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PROFESSOR: I now want to start with a very general initial condition of an
object, little m, 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 = 2a, a being the semi-major axis.
And our object happens to be here, mass little m.
And this distance equals r_0.
Think of it as being time 0.
And at time 0 when it is there, it has a velocity in that ellipse.
Let this be v_0.
And there is an angle between the position vector and
v_0, I call that phi_0.
So I'm giving you M, I'm giving you v_0, I'm giving you r_0,
I'm giving you phi_0.
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 semi-major 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 semi-major 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 1/2 * m * (v_0)^2--
that is when the object is here at location D--
minus little m, capital MG divided by this r_0 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 m * M * G / 2a.
And so you have one equation with one unknown, which is a.
But 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--
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 tried 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 especially 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 much time you want to spend on notes or 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.
For capital M, we take the earth, and that is 6 * 10^24 kilograms.
So that's my M. I promise you, you will know M. I will give you r_0.
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_0 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 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 2 * MG divided
by r_0, 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 9 kilometers per second, you're pretty close already.
So that's the reason why this semi-major 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 OK.
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 * v_0 * r_0 times the sine of phi_0.
This is the situation at D. Why do we have a sine phi_0?
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 1.
So now I simply get m times v_P times the distance QP.
And you can do the same for point A. You can write down m
times v_A times QA.
I'm not doing that.
You will see shortly why I'm not doing that.
Nature is very kind.
Nature is 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_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'll do it at point P--
equals 1/2 * m * (v_P)^2--
that is the kinetic energy--
minus m*M*G divided by the distance QP.
This is the potential energy when the distance between capital M
and little m is QP.
But 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 life 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_P, and so I can
solve for both.
Notice that this second equation is a quadratic equation in v_P.
So you're going to get two solutions, and the two solutions, one v_P will
give you the distance QP.
The other one will be v_A, which gives you the distance QA.
How come we get both solutions?
Well, this is only a stupid equation.
This equation doesn't know that I used a subscript B. 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 v_P and v_A
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 * 10^3, three kilometers.
It means 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 semi-major axis is 50,000 kilometers.
We find 9.3 * 10^4 kilometers, and we find for v_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 of 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 + 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.