00:02
In orbital mechanics, orbital motion, there are two constants that you can be guaranteed of if the force is gravity, and in a more general case, if it is a central force.
00:17
But the two constants are good pals, energy, total energy, that is, and angular momentum.
00:27
Now, energy is, of course, a scalar, and it is the sum of the kinetic energy of the object plus its gravitational potential energy.
00:40
Angular momentum is a little bit trickier because it is a vector.
00:50
I'll call that little l for the time being.
00:53
And that vector is the cross -product between r, the position of the...
01:02
Object that's orbiting in this case, a central body that's big and massive, the cross product of r times v times the mass, or r cross momentum if we want to get momentum into this.
01:18
And in general, those two things are not perpendicular.
01:27
So especially if the object is an elliptical orbit, if it's a circular orbit, we're okay.
01:34
But in general, these are not perpendicular for other types of orbits, including elliptical, parabolic, and hyperbolic types of orbits.
01:50
In any event, there are two special places, the extreme values of the position along the semi -major axis of an elliptical orbit.
02:01
And that's what we are going to consider here.
02:05
In elliptical orbits, r is perpendicular to be at min and max positions.
02:17
In our solar system, they are called perheelian and appelian, but we'll just call them the min and max positions.
02:28
So far away from the point of gravity, gravitational mass.
02:36
That it's interacting with is armax and arm in is closest approach.
02:44
Ok, so our goal here is to work solely with the minimum and maximum positions.
02:53
And at those two positions, we can now turn the angular momentum into a simple algebraic expression.
03:04
The energy is going to look like one -half mv squared minus gmm over r for the gravitational potential energy.
03:19
And we can see that we have two equations in two unknowns, and what we would like to do is try to work with the positions.
03:33
So the unknowns are r and v, and we are going to try to eliminate v and see what we get.
03:53
And it turns out we will wind up getting an equation that we can solve for both positions.
04:08
Since we used a condition on the angular momentum that is true at both of those positions, we should wind up with an equation valid at both positions.
04:19
So let's go ahead and do that, eliminate v.
04:23
So we'll simply take the angular momentum equation and do some algebra on it to get v.
04:33
V is equal to l over mr.
04:40
And we can put that back into equation one, sorry, equation two.
04:52
And we get something that looks a little unwieldy, but not impossible to deal with.
05:09
Now i'm going to multiply it through by r squared and i'm manipulated a little bit.
05:18
Multiply by r squared, algebra, and we'll get it in the form of a quadratic, which will have two solutions, and that is exactly what we need in order to solve for the r -men and the r -mex.
05:38
And when we do that, we get r squared plus, let's see, g .m.
05:48
Little m.
05:50
R over energy minus l squared over 2m .e is equal to zero.
06:09
Now, i'm going to do something a little bit unusual.
06:13
What i am going to do is replace the absolute value of e for negative e.
06:24
So this is understanding that an ellipse is a bound orbit, and that means that the energy is negative.
06:37
The total energy is negative.
06:40
That is the gravitational force is strong enough to keep, things in orbit without having the object orbiting sail back into space with lots and lots of kinetic energy.
06:58
And by doing this, i think the results of the quadratic solution will be a little bit easier to understand.
07:09
So let's go ahead and make this substitution.
07:21
And why this makes things a little bit easier to understand is all the coefficients in this equation are now positive.
07:34
So this is a quadratic equation with all positive coefficients.
07:42
Yeah, so we don't have to worry about their meaning or who's bigger than whom, et cetera.
07:53
And we'll call the first term beta.
07:58
Let me show that a little bit easier.
08:01
And i want to avoid a, b, and c, because we've got some a's for the semi -major axis and b for the semi -minor axis.
08:10
We don't want to get goofed up there.
08:15
So beta is, yeah, i said all the coefficients were positive, but i meant all the things that are symbols.
08:25
In the coefficient are positive.
08:29
Beta is actually negative, and this will call gamma.
08:36
Okay, just so we don't have to keep writing it down, and you'll see why that can be helpful.
08:42
So now we're going to work out solution.
08:46
There'll be two roots, and as i alluded to, one will be a minimum and one will be a maximum, like usual.
08:55
So r is minus beta, so that's the absolute value of beta plus or minus the square root of beta squared minus 4 times gamma over 2.
09:22
So again, we can interpret all the results offhand.
09:27
We don't have to worry about plus minus signs...