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.
It's 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. That is almost infinity, not quite.
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 50,00 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.
So far, so good. 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 on 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 7,000 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] And 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.
