00:01
Here we're going to work out the orbit of a satellite that gets ejected from the space shuttle.
00:08
And what we know is that the shuttle was in a circular orbit when the small satellite was ejected.
00:16
And it was ejected at the position shown my little lozenge there is the shuttle.
00:24
And the satellite is too small to see.
00:27
But it was ejected towards the center of the earth at 100 meters per second, and of course was also being carried along in the tangential paths of the circle.
00:41
So our goal is to figure out some parameters about the orbit of the ejected satellite.
00:48
And what we know for sure is that its angular momentum and its energy are both concerned.
00:58
And that is going to allow us to unravel some things.
01:05
And instead of finding its full angular momentum, i will find what we'll call little h, which is big h divided by the small mass of the satellite.
01:21
So we'll try to do things without the mass of the satellite because we do not know what that is.
01:39
Okay, so little h is the angular momentum divided by the mass.
01:46
And what we know about that angular momentum is it has a magnitude of r cross velocity, but here we want to use the velocity tangent to the circle because that is perpendicular to the position of the satellite.
02:09
So those two quantities, let me kind of show what the tangential looks like, and that is going to come from the circular motion of the shuttle, or from the circular orbit of the shuttle.
02:36
So the way we would find that a tangential speed is to set mv squared over r0 equal to the force of gravity on the shuttle or the satellite.
02:56
It really doesn't matter what you want to think about as traveling in that circle.
03:01
And that is big g times the mass of the earth times the mass of the shuttle or the satellite.
03:11
Whichever you want to think about over r0 squared.
03:18
And so we can solve for the tangential speed, and thankfully mass cancels out from a lot of these calculations.
03:33
And it turns out that this is equal to square root of g times the mass of the earth over the initial radius, with the center of the earth being the position to find the angular momentum about.
03:58
Okay, so we know that that is going to be conserved.
04:02
We also know that energy is conserved.
04:13
And here i will write down the energy of the satellite.
04:17
And i'm going to call it little e with the idea that the little mass gets canceled.
04:24
And that is going to be equal to the kinetic energy.
04:28
Which is 1 .5 vt squared plus the relative velocity squared.
04:41
And the reason why i can separate it out like that is the tangential velocity is in the x direction, the relative velocity shooting towards the center of the earth is in the y direction, and those are perpendicular to each other.
05:00
So imagine taking some of the components squared to get your full v squared.
05:11
And then we have the u -gravity divided by mass, which is big g times the mass of the earth over r0.
05:28
Okay, and then we can use the tangential speed.
05:33
That's one -half times g -m -e over r -0.
05:43
Me is the mass of the earth, by the way, i should indicate that.
05:50
And r -0 is the radius of the earth plus the 200 -kilometer amplitude of the spacecraft.
05:59
And then we have plus 1 -half times 100 squared.
06:03
We'll just put that number in for the time being.
06:07
But notice that i have two terms that look very similar, that i can combine into something smaller for the energy.
06:29
So our energy is equal to minus one -half.
06:43
Big g mass of the earth over r0, plus that little bitty term for the launched speed.
06:58
And that is energy per unit mass.
07:00
So i have divided out the mass.
07:03
Okay.
07:04
Now, what we're trying to do, besides consular momentum and energy, is the first thing to get a grasp of is to find what are called the apogee and the parogy.
07:22
And those are the radii minimum from the earth and maximum.
07:30
And those are special points in the orbit where the speed is entirely, that the velocity vector is entirely tangent to the path.
07:43
Anywhere else along the ellipse and the velocity is not exactly how to say it.
07:57
Perpendicular to the radius, i should say.
07:59
It's always tangent to the path, but it's not perpendicular to the radius out from the center of the earth.
08:09
So these are called turning points, and it is kind of where the satellite is going to kind of turn around and go in a different direction.
08:28
Okay, and so we can write that energy at the turning points.
08:34
In the following way.
08:39
Okay, so the energy is going to be entirely equal to h squared over to r squared, where r is one of those two turning points minus.
09:00
So this is the kinetic energy term, and then we have the gravitational potential energy term.
09:09
And why the kinetic energy looks so simple is that there is no r dot at these points.
09:27
And i can kind of show what i mean by an ellipse over here.
09:30
Let's just draw an ellipse with a center here.
09:35
And the object's going around.
09:38
The apache is the closest point, r minimum, and the velocity at that point is perpendicular to the radius, and it is also perpendicular to the radius.
09:54
At the r max.
09:58
But somewhere in between, the velocity makes some sort of angle to the radius that's not necessarily 90 degrees.
10:13
That drawing may not look very good.
10:15
So, what's call this position, a, b, c, and d.
10:23
So at positions b and c, v is not perpendicular to r, and there is an r dot.
10:50
And that makes things a little bit more complicated with the kinetic energy.
10:57
Anyway, we can now equate the energy to the initial energy.
11:04
So that's the nice thing about conservation, is that once you have your quantities figured out at one point, they become the same at a later point.
11:17
So we'll call this initial energy ei, and we'll call the final energy e final.
11:28
We'll equate those, and we will replace the angular momentum into there.
11:34
And what we get is 1 1โ2 times 100 squared minus 1 half gme over r0 equals.
12:08
Let's put in our h.
12:13
Let me see, h squared.
12:15
Turns out to be gmer.
12:22
R.
12:41
Let me see if i h somewhere.
12:48
Well, that should be r0.
12:55
The initial radius of the satellite from the center of the earth.
13:14
And i believe, let's see if we put in h into that formula.
13:24
Let's make sure i did h up here.
13:30
Let me work that out a little bit more detail.
13:42
So we're taking the tangential velocity and multiplying it by the radius, yes, and that gives us g -m -e -r -0.
13:57
Okay, so this may not be obvious, but what we have essentially is a quadratic equation in the turning points...