could be the Sun-- is at location Q. The ellipse has
a semimajor axis A, so the distance P to A--
perigee to apogee-- is 2a. If M were the Earth, capital M, then we would call the point
of closest approach "perigee," and the point farthest
away from the Earth, we would call that "apogee." If capital M were the Sun, we would call that "aphelion"
and "perihelion." So you see the little m
going in orbit; you see the position
vector r of q. It has a certain velocity v. And so the total mechanical
energy is conserved. The sum of kinetic energy and
potential energy doesn't change. The first term is the kinetic
energy-- one-half mv squared, and the second term
is the potential energy-- no different from equation three
for circular orbits, except that now capital R, which
was a fixed number in a circle, is now a little r, and little r
changes, of course, with time. Also that velocity, v,
in that equation number five will also change with time because it's
an elliptical orbit. It will not change in time
in equation number two and in equation number three. Now I give you a result
which I didn't prove, and that is that
the total mechanical energy, which has these two terms in it
which you do fully understand, also equals minus mMG
divided by 2a, and a is the semimajor axis. And compare number five
with number three, then you see they are
brothers and sisters. The only change is that
what was capital R before is now little a,
the semimajor axis. And if you want to calculate the
time to go around the ellipse, then you get an equation
for T squared, which is almost identical
to number one for the circular orbit, except that now the radius
has to be replaced by a, which is the semimajor axis. And the escape velocity
you can calculate in exactly the same way that you
calculate the escape velocity under equation number four. All you do is you make
the total energy zero, and then you solve equation
three and equation five, and out pops the speed
that you need to make it all the way
out to infinity. And that is in the case
of the circular orbit, square root of two
times the speed in orbit. So these are the numbers that we are going
to use today, the equations. And there's one thing which
is already quite remarkable and very nonintuitive-- very
nonintuitive, to say the least. That if you have various orbits which have the same
semimajor axis, that the period is the same
and the energy is the same, and that's by no means obvious. So, this is one orbit-- think
of it as being an ellipse-- and this is another one. This distance is the same
as this distance. I've just done it that way. That means, according
to equation five and six, that both orbits have exactly
the same mechanical energy, and both orbits have
the same periods. So to go around
this circular orbit will take the same amount of
time as to go around this one, and that is by no means obvious. I now want to start with
a very general initial condition of an object, little m, in
orbit... in an elliptical orbit. And I want to see how we can get all the information
about the ellipse that we would like to find out. So I'm only giving you
the initial conditions. So here is an ellipse,
here is P and here is A. If this is an ellipse
around the Earth, then this would be perigee
and this would be apogee. The mass is capital M,
this is point Q. Let me get a ruler so that
I can draw some nice lines. So the distance AP equals 2a--
a being the semimajor axis-- and our object happens
to be here-- mass little m-- and this distance equals R zero. Think of it as being time zero. And at time zero,
when it is there, it has a velocity
in that ellipse. Let this be v zero. And there is an angle between
the position vector and v zero; I call that phi zero. So I'm giving you M,
I'm giving you v zero, I'm giving you r zero,
I'm giving you phi zero. And now I'm going to ask you, can we find out from
these initial conditions how long it takes for
this object to go around? Can we find out what QP is? Can we find out what
the semimajor axis is? Can we find out what
the velocity is at point P, at closest approach when
this angle is 90 degrees? And can we find out
what the velocity is when the object, little m,
is farthest away-- apogee? Can we find all these things? And the answer is yes. a is the easiest to find--
the semimajor axis. I turn to equation number five, which is the conservation
of mechanical energy. And the conservation
of mechanical energy says that the total energy is the kinetic energy
plus the potential energy equals one-half mv zero squared. That is when the object
is here at location D minus mMG divided by this r zero
at location D. This can never change. This is the same throughout
the whole orbit, so it must also be,
according to equation five, minus mMG divided by 2a. And so you have one equation
with one unknown, which is a, because you know
all the other things: M cancels... M always cancels
when you deal with gravity, so you only have a
as an unknown. So that's done. If the total energy
were positive, then for this to be positive,
a has to be negative. That's physical nonsense,
of course, so this only holds
for bound orbits. So positive values for E total
are not allowed. Once you have a, you use... so this is from
equation number five. If you now apply
equation number six, immediately pops out T,
the orbital periods, because the only thing
you didn't know yet was a, but you know a now. So we also know
how long it takes for the object
to go around in orbit. And I try to be quantitative
with you. Step by step,
as we analyze this further, I will apply this
to a specific case for someone going
around the Earth. Everything I'm
telling you today, including all
numerical examples, are in a handout
which is six pages thick, which I wrote
specially for you. It will be on the Web. We're not going to print it
here-- that's a waste of paper. It's 1999, so that's
what we have the Web for. So you can decide on your own
how many... how much notes, how much time you want
to spend on notes, and to what extent
you want to concentrate and try to follow the steps. It's up to you. But everything is there-- literally everything,
every numerical example. We take for capital M...
we take the Earth, and that is six times ten
to the 24 kilograms. So that's my M. I promised you you will know M. I will give you r zero. That is 9,000 kilometers. That's the location at point D. I give you the conditions at D. The speed at point D
is 9.0 kilometers per second, and I'll give you phi zero
is 120 degrees. Everything else we should
be able to calculate now from these numbers. First of all,
with equation five, you can convince yourself
sticking in these numbers that the total energy
is indeed negative. Of course, if I make
the total energy positive, it's not an ellipse,
so then it's all over. It is negative;
it is an ellipse. So with equation number five,
I then pop out a. Right? Because that's one
equation with one unknown, and I put in the numbers-- you can confirm them
and check them at home-- and I find that a
is quite large. a is about 50,000 kilometers. That's huge. Remember, it starts off
at 9,000 kilometers, but a is 50,000 kilometers. That means 2a
is 100,000 kilometers. Why is that so large? Well, the answer lies in
evaluating the escape velocity. The escape velocity
of this little mass when it is at position D, for which these
are the input parameters, is the square root
of 2MG divided by r zero, and that is
9.4 kilometers per second. Well, if you need
9.4 kilometers per second to make it out to infinity, and you have
nine kilometers per second, you're pretty close already. So that's the reason
why this semimajor axis is indeed such
a horrendous number. It's no surprise. If now I use
equation number six, then I find the period, and I find that it takes
about 31 hours for this object
to go around the Earth. So far, so good. Now we want to know
what the situation is with perigee and with apogee. Can we calculate
the distance QP? Can we calculate the speed at
location P and at location A? And now comes
our superior knowledge. Now we're going to apply for the
first time in systems like this, the conservation
of angular momentum. Angular momentum is conserved
about this point Q, but only about that point Q. It is not conserved about any
other point, but that's okay. All I want is that point Q. That is where capital M
is located. What is the magnitude
of that angular momentum? Well, let's first take point D. When the object is at D, the magnitude
of the angular momentum is m times v zero times r zero
times the sine of phi zero. This is the situation at D. Why do we have a sine phi zero? Because we have a cross
between r and v, and with a cross product,
you have the sine of the angle. So that's the situation
at point D. What is the situation
at point P? Well, at point P,
the velocity vector is perpendicular to the line QP, so the sine of that angle
is one. So now I simply get m times
v of P times the distance QP. And you can do the same
for point A. You can write down
m times v of A times QA. I'm not doing that. You will see shortly
why I'm not doing that. Nature is very kind. Nature's going to give me
that last part for free. This, by the way, is the
conservation of angular momentum about that point Q
where the mass is located. I have here one equation with
two unknowns-- v of P and QP-- so I can't solve. So I need another equation. Well, of course,
there is another one. We have also the conservation
of mechanical energy. So now we can say that the total
energy must be conserved, and the total energy
is one-half... I do it at point P-- equals one-half mvP-squared--
that is the kinetic energy-- minus mMG divided
by the distance, QP. This is the potential energy when the distance between
capital M and little m is QP. With this number, we know... because that is
minus MG divided by 2a. That's our equation number five. Oops! I slipped up here. You may have noticed it. I dropped a little m
which should be in here. Sorry for that. And so now we have here
a big moment in our lives that we have applied both laws. This is the conservation
of mechanical energy. And now I have two equations
with two unknowns-- QP and v of P--
and so I can solve for both. Notice that this second equation is a quadratic equation
in v of P. So you're going
to get two solutions, and the two solutions-- one, v of P will give you
the distance QP. The other one will be vA,
which gives you the distance QA. How come that we get
both solutions? Well, this is only
a stupid equation. This equation doesn't know
that I used a subscript P. I could have used a subscript A
here and put in here QA. That's the term that I left out. And therefore,
when I solve the equations, I get both vP and vA because
those are the situations that the velocity vector is perpendicular
to the position vector. And if I use now
our numerical results, and I solve for you
that quadratic equation-- two equations
with two unknowns-- then I find that QP-- you may
want to check that at home-- is about 6.6 times ten to
the third... three kilometers. It means that it's
only 200 kilometers above the Earth's surface. At that low altitude, this
orbit will not last very long, and the satellite will reenter
into the Earth's atmosphere. And it leads to a speed
at point P, at perigee, of 10.7 kilometers per second. My second solution then
is that QA turns out to be huge. No surprise because we know
that the semimajor axis is 50,000 kilometers. We find 9.3 times ten
to the four kilometers, and we find for v of A, this value is 14 times larger
than this one, and so the velocity
will be 14 times smaller. I think it's 0.75 kilometers
per second. Yes, that's what it is. Immediate result-- the conservation
of angular momentum-- that the product of QP and vP
must be the same as QA times vA. That's immediate consequence of the conservation
of angular momentum. And when I add this up,
QA plus QP, I better find 2a, which in our case
is about 100,000 kilometers, because a was 50,000 kilometers. So when you add these two up, you must find very close
to 100,000 and indeed you do. So now we know everything
there is to be known about this ellipse, and that
came from the initial conditions from the four numbers
that I gave you. We know the period,
we know where apogee is, we know where perigee is,
we know the orbital period-- anything we want to know. Now I want to get into a subject
which is quite difficult, and it has to do
with change of orbits. Burning a rocket
when you are in orbit and your orbit will change. And I will do it only
for some simplified situations. I start off
with a circular orbit, and I will fire the rocket
in such a way... that I will only fire it in such
a way that my velocity will either increase exactly
tangentially to the orbit, so that it will increase
in this direction or that it will decrease
in this direction. So if I'm going in orbit
like this, I either fire my rocket
like this, or I fire my rocket like this, but that is difficult enough
what we do now. So this is our circular orbit
with radius r, and at location X, at 12:00,
that is where I fire my rocket. The first thing I do,
I increase the kinetic energy. So I fire my rocket, I blast my
rocket, we go in this direction. I blast my rocket
in this direction, and so the speed which was
originally this in orbit-- the speed will now increase. I add kinetic energy, and now I have a new speed
which is higher. If my speed is higher, then
my total energy has increased. I increased the kinetic energy. The burn of the rocket
is so short that I can consider
after the burn that the object is still at X. It's a very brief burn. So the kinetic energy
has increased; the potential energy
is the same, so the total energy is up. And therefore, the total energy
now is larger than the total energy
that I had in my circular orbit. But if that's the case, then clearly 2a
must be larger than 2R. I now go into an elliptical
orbit because the new velocity is no longer the right velocity
for a circular orbit. And so what's going to happen-- I'm going to get
an elliptical orbit like so, whereby 2a must be larger
than 2R because my total energy
is larger. And you see that immediately when you go
to equation number five. If you increase the total
energy, then your a will go up. Okay, so 2a is larger than 2R. That also means
that the period T must be larger than the period
in your circular orbit. Fine. My other option is that
I'm going to fire the rocket when I spew out gas
in this direction, so I take kinetic energy out. So after the burn,
my speed is lower. My speed is now lower. I have taken kinetic energy out. When I take kinetic energy out, the total energy is going to be
less than the circular energy, 2a will be less than 2R,
and the orbital period will be less than
the circular orbital period, and therefore, my new ellipse
will look like this. And so these are
the three situations that I want you
to carefully look at because I'm going to need them
in the next very dramatic story which has to do with the romance
between Peter and Mary. Peter and Mary
are two astronauts, and they are both in orbit in one and the same orbit
around the Earth. This is where Peter is at this
moment, at that location X, and this is where Mary is. They are in exactly the same
orbit, but different satellites. They go around like this, and they are at a distance from
each other which I will express in terms of a fraction F
of the total circumference, so that this arc
equals F times 2 pi R. That's how far they are apart. And that means for Mary to make
it all the way back to point X would be one minus F
times 2 pi R. So far, so good. Mary forgot her lunch,
radios Peter and says, "Peter,
I have no food." Peter feels very sorry for her,
says, "No sweat. I will throw you
a ham sandwich." So Peter prepares a ham sandwich
and wants to throw it to Mary in such a way that Mary
can make the catch. How can Peter possibly do this? Well, the best way, the most
obvious way to do it is to make an orbit
for the ham sandwich whose orbital period
is exactly the same as this time for Mary
to make it back to X. And I will be more specific
by giving you some numbers. Then you can digest that better. Suppose they are in an orbit with a radius
of 7,000 kilometers. And suppose F equals .05, so the separation between Peter
and Mary is 2,200 kilometers. So that is F times 2 pi R. If you know the radius R,
then, of course, the velocity of the astronauts
follows immediately. You have all the tools there. So with R equals 7
kilometers, the astronauts-- a now stands
for astronauts-- is a given, nonnegotiable, and that is about 7.55
kilometers per second. 7.55 kilometers per second. And what is also nonnegotiable is the period to go around,
which is 97 minutes. All of that follows from this R. Okay, if it takes 97 minutes
to go around, then five percent of 97 minutes
is five minutes, if I round it off. So this takes
five minutes to go. 95% of 97 minutes is 92 minutes. So for Mary to go around
and go back to point X is 92 minutes,
rounded-off numbers. So if I can give
my sandwich an orbit which has a period
of 92 minutes, I've got it made
because after 92 minutes, the sandwich will come back to X
and Mary is at X. It's important
that you get that idea. If you get that idea,
then all the rest will follow. So the period of the sandwich
after the throw of Peter-- maybe he has
to throw backwards. If that period is 92 minutes, when Mary is here,
she will catch the sandwich. And so the necessary condition
for this first solution, which is an obvious one, is to make the period
of the sandwich-- S stands for sandwich--
to make that one minus F of the period of the astronauts
in orbit. This is the 97 minutes;
this is .95, so this is 92 minutes
and this is 92 minutes. So then Mary will be back
at point X. What is the orbital period of
the sandwich after the throw? Well, that is 4 pi squared-- you can find that
in equation number six-- times a to the third divided
by MG to the power one-half. That must be equal one minus f times the orbital period
of the astronauts who are in circular orbit. So I take equation number one, and that is four
pi squared, R cubed divided by GM
to the power one-half. That is a necessary condition. Look, we lose M, we lose G,
we lose four, we lose pi. What don't we lose? Well, what we don't lose is a
to the power three-halves equals one minus f times R
to the power three-halves. So a equals R times one minus f
to the power two-thirds. And this is an amazingly
simple result. It means if you know the orbit
of Peter and Mary, which is R, and if you know how far
the two lovers are apart, which is expressed in this f,
then you know what the semimajor axis is
of the sandwich orbit. That comes immediately out
of this equation. But once you know
the semimajor axis, you can immediately calculate,
with equation five, the speed of the sandwich. Because if equation number five
will tell you that minus mMG divided by 2a-- a being now the semimajor axis
of the sandwich orbit-- equals one-half m times
the velocity of the sandwich squared. This happens at location X
after the burn-- after the burn means
after Peter has thrown-- minus mMG divided by capital R, because the sandwich is still
at location capital R, but Peter has changed the speed
to vs. And so once you know a, this equation immediately gives
you vs, and once you know vs, then you know with what speed
Peter should throw. Well, let us work it out
in detail in the example
that we have there. If we calculate a with
the numbers that we have there, which you can easily confirm because you can apply this
equation for yourself-- you know what f is,
you know what R is. Then I find that a
is 6,765 kilometers. Notice that this
is smaller than R. It better be, because it's clear that after the sandwich
is thrown, that we get a green ellipse. We want this time to go around to be less time
than Peter to go around. And if this time is less than the time that it takes
Peter to go around, he has to throw
the sandwich backwards, and therefore, you expect
that the semimajor axis will be smaller than R,
and it is. The speed of the sandwich, which follows then
from equation number six, which was the conservation
of mechanical energy, is 7.42 kilometers per second. Now, what matters is not so much what the speed
of the sandwich is, but what matters for Peter
is what is the speed that he will have
to give the sandwich, which is v of s minus v of a, and that you have
to subtract the speed-- v of s is the speed
of the sandwich, v of a is the speed
of the astronauts in orbit. This is minus 0.13 kilometers
per seconds. And the minus sign indicates that he has to throw
the sandwich backwards. So quite amazing. He is seeing Mary all the way
in the distance, and in order to get the sandwich
to Mary, he doesn't do this,
but he does this-- (whooshes) And the sandwich then will go
into this new orbit, going still forward. 92 minutes later, it's here,
and Mary... Oh, we were here. So he goes forward. 92 minutes later,
the sandwich is here. 92 minutes later,
Mary is right at that point and can make the catch. Now, .13 kilometers per second
is 300 miles per hour, which is a little tough,
even for Peter. And so we have to look
for different solutions. This won't work. This was an easy one,
but it doesn't work. Well, there is
no reason to rush. We can make the sandwich go
around the Earth two times and Mary three times, or Mary two times
and the sandwich only once. As long as they meet at point X,
there is no problem. So we have a whole family
of solutions. We can have Mary pass
that point X na times, and we can have the sandwich pass that point X
n-sandwich times. As long as these are integers,
that's perfectly fine. Then ultimately, if they
have enough patience, they will meet at point X. And if you take these...
this new concept into account that you can wait a certain
number of passages through X, then the equation
that you see there-- the relation
between a and R-- changes only slightly. You now get that a equals R
times n of a minus f divided by ns
to the power two-thirds. And if you substitute
in here a one and a one, which is the case that they make
the catch right away, then you see indeed
you get R (one minus f) to the power two-thirds. So that's exactly
what you have there. Not all solutions that you try
will work. One solution that won't work is na equals one and ns equals
three has no solution. And I'll leave you with the
thought why that is the case. Has no solution. In 1990, when I lectured 8.01
for the first time, I asked my friend and colleague
George Clark to write a program so that
I could show the class this toss of the sandwich
with Mary and Peter in orbit and the sandwich orbit
and everything and the catch, and he did. It was a wonderful program, but that program no longer works
because that's called progress. The computers have changed and so, my right hand,
Dave Pooley, offered to rewrite the program
so that it works on Athena, and we were going
to demonstrate it to you, and you can play
with it yourself. It is available on the homepage, so whatever Dave is going to
show you, you can do yourself. The input parameters that
we need for this program are the radius R, our f
and our n of a and our n of s. And the program
will do all the rest, so you can specify
how many times you want Mary to go
through point X, how many times you want the
sandwich to go through point X. The program will then
calculate for you the speed of the sandwich. It will also give you
vs minus va, which is really the speed with
which Peter has to throw it, but very cleverly,
the program works with a dimensionless parameter
which is this value. And this value, which is
vs divided by va minus one is a number that
is quite unique, because you get solutions
which turn out to be independent of capital G
and independent of capital M, and I'll give you an example. Suppose you will find
for this... for this dimensionless number,
suppose you find minus 0.0175, which is the solution
for n one... na is one and for ns is one. So you'll see that. The computer will generate
that number for us. If now we take our orbit
of 7,000 kilometers... we know what v of a is,
and so we can calculate now that vs minus va
is the va times that number. But we know what the va is,
it's 7.5, so you get minus 0.175
times 7.55 kilometers, and lo and behold, that is our minus 0.13
kilometers per second. Of course! It has to be that number, because that's
what we calculated for our case, n equals one. There it is. And so this dimensionless number
is very transparent, and we will show you
some examples. This would be
the 300 miles per hour, which, of course,
is not very doable. Dave, why don't you demonstrate
the program? And then you'll see what we can
do with that program. We can substitute in there
quite a few parameters that you will find
no doubt interesting. Give us an explanation, Dave, of what the students
can do with this. Oh, let me show them an overhead
here which would help you in following what Dave
will be telling you. You see there the f value? It's always 5% we took, and you're going to see here
the numbers for... the number of times that Mary
passes through point X and the number of times
that the sandwich will pass through point X. This is that very first case
that we worked on together. Here, you see that number
minus 0.017, and you see indeed
that it is a successful catch. So let's work at that first. David, explain how it works. DAVID:
Okay, well, you can see in the middle of the screen
is planet Earth, and these two triangles
represent the astronauts. The yellow one is Mary
and the red one is Peter, and he's holding the sandwich
right in the middle. They start off at a radius
of 22,000 kilometers from the center of the Earth. That's the default, but you
can change that if you'd like. And we set na and ns through
these pulldown menus, so we'll set them both
to one and one now. And we ask the program
to calculate the value for us of this dimensionless parameter, and it comes up with it,
and we want to use that value. And so we have everything set... LEWIN:
That's this number, right, Dave? This minus 0.0175, etc. DAVID:
Yep, it's right over here. So then we ask the program
to prepare the toss. We click this button down here, and when it's ready, the green
play button will become active. And when that happens,
we can click on that, and it'll play the toss for us. LEWIN:
Peter always throws at X, 12:00. DAVID:
Always at 12:00. LEWIN:
There goes the sandwich. You see the sandwich? Great sandwich. Big sandwich. So notice that the sandwich
makes it around exactly at the same time--
(exclaims) for Mary to be happy. Now there's no reason why
we shouldn't try one, too. So that means Mary
reaches point X, but the sandwich went
twice around the Earth. There is no problem with
this solution in principle, but you have to be quite
far away from the Earth. If you're too close to the Earth
like Dave's orbit, which has a radius
of only 22,000 kilometers, something very catastrophic
will happen. DAVID: Yeah, okay, so now we
want to set ns to two, and we do that
with the pulldown menu. And we ask it to prepare
the toss again, and it goes through
its numerical calculations of the orbit. And when it's ready,
it'll let us know. And now we can watch the toss. LEWIN:
So there goes the sandwich. It wants to go around the Earth
twice, but it hits the Earth. That's too bad. If you make this dimensionless
parameter minus one, then v of s is zero. And what does it mean
that v of s is zero. That the sandwich stands still,
has no speed in orbit anymore. And so what happens
with the ellipse, that becomes radial infall. Dave? DAVE: Okay, so if we want
to use our own value for this dimensionless
parameter, then we can go
to this box right here and put in whatever we want,
so we'll put in minus one. And we make sure
that the program is going to use our value
instead of the calculated value. In this case, these numbers
don't matter-- na and ns. They're irrelevant, because the program is going
to use our value. We ask it to prepare... LEWIN: The minus one overrides
everything else now. DAVID:
Yes. It's going through
its calculations, and now we can see what happens. LEWIN:
12:00, there goes the sandwich. Now, Peter decides at one point that instead of throwing
the sandwich backwards, he can also throw
the sandwich forward, because, look, we have here
the red ellipse. There's no reason why Mary
couldn't go twice through X-- one... twice. And then the sandwich would
make a larger ellipse and meet here when Mary
has gone around twice. Then, of course, the sandwich
has to be thrown forward. And so Peter makes a calculation for what we call
the 2/1 situation. Mary goes twice through X; the sandwich goes
once through X. But Peter made a mistake. Peter got nervous, and he puts
in the wrong parameters, and you will see what happens. Dave will first show you
the right parameters. DAVID: Okay, so if we would
have asked the program to calculate it
for the case of 2/1, it would have come up
with a value for this dimensionless parameter of .1659
or something thereabouts. But, you know, Peter
made his miscalculation, and he wants to use .164, so this is what we'll put
into the program, and we'll prepare the toss and see what happens
with this value. Okay, now it's ready. LEWIN:
Poor Mary must be hungry by now. There we go. Now we go forward. You can see that. You see, it goes forward. It goes a very large ellipse,
and Mary will go around twice. When Mary is here, see,
the sandwich is only halfway. And if Peter had
only done it right, Mary's troubles
would now soon be over. But Peter made this small
mistake, and... (students laugh) Mary cannot catch it. If you make this dimensionless
parameter plus .42, then it's very easy
to convince yourself that the sandwich--
must be a plus-- will have the escape velocity
from the orbit. Maybe Peter got angry
at one point at Mary-- you never know
about these situations-- and he threw it very fast, and Dave will show you
what happens then. DAVID:
Okay. LEWIN:
And it goes to infinity, and it won't be fresh anymore
when it gets there. Okay, see you Friday.
