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Today, I will talk to you

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about elliptical orbits
and Kepler's famous laws.

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I first want to review
with you briefly

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what we know about
circular orbits,

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so I wrote on the blackboard
everything we know

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about circular orbits.

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There's an object mass little m

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going in a circle
around capital M.

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This could be the Sun; 
it could be the Earth.

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It has radius R, circular.

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We know there in equation one

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how to derive the time
that it takes to go around.

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The way we found that

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was by setting the centripetal
force onto little m

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the same as
the gravitational force.

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Also, the velocity in orbit--

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maybe I should say
speed in orbit--

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also follows through
the same kind of reasoning.

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Then we have the conservation
of mechanical energy--

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the sum of kinetic energy
and potential energy.

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It's a constant; 
it's not changing.

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You see there first the
component of the kinetic energy,

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which is
the one-half mv-squared,

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and then you see the term
which is the potential energy.

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We have defined potential energy
to be zero at infinity,

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and that is why all bound orbits
have negative total energy.

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If the total energy is positive,
the orbit is not bound.

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And when you add these two up,

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you have an amazing coincidence
that we have discussed before.

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We get here
a very simple answer.

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The escape velocity you find by
setting this E total to be zero,

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so this part of the equation
is zero.

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Out pops that speed
with which you can escape

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the gravitational pull
of capital M,

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which is the square root
of two times larger than this V.

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And I want to remind you
that for near Earth orbits,

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the period to go around
the Earth is about 90 minutes,

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and the speed--
this velocity, then,

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that you see in equation two--

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is about eight kilometers
per second,

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and the escape velocity
from that orbit

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would be about
11.2 kilometers per second.

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And for the Earth
going around the Sun,

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the period would be
about 365 days,

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and the speed
of the Earth in orbit

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is about 30 kilometers
per second,

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just to refresh your memory.

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Now, circular orbits
are special.

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In general, bound orbits
are ellipses,

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even though I must add to it

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that most orbits of our planets
in our solar system--

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very close to circular,
but not precisely circular.

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But the general solutions call
for a elliptical orbit.

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And I first want
to discuss with you

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the three famous laws by Kepler
from the early 17th century.

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These were brilliant statements
that he made.

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The interesting thing is that

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before he made these
brilliant statements,

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he published more nonsense
than anyone else.

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But finally he arrived
at two... three golden eggs.

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And the first golden egg then is
that the orbits are ellipses--

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he talked always about planets--
and the Sun is at one focus.

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That's Kepler's law number one.

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These are from
around 1618 or so.

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The second...
Kepler's second law is--

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quite bizarre how he found that
out, an amazing accomplishment.

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If you take an ellipse, and you
put the Sun here at a focus--

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this is highly exaggerated

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because I told you that most
orbits look sort of circular--

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and the planet goes
from here to here

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in a certain amount of time,

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and you compare that with the
planet going from here to here

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in a certain amount of time,

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then Kepler found out
that if this area here

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is the same as that area here,

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that the time to go
from here to here

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is the same as to go
from there to there.

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An amazing accomplishment
to come up with that idea.

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And this is called
"equal areas, equal times."

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Somehow, it has the smell

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of some conservation
of angular momentum.

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And then his third law was that

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if you take the orbital period
of an ellipse,

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that is proportional
to the third power

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of the mean distance to the Sun.

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And he was so pleased
with that result

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that he wrote jubilantly
about it.

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I will show you here the data

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that Kepler had available
in 1618,

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largely from the work done
by, of course, astronomers,

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observers like Tycho Brahe
and others.

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You see here the six planets
that were known at the time,

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and the mean distance
to the Sun.

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For the Earth, it is one because
we work in astronomical units.

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Everything is referenced
to the distance of the Earth.

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This is 150 million kilometers.

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And it takes the Earth 365 days
to go around the Sun; 

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Jupiter, about 12 years; 
and Saturn, about 30 years.

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And then when he takes this
number to the power of three

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and this number squared,
and he divides the two,

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then he gets numbers
which are amazingly constant.

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And that is his third law.

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The third law leads immediately

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to the inverse square dependence
of gravity,

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which he was not aware of,

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but Newton later
put that all together.

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But he very jubilantly writes:

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And he wrote that in 1619.

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So, orbits in general
are ellipses.

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And now I want
to review with you

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what I have there on
the blackboard about ellipses.

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You see an ellipse there?

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Capital M-- could be the Earth,
could be the Sun--

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is at location Q.

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The ellipse has
a semimajor axis A,

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so the distance P to A--
perigee to apogee-- is 2a.

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If M were the Earth, capital M,

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then we would call the point
of closest approach "perigee,"

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and the point farthest
away from the Earth,

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we would call that "apogee."

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If capital M were the Sun,

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we would call that "aphelion"
and "perihelion."

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So you see the little m
going in orbit; 

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you see the position
vector r of q.

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It has a certain velocity v.

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And so the total mechanical
energy is conserved.

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The sum of kinetic energy and
potential energy doesn't change.

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The first term is the kinetic
energy-- one-half mv squared,

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and the second term
is the potential energy--

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no different from equation three
for circular orbits,

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except that now capital R, which
was a fixed number in a circle,

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is now a little r, and little r
changes, of course, with time.

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Also that velocity, v,
in that equation number five

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will also change with time

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because it's
an elliptical orbit.

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It will not change in time
in equation number two

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and in equation number three.

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Now I give you a result
which I didn't prove,

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and that is that
the total mechanical energy,

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which has these two terms in it
which you do fully understand,

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also equals minus mMG
divided by 2a,

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and a is the semimajor axis.

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And compare number five
with number three,

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then you see they are
brothers and sisters.

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The only change is that
what was capital R before

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is now little a,
the semimajor axis.

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And if you want to calculate the
time to go around the ellipse,

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then you get an equation
for T squared,

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which is almost identical
to number one

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for the circular orbit,

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except that now the radius
has to be replaced by a,

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which is the semimajor axis.

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And the escape velocity
you can calculate

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in exactly the same way that you
calculate the escape velocity

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under equation number four.

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All you do is you make
the total energy zero,

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and then you solve equation
three and equation five,

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and out pops the speed
that you need

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to make it all the way
out to infinity.

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And that is in the case
of the circular orbit,

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square root of two
times the speed in orbit.

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So these are the numbers

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that we are going
to use today, the equations.

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And there's one thing which
is already quite remarkable

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and very nonintuitive-- very
nonintuitive, to say the least.

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That if you have various orbits

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which have the same
semimajor axis,

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that the period is the same
and the energy is the same,

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and that's by no means obvious.

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So, this is one orbit-- think
of it as being an ellipse--

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and this is another one.

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This distance is the same
as this distance.

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I've just done it that way.

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That means, according
to equation five and six,

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that both orbits have exactly
the same mechanical energy,

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and both orbits have
the same periods.

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So to go around
this circular orbit

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will take the same amount of
time as to go around this one,

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and that is by no means obvious.

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I now want to start with
a very general initial condition

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of an object, little m, in
orbit... in an elliptical orbit.

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And I want to see how we can get

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all the information
about the ellipse

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that we would like to find out.

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So I'm only giving you
the initial conditions.

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So here is an ellipse,
here is P and here is A.

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If this is an ellipse
around the Earth,

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then this would be perigee
and this would be apogee.

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The mass is capital M,
this is point Q.

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Let me get a ruler so that
I can draw some nice lines.

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So the distance AP equals 2a--
a being the semimajor axis--

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and our object happens
to be here-- mass little m--

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and this distance equals R zero.

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Think of it as being time zero.

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And at time zero,
when it is there,

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it has a velocity
in that ellipse.

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Let this be v zero.

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And there is an angle between
the position vector and v zero;

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I call that phi zero.

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So I'm giving you M,
I'm giving you v zero,

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I'm giving you r zero,
I'm giving you phi zero.

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And now I'm going to ask you,

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can we find out from
these initial conditions

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how long it takes for
this object to go around?

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Can we find out what QP is?

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Can we find out what
the semimajor axis is?

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Can we find out what
the velocity is at point P,

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at closest approach when
this angle is 90 degrees?

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And can we find out
what the velocity is

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when the object, little m,
is farthest away-- apogee?

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Can we find all these things?

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And the answer is yes.

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a is the easiest to find--
the semimajor axis.

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I turn to equation number five,

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which is the conservation
of mechanical energy.

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And the conservation
of mechanical energy

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says that the total energy

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is the kinetic energy
plus the potential energy

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equals one-half mv zero squared.

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That is when the object
is here at location D

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minus mMG divided by this r zero
at location D.

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This can never change.

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This is the same throughout
the whole orbit,

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so it must also be,
according to equation five,

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minus mMG divided by 2a.

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And so you have one equation
with one unknown, which is a,

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because you know
all the other things:

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M cancels... M always cancels
when you deal with gravity,

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so you only have a
as an unknown.

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So that's done.

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If the total energy
were positive,

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then for this to be positive,
a has to be negative.

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That's physical nonsense,
of course,

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so this only holds
for bound orbits.

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So positive values for E total
are not allowed.

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Once you have a, you use...

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so this is from
equation number five.

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If you now apply
equation number six,

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immediately pops out T,
the orbital periods,

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because the only thing
you didn't know yet was a,

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but you know a now.

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So we also know
how long it takes

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for the object
to go around in orbit.

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And I try to be quantitative
with you.

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Step by step,
as we analyze this further,

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I will apply this
to a specific case

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for someone going
around the Earth.

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Everything I'm
telling you today,

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including all
numerical examples,

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are in a handout
which is six pages thick,

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which I wrote
specially for you.

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It will be on the Web.

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We're not going to print it
here-- that's a waste of paper.

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It's 1999, so that's
what we have the Web for.

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So you can decide on your own
how many... how much notes,

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how much time you want
to spend on notes,

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and to what extent
you want to concentrate

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and try to follow the steps.

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It's up to you.

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But everything is there--

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literally everything,
every numerical example.

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We take for capital M...
we take the Earth,

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and that is six times ten
to the 24 kilograms.

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So that's my M.

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I promised you you will know M.

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I will give you r zero.

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That is 9,000 kilometers.

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That's the location at point D.

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I give you the conditions at D.

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The speed at point D
is 9.0 kilometers per second,

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and I'll give you phi zero
is 120 degrees.

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Everything else we should
be able to calculate now

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from these numbers.

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First of all,
with equation five,

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you can convince yourself
sticking in these numbers

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that the total energy
is indeed negative.

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00:15:43 --> 00:15:49
Of course, if I make
the total energy positive,

288
00:15:46 --> 00:15:52
it's not an ellipse,
so then it's all over.

289
00:15:48 --> 00:15:54
It is negative; 
it is an ellipse.

290
00:15:51 --> 00:15:57
So with equation number five,
I then pop out a.

291
00:15:54 --> 00:16:00
Right? Because that's one
equation with one unknown,

292
00:15:58 --> 00:16:04
and I put in the numbers--

293
00:16:00 --> 00:16:06
you can confirm them
and check them at home--

294
00:16:04 --> 00:16:10
and I find that a
is quite large.

295
00:16:07 --> 00:16:13
a is about 50,000 kilometers.

296
00:16:10 --> 00:16:16
That's huge.

297
00:16:11 --> 00:16:17
That is almost infinity,
not quite.

298
00:16:14 --> 00:16:20
Remember, it starts off
at 9,000 kilometers,

299
00:16:19 --> 00:16:25
but a is 50,000 kilometers.

300
00:16:21 --> 00:16:27
That means 2a
is 100,000 kilometers.

301
00:16:25 --> 00:16:31
Why is that so large?

302
00:16:27 --> 00:16:33
Well, the answer lies in
evaluating the escape velocity.

303
00:16:33 --> 00:16:39
The escape velocity
of this little mass

304
00:16:38 --> 00:16:44
when it is at position D,

305
00:16:41 --> 00:16:47
for which these
are the input parameters,

306
00:16:46 --> 00:16:52
is the square root
of 2MG divided by r zero,

307
00:16:51 --> 00:16:57
and that is
9.4 kilometers per second.

308
00:16:56 --> 00:17:02
Well, if you need
9.4 kilometers per second

309
00:16:59 --> 00:17:05
to make it out to infinity,

310
00:17:01 --> 00:17:07
and you have
nine kilometers per second,

311
00:17:03 --> 00:17:09
you're pretty close already.

312
00:17:05 --> 00:17:11
So that's the reason
why this semimajor axis

313
00:17:08 --> 00:17:14
is indeed such
a horrendous number.

314
00:17:10 --> 00:17:16
It's no surprise.

315
00:17:11 --> 00:17:17
If now I use
equation number six,

316
00:17:14 --> 00:17:20
then I find the period,

317
00:17:16 --> 00:17:22
and I find that it takes
about 31 hours

318
00:17:20 --> 00:17:26
for this object
to go around the Earth.

319
00:17:23 --> 00:17:29
So far, so good.

320
00:17:25 --> 00:17:31
Now we want to know
what the situation is

321
00:17:28 --> 00:17:34
with perigee and with apogee.

322
00:17:31 --> 00:17:37
Can we calculate
the distance QP?

323
00:17:34 --> 00:17:40
Can we calculate the speed at
location P and at location A?

324
00:17:40 --> 00:17:46
And now comes
our superior knowledge.

325
00:17:44 --> 00:17:50
Now we're going to apply for the
first time in systems like this,

326
00:17:49 --> 00:17:55
the conservation
of angular momentum.

327
00:17:52 --> 00:17:58
Angular momentum is conserved
about this point Q,

328
00:17:56 --> 00:18:02
butonly about that point Q.

329
00:17:59 --> 00:18:05
It is not conserved about any
other point, but that's okay.

330
00:18:03 --> 00:18:09
All I want is that point Q.

331
00:18:04 --> 00:18:10
That is where capital M
is located.

332
00:18:07 --> 00:18:13
What is the magnitude
of that angular momentum?

333
00:18:11 --> 00:18:17
Well, let's first take point D.

334
00:18:15 --> 00:18:21
When the object is at D,

335
00:18:17 --> 00:18:23
the magnitude
of the angular momentum

336
00:18:20 --> 00:18:26
is m times v zero times r zero
times the sine of phi zero.

337
00:18:25 --> 00:18:31
This is the situation at D.

338
00:18:27 --> 00:18:33
Why do we have a sine phi zero?

339
00:18:30 --> 00:18:36
Because we have a cross
between r and v,

340
00:18:33 --> 00:18:39
and with a cross product,
you have the sine of the angle.

341
00:18:36 --> 00:18:42
So that's the situation
at point D.

342
00:18:38 --> 00:18:44
What is the situation
at point P?

343
00:18:42 --> 00:18:48
Well, at point P,
the velocity vector

344
00:18:45 --> 00:18:51
is perpendicular to the line QP,

345
00:18:48 --> 00:18:54
so the sine of that angle
is one.

346
00:18:51 --> 00:18:57
So now I simply get m times
v of P times the distance QP.

347
00:18:59 --> 00:19:05
And you can do the same
for point A.

348
00:19:02 --> 00:19:08
You can write down
m times v of A times QA.

349
00:19:07 --> 00:19:13
I'm not doing that.

350
00:19:08 --> 00:19:14
You will see shortly
why I'm not doing that.

351
00:19:11 --> 00:19:17
Nature is very kind.

352
00:19:13 --> 00:19:19
Nature's going to give me
that last part for free.

353
00:19:16 --> 00:19:22
This, by the way, is the
conservation of angular momentum

354
00:19:25 --> 00:19:31
about that point Q
where the mass is located.

355
00:19:30 --> 00:19:36
I have here one equation with
two unknowns-- v of P and QP--

356
00:19:34 --> 00:19:40
so I can't solve.

357
00:19:35 --> 00:19:41
So I need another equation.

358
00:19:37 --> 00:19:43
Well, of course,
there is another one.

359
00:19:39 --> 00:19:45
We have also the conservation
of mechanical energy.

360
00:19:42 --> 00:19:48
So now we can say that the total
energy must be conserved,

361
00:19:49 --> 00:19:55
and the total energy
is one-half...

362
00:19:52 --> 00:19:58
I do it at point P--

363
00:19:54 --> 00:20:00
equals one-half mvP-squared--
that is the kinetic energy--

364
00:20:03 --> 00:20:09
minus mMG divided
by the distance, QP.

365
00:20:08 --> 00:20:14

366
00:20:10 --> 00:20:16
This is the potential energy

367
00:20:12 --> 00:20:18
when the distance between
capital M and little m is QP.

368
00:20:16 --> 00:20:22
With this number, we know...

369
00:20:19 --> 00:20:25
because that is
minus MG divided by 2a.

370
00:20:23 --> 00:20:29
That's our equation number five.

371
00:20:26 --> 00:20:32

372
00:20:28 --> 00:20:34
Oops! I slipped up here.

373
00:20:31 --> 00:20:37
You may have noticed it.

374
00:20:33 --> 00:20:39
I dropped a little m
which should be in here.

375
00:20:37 --> 00:20:43
Sorry for that.

376
00:20:38 --> 00:20:44

377
00:20:40 --> 00:20:46
And so now we have here
a big moment in our lives

378
00:20:44 --> 00:20:50
that we have applied both laws.

379
00:20:46 --> 00:20:52
This is the conservation
of mechanical energy.

380
00:20:55 --> 00:21:01
And now I have two equations
with two unknowns--

381
00:20:59 --> 00:21:05
QP and v of P--
and so I can solve for both.

382
00:21:02 --> 00:21:08
Notice that this second equation

383
00:21:05 --> 00:21:11
is a quadratic equation
in v of P.

384
00:21:08 --> 00:21:14
So you're going
to get two solutions,

385
00:21:11 --> 00:21:17
and the two solutions--

386
00:21:14 --> 00:21:20
one, v of P will give you
the distance QP.

387
00:21:20 --> 00:21:26
The other one will be vA,
which gives you the distance QA.

388
00:21:25 --> 00:21:31
How come that we get
both solutions?

389
00:21:28 --> 00:21:34
Well, this is only
a stupid equation.

390
00:21:31 --> 00:21:37
This equation doesn't know
that I used a subscript P.

391
00:21:36 --> 00:21:42
I could have used a subscript A
here and put in here QA.

392
00:21:40 --> 00:21:46
That's the term that I left out.

393
00:21:42 --> 00:21:48
And therefore,
when I solve the equations,

394
00:21:46 --> 00:21:52
I get both vP and vA because
those are the situations

395
00:21:50 --> 00:21:56
that the velocity vector

396
00:21:52 --> 00:21:58
is perpendicular
to the position vector.

397
00:21:56 --> 00:22:02
And if I use now
our numerical results,

398
00:22:00 --> 00:22:06
and I solve for you
that quadratic equation--

399
00:22:05 --> 00:22:11
two equations
with two unknowns--

400
00:22:08 --> 00:22:14
then I find that QP-- you may
want to check that at home--

401
00:22:14 --> 00:22:20
is about 6.6 times ten to
the third... three kilometers.

402
00:22:18 --> 00:22:24
It means that it's
only 200 kilometers

403
00:22:21 --> 00:22:27
above the Earth's surface.

404
00:22:23 --> 00:22:29
At that low altitude, this
orbit will not last very long,

405
00:22:27 --> 00:22:33
and the satellite will reenter
into the Earth's atmosphere.

406
00:22:30 --> 00:22:36
And it leads to a speed
at point P, at perigee,

407
00:22:35 --> 00:22:41
of 10.7 kilometers per second.

408
00:22:39 --> 00:22:45
My second solution then
is that QA turns out to be huge.

409
00:22:44 --> 00:22:50
No surprise because we know
that the semimajor axis

410
00:22:49 --> 00:22:55
is 50,000 kilometers.

411
00:22:51 --> 00:22:57
We find 9.3 times ten
to the four kilometers,

412
00:22:55 --> 00:23:01
and we find for v of A,

413
00:22:58 --> 00:23:04
this value is 14 times larger
than this one,

414
00:23:02 --> 00:23:08
and so the velocity
will be 14 times smaller.

415
00:23:06 --> 00:23:12
I think it's 0.75 kilometers
per second.

416
00:23:09 --> 00:23:15
Yes, that's what it is.

417
00:23:10 --> 00:23:16

418
00:23:14 --> 00:23:20
Immediate result--

419
00:23:15 --> 00:23:21
the conservation
of angular momentum--

420
00:23:19 --> 00:23:25
that the product of QP and vP
must be the same as QA times vA.

421
00:23:25 --> 00:23:31
That's immediate consequence

422
00:23:28 --> 00:23:34
of the conservation
of angular momentum.

423
00:23:32 --> 00:23:38
And when I add this up,
QA plus QP, I better find 2a,

424
00:23:37 --> 00:23:43
which in our case
is about 100,000 kilometers,

425
00:23:41 --> 00:23:47
because a was 50,00 kilometers.

426
00:23:44 --> 00:23:50
So when you add these two up,

427
00:23:47 --> 00:23:53
you must find very close
to 100 -->  and indeed you do.

428
00:23:54 --> 00:24:00
So now we know everything
there is to be known

429
00:23:57 --> 00:24:03
about this ellipse, and that
came from the initial conditions

430
00:24:01 --> 00:24:07
from the four numbers
that I gave you.

431
00:24:03 --> 00:24:09
We know the period,
we know where apogee is,

432
00:24:06 --> 00:24:12
we know where perigee is,
we know the orbital period--

433
00:24:10 --> 00:24:16
anything we want to know.

434
00:24:13 --> 00:24:19
Now I want to get into a subject
which is quite difficult,

435
00:24:19 --> 00:24:25
and it has to do
with change of orbits.

436
00:24:24 --> 00:24:30
Burning a rocket
when you are in orbit

437
00:24:29 --> 00:24:35
and your orbit will change.

438
00:24:32 --> 00:24:38
And I will do it only
for some simplified situations.

439
00:24:38 --> 00:24:44
I start off
with a circular orbit,

440
00:24:42 --> 00:24:48
and I will fire the rocket
in such a way...

441
00:24:48 --> 00:24:54
that I will only fire it in such
a way that my velocity

442
00:24:55 --> 00:25:01
will either increase exactly
tangentially to the orbit,

443
00:25:02 --> 00:25:08
so that it will increase
in this direction

444
00:25:04 --> 00:25:10
or that it will decrease
in this direction.

445
00:25:07 --> 00:25:13
So if I'm going in orbit
like this,

446
00:25:09 --> 00:25:15
I either fire my rocket
like this,

447
00:25:11 --> 00:25:17
or I fire my rocket like this,

448
00:25:13 --> 00:25:19
but that is difficult enough
what we do now.

449
00:25:15 --> 00:25:21
So this is our circular orbit
with radius r,

450
00:25:20 --> 00:25:26
and at location X, at 12:00,
that is where I fire my rocket.

451
00:25:26 --> 00:25:32

452
00:25:30 --> 00:25:36
The first thing I do,
I increase the kinetic energy.

453
00:25:33 --> 00:25:39
So I fire my rocket, I blast my
rocket, we go in this direction.

454
00:25:37 --> 00:25:43
I blast my rocket
in this direction,

455
00:25:40 --> 00:25:46
and so the speed which was
originally this in orbit--

456
00:25:43 --> 00:25:49
the speed will now increase.

457
00:25:45 --> 00:25:51
I add kinetic energy,

458
00:25:48 --> 00:25:54
and now I have a new speed
which is higher.

459
00:25:52 --> 00:25:58
If my speed is higher, then
my total energy has increased.

460
00:25:56 --> 00:26:02
I increased the kinetic energy.

461
00:25:59 --> 00:26:05
The burn of the rocket
is so short

462
00:26:01 --> 00:26:07
that I can consider
after the burn

463
00:26:04 --> 00:26:10
that the object is still at X.

464
00:26:06 --> 00:26:12
It's a very brief burn.

465
00:26:08 --> 00:26:14
So the kinetic energy
has increased; 

466
00:26:10 --> 00:26:16
the potential energy
is the same,

467
00:26:12 --> 00:26:18
so the total energy is up.

468
00:26:13 --> 00:26:19
And therefore, the total energy
now is larger

469
00:26:18 --> 00:26:24
than the total energy
that I had in my circular orbit.

470
00:26:24 --> 00:26:30
But if that's the case,

471
00:26:27 --> 00:26:33
then clearly 2a
must be larger than 2R.

472
00:26:32 --> 00:26:38
I now go into an elliptical
orbit because the new velocity

473
00:26:35 --> 00:26:41
is no longer the right velocity
for a circular orbit.

474
00:26:38 --> 00:26:44
And so what's going to happen--

475
00:26:41 --> 00:26:47
I'm going to get
an elliptical orbit like so,

476
00:26:45 --> 00:26:51
whereby 2a must be larger
than 2R

477
00:26:48 --> 00:26:54
because my total energy
is larger.

478
00:26:52 --> 00:26:58
And you see that immediately

479
00:26:54 --> 00:27:00
when you go
to equation number five.

480
00:26:57 --> 00:27:03
If you increase the total
energy, then your a will go up.

481
00:27:03 --> 00:27:09
Okay, so 2a is larger than 2R.

482
00:27:06 --> 00:27:12
That also means
that the period T

483
00:27:09 --> 00:27:15
must be larger than the period
in your circular orbit.

484
00:27:14 --> 00:27:20

485
00:27:16 --> 00:27:22
Fine.

486
00:27:17 --> 00:27:23
So far, so good.

487
00:27:20 --> 00:27:26
My other option is that
I'm going to fire the rocket

488
00:27:23 --> 00:27:29
when I spew out gas
in this direction,

489
00:27:26 --> 00:27:32
so I take kinetic energy out.

490
00:27:28 --> 00:27:34
So after the burn,
my speed is lower.

491
00:27:32 --> 00:27:38
My speed is now lower.

492
00:27:36 --> 00:27:42
I have taken kinetic energy out.

493
00:27:38 --> 00:27:44
When I take kinetic energy out,

494
00:27:40 --> 00:27:46
the total energy is going to be
less than the circular energy,

495
00:27:44 --> 00:27:50
2a will be less than 2R,
and the orbital period

496
00:27:46 --> 00:27:52
will be less than
the circular orbital period,

497
00:27:49 --> 00:27:55
and therefore, my new ellipse
will look like this.

498
00:27:53 --> 00:27:59

499
00:27:57 --> 00:28:03
And so these are
the three situations

500
00:27:59 --> 00:28:05
that I want you
to carefully look at

501
00:28:01 --> 00:28:07
because I'm going to need them
in the next very dramatic story

502
00:28:06 --> 00:28:12
which has to do with the romance
between Peter and Mary.

503
00:28:09 --> 00:28:15
Peter and Mary
are two astronauts,

504
00:28:12 --> 00:28:18
and they are both in orbit

505
00:28:15 --> 00:28:21
in one and the same orbit
around the Earth.

506
00:28:19 --> 00:28:25

507
00:28:21 --> 00:28:27
This is where Peter is at this
moment, at that location X,

508
00:28:25 --> 00:28:31
and this is where Mary is.

509
00:28:27 --> 00:28:33

510
00:28:29 --> 00:28:35
They are in exactly the same
orbit, but different satellites.

511
00:28:33 --> 00:28:39
They go around like this,

512
00:28:36 --> 00:28:42
and they are at a distance from
each other which I will express

513
00:28:43 --> 00:28:49
in terms of a fraction F
of the total circumference,

514
00:28:49 --> 00:28:55
so that this arc
equals F times 2 pi R.

515
00:28:53 --> 00:28:59
That's how far they are apart.

516
00:28:56 --> 00:29:02
And that means for Mary to make
it all the way back to point X

517
00:29:03 --> 00:29:09
would be one minus F
times 2 pi R.

518
00:29:07 --> 00:29:13

519
00:29:10 --> 00:29:16
So far, so good.

520
00:29:12 --> 00:29:18

521
00:29:15 --> 00:29:21
Mary forgot her lunch,
radios Peter

522
00:29:17 --> 00:29:23
and says, "Peter,
I have on food."

523
00:29:20 --> 00:29:26
Peter feels very sorry for her,
says, "No sweat.

524
00:29:24 --> 00:29:30
I will throw you
a ham sandwich."

525
00:29:27 --> 00:29:33
So Peter prepares a ham sandwich
and wants to throw it to Mary

526
00:29:30 --> 00:29:36
in such a way that Mary
can make the catch.

527
00:29:33 --> 00:29:39
How can Peter possibly do this?

528
00:29:39 --> 00:29:45
Well, the best way, the most
obvious way to do it

529
00:29:47 --> 00:29:53
is to make an orbit
for the ham sandwich

530
00:29:55 --> 00:30:01
whose orbital period
is exactly the same

531
00:29:59 --> 00:30:05
as this time for Mary
to make it back to X.

532
00:30:02 --> 00:30:08
And I will be more specific
by giving you some numbers.

533
00:30:05 --> 00:30:11
Then you can digest that better.

534
00:30:08 --> 00:30:14
Suppose they are in an orbit

535
00:30:11 --> 00:30:17
with a radius
of 7,000 kilometers.

536
00:30:16 --> 00:30:22
And suppose F equals .05,

537
00:30:19 --> 00:30:25
so the separation between Peter
and Mary is 2,200 kilometers.

538
00:30:26 --> 00:30:32
So that is F times 2 pi R.

539
00:30:31 --> 00:30:37
If you know the radius R,
then, of course,

540
00:30:34 --> 00:30:40
the velocity of the astronauts
follows immediately.

541
00:30:37 --> 00:30:43
You have all the tools there.

542
00:30:39 --> 00:30:45
So with R equals 7
kilometers,

543
00:30:42 --> 00:30:48
the astronauts-- a now stands
for astronauts--

544
00:30:45 --> 00:30:51
is a given, nonnegotiable,

545
00:30:47 --> 00:30:53
and that is about 7.55
kilometers per second.

546
00:30:53 --> 00:30:59
7.55 kilometers per second.

547
00:30:59 --> 00:31:05
And what is also nonnegotiable

548
00:31:01 --> 00:31:07
is the period to go around,
which is 97 minutes.

549
00:31:04 --> 00:31:10
All of that follows from this R.

550
00:31:07 --> 00:31:13
Okay, if it takes 97 minutes
to go around,

551
00:31:11 --> 00:31:17
then five percent of 97 minutes
is five minutes,

552
00:31:15 --> 00:31:21
if I round it off.

553
00:31:17 --> 00:31:23
So this takes
five minutes to go.

554
00:31:21 --> 00:31:27
95% of 97 minutes is 92 minutes.

555
00:31:26 --> 00:31:32
So for Mary to go around
and go back to point X

556
00:31:30 --> 00:31:36
is 92 minutes,
rounded-off numbers.

557
00:31:34 --> 00:31:40
So if I can give
my sandwich an orbit

558
00:31:37 --> 00:31:43
which has a period
of 92 minutes,

559
00:31:40 --> 00:31:46
I've got it made
because after 92 minutes,

560
00:31:43 --> 00:31:49
the sandwich will come back to X
and Mary is at X.

561
00:31:48 --> 00:31:54
It's important
that you get that idea.

562
00:31:49 --> 00:31:55
If you get that idea,
then all the rest will follow.

563
00:31:52 --> 00:31:58
So the period of the sandwich
after the throw of Peter--

564
00:31:56 --> 00:32:02
maybe he has
to throw backwards.

565
00:31:58 --> 00:32:04
If that period is 92 minutes,

566
00:32:00 --> 00:32:06
when Mary is here,
she will catch the sandwich.

567
00:32:03 --> 00:32:09
And so the necessary condition
for this first solution,

568
00:32:08 --> 00:32:14
which is an obvious one,

569
00:32:10 --> 00:32:16
is to make the period
of the sandwich--

570
00:32:13 --> 00:32:19
S stands for sandwich--
to make that one minus F

571
00:32:16 --> 00:32:22
of the period of the astronauts
in orbit.

572
00:32:20 --> 00:32:26
This is the 97 minutes; 
this is .95,

573
00:32:24 --> 00:32:30
so this is 92 minutes
and this is 92 minutes.

574
00:32:28 --> 00:32:34
So then Mary will be back
at point X.

575
00:32:32 --> 00:32:38
What is the orbital period of
the sandwich after the throw?

576
00:32:39 --> 00:32:45
Well, that is 4 pi squared--

577
00:32:42 --> 00:32:48
you can find that
in equation number six--

578
00:32:47 --> 00:32:53
times a to the third divided
by MG to the power one-half.

579
00:32:54 --> 00:33:00

580
00:32:56 --> 00:33:02
That must be equal one minus f

581
00:32:58 --> 00:33:04
times the orbital period
of the astronauts

582
00:33:01 --> 00:33:07
who are in circular orbit.

583
00:33:03 --> 00:33:09
So I take equation number one,

584
00:33:04 --> 00:33:10
and that is four
pi squared, R cubed

585
00:33:10 --> 00:33:16
divided by GM
to the power one-half.

586
00:33:15 --> 00:33:21
That is a necessary condition.

587
00:33:18 --> 00:33:24
Look, we lose M, we lose G,
we lose four, we lose pi.

588
00:33:22 --> 00:33:28
What don't we lose?

589
00:33:24 --> 00:33:30
Well, what we don't lose is a
to the power three-halves

590
00:33:30 --> 00:33:36
equals one minus f times R
to the power three-halves.

591
00:33:37 --> 00:33:43
So a equals R times one minus f
to the power two-thirds.

592
00:33:44 --> 00:33:50
And this is an amazingly
simple result.

593
00:33:50 --> 00:33:56
It means if you know the orbit
of Peter and Mary, which is R,

594
00:33:55 --> 00:34:01
and if you know how far
the two lovers are apart,

595
00:33:58 --> 00:34:04
which is expressed in this f,
then you know

596
00:34:01 --> 00:34:07
what the semimajor axis is
of the sandwich orbit.

597
00:34:05 --> 00:34:11
That comes immediately out
of this equation.

598
00:34:09 --> 00:34:15
But once you know
the semimajor axis,

599
00:34:12 --> 00:34:18
you can immediately calculate,
with equation five,

600
00:34:15 --> 00:34:21
the speed of the sandwich.

601
00:34:17 --> 00:34:23
Because if equation number five
will tell you

602
00:34:23 --> 00:34:29
that minus mMG divided by 2a--

603
00:34:30 --> 00:34:36
a being now the semimajor axis
of the sandwich orbit--

604
00:34:34 --> 00:34:40
equals one-half m times
the velocity

605
00:34:39 --> 00:34:45
of the sandwich squared.

606
00:34:41 --> 00:34:47
This happens at location X
after the burn--

607
00:34:44 --> 00:34:50
after the burn means
after Peter has thrown--

608
00:34:48 --> 00:34:54
minus mMG divided by capital R,

609
00:34:51 --> 00:34:57
because the sandwich is still
at location capital R,

610
00:34:58 --> 00:35:04
but Peter has changed the speed
to vs.

611
00:35:02 --> 00:35:08
And so once you know a,

612
00:35:04 --> 00:35:10
this equation immediately gives
you vs, and once you know vs,

613
00:35:08 --> 00:35:14
then you know with what speed
Peter should throw.

614
00:35:11 --> 00:35:17
Well, let us work it out
in detail

615
00:35:15 --> 00:35:21
in the example
that we have there.

616
00:35:18 --> 00:35:24
If we calculate a with
the numbers that we have there,

617
00:35:24 --> 00:35:30
which you can easily confirm

618
00:35:26 --> 00:35:32
because you can apply this
equation for yourself--

619
00:35:29 --> 00:35:35
you know what f is,
you know what R is.

620
00:35:31 --> 00:35:37
Then I find that a
is 6,765 kilometers.

621
00:35:38 --> 00:35:44

622
00:35:42 --> 00:35:48
Notice that this
is smaller than R.

623
00:35:45 --> 00:35:51
It better be, because it's clear

624
00:35:48 --> 00:35:54
that after the sandwich
is thrown,

625
00:35:51 --> 00:35:57
that we get a green ellipse.

626
00:35:54 --> 00:36:00
We want this time to go around

627
00:35:57 --> 00:36:03
to be less time
than Peter to go around.

628
00:36:00 --> 00:36:06
And if this time is less

629
00:36:02 --> 00:36:08
than the time that it takes
Peter to go around,

630
00:36:06 --> 00:36:12
he has to throw
the sandwich backwards,

631
00:36:09 --> 00:36:15
and therefore, you expect
that the semimajor axis

632
00:36:13 --> 00:36:19
will be smaller than R,
and it is.

633
00:36:16 --> 00:36:22
The speed of the sandwich,

634
00:36:18 --> 00:36:24
which follows then
from equation number six,

635
00:36:22 --> 00:36:28
which was the conservation
of mechanical energy,

636
00:36:26 --> 00:36:32
is 7.42 kilometers per second.

637
00:36:29 --> 00:36:35

638
00:36:30 --> 00:36:36
Now, what matters is not so much

639
00:36:33 --> 00:36:39
what the speed
of the sandwich is,

640
00:36:35 --> 00:36:41
but what matters for Peter
is what is the speed

641
00:36:39 --> 00:36:45
that he will have
to give the sandwich,

642
00:36:42 --> 00:36:48
which is v of s minus v of a,

643
00:36:44 --> 00:36:50
and that you have
to subtract the speed--

644
00:36:47 --> 00:36:53
v of s is the speed
of the sandwich,

645
00:36:49 --> 00:36:55
v of a is the speed
of the astronauts in orbit.

646
00:36:52 --> 00:36:58
This is minus 0.13 kilometers
per seconds.

647
00:36:55 --> 00:37:01
And the minus sign indicates

648
00:36:57 --> 00:37:03
that he has to throw
the sandwich backwards.

649
00:37:00 --> 00:37:06
So quite amazing.

650
00:37:02 --> 00:37:08
He is seeing Mary all the way
in the distance,

651
00:37:05 --> 00:37:11
and in order to get the sandwich
to Mary,

652
00:37:07 --> 00:37:13
he doesn't do this,
but he does this-- (whooshes )

653
00:37:10 --> 00:37:16
And the sandwich then will go
into this new orbit,

654
00:37:16 --> 00:37:22
going still forward.

655
00:37:18 --> 00:37:24
92 minutes later, it's here,
and Mary...

656
00:37:21 --> 00:37:27
Oh, we were here.

657
00:37:22 --> 00:37:28
So he goes forward.

658
00:37:24 --> 00:37:30
92 minutes later,
the sandwich is here.

659
00:37:27 --> 00:37:33
92 minutes later,
Mary is right at that point

660
00:37:30 --> 00:37:36
and can make the catch.

661
00:37:32 --> 00:37:38
Now, .13 kilometers per second
is 300 miles per hour,

662
00:37:37 --> 00:37:43
which is a little tough,
even for Peter.

663
00:37:40 --> 00:37:46
And so we have to look
for different solutions.

664
00:37:43 --> 00:37:49
This won't work.

665
00:37:44 --> 00:37:50
This was an easy one,
but it doesn't work.

666
00:37:47 --> 00:37:53
Well, there is
no reason to rush.

667
00:37:50 --> 00:37:56
We can make the sandwich go
around the Earth two times

668
00:37:54 --> 00:38:00
and Mary three times,

669
00:37:56 --> 00:38:02
or Mary two times
and the sandwich only once.

670
00:38:00 --> 00:38:06
As long as they meet at point X,
there is no problem.

671
00:38:05 --> 00:38:11
So we have a whole family
of solutions.

672
00:38:09 --> 00:38:15
We can have Mary pass
that point X na times,

673
00:38:15 --> 00:38:21
and we can have the sandwich

674
00:38:20 --> 00:38:26
pass that point X
n-sandwich times.

675
00:38:25 --> 00:38:31
As long as these are integers,
that's perfectly fine.

676
00:38:29 --> 00:38:35
Then ultimately, if they
have enough patience,

677
00:38:32 --> 00:38:38
they will meet at point X.

678
00:38:34 --> 00:38:40
And if you take these...
this new concept into account

679
00:38:37 --> 00:38:43
that you can wait a certain
number of passages through X,

680
00:38:41 --> 00:38:47
then the equation
that you see there--

681
00:38:44 --> 00:38:50
the relation
between a and R--

682
00:38:46 --> 00:38:52
changes only slightly.

683
00:38:47 --> 00:38:53
You now get that a equals R
times n of a minus f

684
00:38:55 --> 00:39:01
divided by ns
to the power two-thirds.

685
00:39:00 --> 00:39:06
And if you substitute
in here a one and a one,

686
00:39:03 --> 00:39:09
which is the case that they make
the catch right away,

687
00:39:06 --> 00:39:12
then you see indeed
you get R (one minus f)

688
00:39:09 --> 00:39:15
to the power two-thirds.

689
00:39:10 --> 00:39:16
So that's exactly
what you have there.

690
00:39:13 --> 00:39:19
Not all solutions that you try
will work.

691
00:39:17 --> 00:39:23
One solution that won't work

692
00:39:19 --> 00:39:25
is na equals one and ns equals
three has no solution.

693
00:39:25 --> 00:39:31
And I'll leave you with the
thought why that is the case.

694
00:39:29 --> 00:39:35
Has no solution.

695
00:39:30 --> 00:39:36

696
00:39:32 --> 00:39:38
In 1990, when I lectured 8.01
for the first time,

697
00:39:35 --> 00:39:41
I asked my friend and colleague
George Clark

698
00:39:38 --> 00:39:44
to write a program so that
I could show the class

699
00:39:41 --> 00:39:47
this toss of the sandwich
with Mary and Peter in orbit

700
00:39:45 --> 00:39:51
and the sandwich orbit
and everything

701
00:39:47 --> 00:39:53
and the catch, and he did.

702
00:39:49 --> 00:39:55
It was a wonderful program,

703
00:39:50 --> 00:39:56
but that program no longer works
because that's called progress.

704
00:39:54 --> 00:40:00
The computers have changed

705
00:39:55 --> 00:40:01
and so, my right hand,
Dave Pooley, offered

706
00:39:58 --> 00:40:04
to rewrite the program
so that it works on Athena,

707
00:40:00 --> 00:40:06
and we were going
to demonstrate it to you,

708
00:40:03 --> 00:40:09
and you can play
with it yourself.

709
00:40:05 --> 00:40:11
It is available on the homepage,

710
00:40:08 --> 00:40:14
so whatever Dave is going to
show you, you can do yourself.

711
00:40:13 --> 00:40:19
The input parameters that
we need for this program

712
00:40:18 --> 00:40:24
are the radius R, our f
and our n of a and our n of s.

713
00:40:23 --> 00:40:29
And the program
will do all the rest,

714
00:40:25 --> 00:40:31
so you can specify
how many times

715
00:40:27 --> 00:40:33
you want Mary to go
through point X,

716
00:40:29 --> 00:40:35
how many times you want the
sandwich to go through point X.

717
00:40:32 --> 00:40:38
The program will then
calculate for you

718
00:40:35 --> 00:40:41
the speed of the sandwich.

719
00:40:38 --> 00:40:44
It will also give you
vs minus va,

720
00:40:40 --> 00:40:46
which is really the speed with
which Peter has to throw it,

721
00:40:45 --> 00:40:51
but very cleverly,
the program works

722
00:40:48 --> 00:40:54
with a dimensionless parameter
which is this value.

723
00:40:52 --> 00:40:58
And this value, which is
vs divided by va minus one

724
00:41:00 --> 00:41:06
is a number that
is quite unique,

725
00:41:05 --> 00:41:11
because you get solutions
which turn out

726
00:41:08 --> 00:41:14
to be independent of capital G
and independent of capital M,

727
00:41:12 --> 00:41:18
and I'll give you an example.

728
00:41:14 --> 00:41:20
Suppose you will find
for this...

729
00:41:18 --> 00:41:24
for this dimensionless number,
suppose you find minus 0.0175,

730
00:41:26 --> 00:41:32
which is the solution
for n one...

731
00:41:28 --> 00:41:34
na is one and for ns is one.

732
00:41:30 --> 00:41:36
So you'll see that.

733
00:41:32 --> 00:41:38
The computer will generate
that number for us.

734
00:41:35 --> 00:41:41
If now we take our orbit
of 7,000 kilometers...

735
00:41:39 --> 00:41:45
we know what v of a is,
and so we can calculate now

736
00:41:44 --> 00:41:50
that vs minus va
is the va times that number.

737
00:41:48 --> 00:41:54
But we know what the va is,
it's 7.5,

738
00:41:51 --> 00:41:57
so you get minus 0.175
times 7.55 kilometers,

739
00:41:56 --> 00:42:02
and lo and behold,

740
00:41:58 --> 00:42:04
that is our minus 0.13
kilometers per second.

741
00:42:02 --> 00:42:08
Of course!

742
00:42:03 --> 00:42:09
It has to be that number,

743
00:42:05 --> 00:42:11
because that's
what we calculated

744
00:42:08 --> 00:42:14
for our case, n equals one.

745
00:42:09 --> 00:42:15
There it is.

746
00:42:11 --> 00:42:17

747
00:42:14 --> 00:42:20
And so this dimensionless number
is very transparent,

748
00:42:17 --> 00:42:23
and we will show you
some examples.

749
00:42:20 --> 00:42:26
This would be
the 300 miles per hour,

750
00:42:22 --> 00:42:28
which, of course,
is not very doable.

751
00:42:24 --> 00:42:30

752
00:42:26 --> 00:42:32
Dave, why don't you demonstrate
the program?

753
00:42:30 --> 00:42:36
And then you'll see what we can
do with that program.

754
00:42:35 --> 00:42:41
We can substitute in there
quite a few parameters

755
00:42:38 --> 00:42:44
that you will find
no doubt interesting.

756
00:42:42 --> 00:42:48
Give us an explanation, Dave,

757
00:42:44 --> 00:42:50
of what the students
can do with this.

758
00:42:46 --> 00:42:52
Oh, let me show them an overhead
here which would help you

759
00:42:50 --> 00:42:56
in following what Dave
will be telling you.

760
00:42:53 --> 00:42:59
You see there the f value?

761
00:42:55 --> 00:43:01
It's always 5% we took,

762
00:42:57 --> 00:43:03
and you're going to see here
the numbers for...

763
00:43:00 --> 00:43:06
the number of times that Mary
passes through point X

764
00:43:04 --> 00:43:10
and the number of times
that the sandwich

765
00:43:07 --> 00:43:13
will pass through point X.

766
00:43:09 --> 00:43:15
This is that very first case
that we worked on together.

767
00:43:13 --> 00:43:19
Here, you see that number
minus 0.017,

768
00:43:15 --> 00:43:21
and you see indeed
that it is a successful catch.

769
00:43:18 --> 00:43:24
So let's work at that first.

770
00:43:20 --> 00:43:26
David, explain how it works.

771
00:43:23 --> 00:43:29
DAVID:
Okay, well, you can see

772
00:43:24 --> 00:43:30
in the middle of the screen
is planet Earth,

773
00:43:27 --> 00:43:33
and these two triangles
represent the astronauts.

774
00:43:30 --> 00:43:36
The yellow one is Mary
and the red one is Peter,

775
00:43:32 --> 00:43:38
and he's holding the sandwich
right in the middle.

776
00:43:35 --> 00:43:41
They start off at a radius
of 22,000 kilometers

777
00:43:38 --> 00:43:44
from the center of the Earth.

778
00:43:40 --> 00:43:46
That's the default, but you
can change that if you'd like.

779
00:43:43 --> 00:43:49
And we set na and ns through
these pulldown menus,

780
00:43:46 --> 00:43:52
so we'll set them both
to one and one now.

781
00:43:49 --> 00:43:55
And we ask the program
to calculate the value for us

782
00:43:52 --> 00:43:58
of this dimensionless parameter,

783
00:43:54 --> 00:44:00
and it comes up with it,
and we want to use that value.

784
00:43:57 --> 00:44:03
And so we have everything set...

785
00:43:59 --> 00:44:05
LEWIN:
That's this number, right, Dave?

786
00:44:02 --> 00:44:08
This minus 0.0175, etc.

787
00:44:03 --> 00:44:09
DAVID:
Yep, it's right over here.

788
00:44:05 --> 00:44:11
So then we ask the program
to prepare the toss.

789
00:44:08 --> 00:44:14
We click this button down here,

790
00:44:10 --> 00:44:16
and when it's ready, the green
play button will become active.

791
00:44:13 --> 00:44:19
And when that happens,
we can click on that,

792
00:44:15 --> 00:44:21
and it'll play the toss for us.

793
00:44:17 --> 00:44:23
LEWIN:
Peter always throws at X, 12:00.

794
00:44:20 --> 00:44:26
DAVID:
Always at 12:00.

795
00:44:21 --> 00:44:27
LEWIN:
There goes the sandwich.

796
00:44:23 --> 00:44:29
You see the sandwich?

797
00:44:24 --> 00:44:30
Great sandwich.

798
00:44:25 --> 00:44:31
Big sandwich.

799
00:44:26 --> 00:44:32

800
00:44:28 --> 00:44:34
So notice that the sandwich
makes it around

801
00:44:31 --> 00:44:37
exactly at the same time--
(exclaims )

802
00:44:35 --> 00:44:41
for Mary to be happy.

803
00:44:36 --> 00:44:42
Now there's no reason why
we shouldn't try one, too.

804
00:44:40 --> 00:44:46
So that means Mary
reaches point X,

805
00:44:43 --> 00:44:49
but the sandwich went
twice around the Earth.

806
00:44:46 --> 00:44:52
There is no problem with
this solution in principle,

807
00:44:50 --> 00:44:56
but you have to be quite
far away from the Earth.

808
00:44:54 --> 00:45:00
If you're too close to the Earth
like Dave's orbit,

809
00:44:58 --> 00:45:04
which has a radius
of only 22,000 kilometers,

810
00:45:01 --> 00:45:07
something very catastrophic
will happen.

811
00:45:04 --> 00:45:10
DAVID: Yeah, okay, so now we
want to set ns to two,

812
00:45:07 --> 00:45:13
and we do that
with the pulldown menu.

813
00:45:09 --> 00:45:15
And we ask it to prepare
the toss again,

814
00:45:11 --> 00:45:17
and it goes through
its numerical calculations

815
00:45:13 --> 00:45:19
of the orbit.

816
00:45:14 --> 00:45:20
And when it's ready,
it'll let us know.

817
00:45:18 --> 00:45:24
And now we can watch the toss.

818
00:45:21 --> 00:45:27

819
00:45:24 --> 00:45:30
LEWIN:
So there goes the sandwich.

820
00:45:27 --> 00:45:33
It wants to go around the Earth
twice, but it hits the Earth.

821
00:45:31 --> 00:45:37
That's too bad.

822
00:45:32 --> 00:45:38
If you make this dimensionless
parameter minus one,

823
00:45:41 --> 00:45:47
then v of s is zero.

824
00:45:42 --> 00:45:48
And what does it mean
that v of s is zero.

825
00:45:45 --> 00:45:51
That the sandwich stands still,
has no speed in orbit anymore.

826
00:45:51 --> 00:45:57
And so what happens
with the ellipse,

827
00:45:54 --> 00:46:00
that becomes radial infall.

828
00:45:56 --> 00:46:02
Dave?

829
00:45:57 --> 00:46:03
DAVE: Okay, so if we want
to use our own value

830
00:45:59 --> 00:46:05
for this dimensionless
parameter,

831
00:46:01 --> 00:46:07
then we can go
to this box right here

832
00:46:03 --> 00:46:09
and put in whatever we want,
so we'll put in minus one.

833
00:46:06 --> 00:46:12
And we make sure
that the program

834
00:46:07 --> 00:46:13
is going to use our value
instead of the calculated value.

835
00:46:10 --> 00:46:16
In this case, these numbers
don't matter-- na and ns.

836
00:46:12 --> 00:46:18
They're irrelevant,

837
00:46:13 --> 00:46:19
because the program is going
to use our value.

838
00:46:16 --> 00:46:22
We ask it to prepare...

839
00:46:17 --> 00:46:23
LEWIN: The minus one overrides
everything else now.

840
00:46:19 --> 00:46:25
DAVID:
Yes.

841
00:46:20 --> 00:46:26
It's going through
its calculations,

842
00:46:22 --> 00:46:28
and now we can see what happens.

843
00:46:25 --> 00:46:31

844
00:46:28 --> 00:46:34
LEWIN:
12:00, there goes the sandwich.

845
00:46:31 --> 00:46:37
Now, Peter decides at one point

846
00:46:33 --> 00:46:39
that instead of throwing
the sandwich backwards,

847
00:46:37 --> 00:46:43
he can also throw
the sandwich forward,

848
00:46:40 --> 00:46:46
because, look, we have here
the red ellipse.

849
00:46:43 --> 00:46:49
There's no reason why Mary
couldn't go twice through X--

850
00:46:52 --> 00:46:58
one... twice.

851
00:46:54 --> 00:47:00
And then the sandwich would
make a larger ellipse

852
00:46:58 --> 00:47:04
and meet here when Mary
has gone around twice.

853
00:47:01 --> 00:47:07
Then, of course, the sandwich
has to be thrown forward.

854
00:47:06 --> 00:47:12
And so Peter makes a calculation

855
00:47:08 --> 00:47:14
for what we call
the 2/1 situation.

856
00:47:11 --> 00:47:17
Mary goes twice through X; 

857
00:47:13 --> 00:47:19
the sandwich goes
once through X.

858
00:47:16 --> 00:47:22
But Peter made a mistake.

859
00:47:17 --> 00:47:23
Peter got nervous, and he puts
in the wrong parameters,

860
00:47:21 --> 00:47:27
and you will see what happens.

861
00:47:22 --> 00:47:28
Dave will first show you
the right parameters.

862
00:47:25 --> 00:47:31
DAVID: Okay, so if we would
have asked the program

863
00:47:27 --> 00:47:33
to calculate it
for the case of 2/1,

864
00:47:29 --> 00:47:35
it would have come up
with a value

865
00:47:31 --> 00:47:37
for this dimensionless parameter

866
00:47:33 --> 00:47:39
of .1659
or something thereabouts.

867
00:47:35 --> 00:47:41
But, you know, Peter
made his miscalculation,

868
00:47:39 --> 00:47:45
and he wants to use .164,

869
00:47:40 --> 00:47:46
so this is what we'll put
into the program,

870
00:47:44 --> 00:47:50
and we'll prepare the toss

871
00:47:46 --> 00:47:52
and see what happens
with this value.

872
00:47:48 --> 00:47:54

873
00:47:53 --> 00:47:59
Okay, now it's ready.

874
00:47:55 --> 00:48:01
LEWIN:
Poor Mary must be hungry by now.

875
00:47:57 --> 00:48:03
There we go.

876
00:47:58 --> 00:48:04
Now we go forward.

877
00:47:59 --> 00:48:05
You can see that.

878
00:48:01 --> 00:48:07
You see, it goes forward.

879
00:48:02 --> 00:48:08
It goes a very large ellipse,
and Mary will go around twice.

880
00:48:06 --> 00:48:12
When Mary is here, see,
the sandwich is only halfway.

881
00:48:13 --> 00:48:19
And if Peter had
only done it right,

882
00:48:16 --> 00:48:22
Mary's troubles
would now soon be over.

883
00:48:20 --> 00:48:26
But Peter made this small
mistake, and...

884
00:48:24 --> 00:48:30
(students laugh )

885
00:48:26 --> 00:48:32
And Mary cannot catch it.

886
00:48:30 --> 00:48:36
If you make this dimensionless
parameter plus .42,

887
00:48:33 --> 00:48:39
then it's very easy
to convince yourself

888
00:48:36 --> 00:48:42
that the sandwich--
must be a plus--

889
00:48:39 --> 00:48:45
will have the escape velocity
from the orbit.

890
00:48:42 --> 00:48:48
Maybe Peter got angry
at one point at Mary--

891
00:48:45 --> 00:48:51
you never know
about these situations--

892
00:48:47 --> 00:48:53
and he threw it very fast,

893
00:48:48 --> 00:48:54
and Dave will show you
what happens then.

894
00:48:51 --> 00:48:57
DAVID:
Okay.

895
00:48:52 --> 00:48:58
LEWIN:
And it goes to infinity,

896
00:48:55 --> 00:49:01
and it won't be fresh anymore
when it gets there.

897
00:49:00 --> 00:49:06
Okay, see you Friday.

898
00:49:02 --> 00:49:08

899
00:49:08 --> 00:49:14.000