Laws of Newton -
More on Ellipse Characteristics
Let's get to more details about properties of an ellipse. It's important for our future discussion of Kepler's Laws described in the next few lectures of this part of a course.
Axes in Polar Coordinates
The equation in polar coordinates (r,θ) with an origin at one of the ellipse' foci and a base axis coinciding with the line between the foci is
r = a·(1−e²)/[1−e·cos(θ)]
where a is half of a major axis,
c is half of a distance between foci,
the ratio e=c/a is called eccentricity of an ellipse and it's always less than 1.
For an ellipse described above, the distance from a focus at the origin of a polar system to a further end of an ellipse along X-axis should be equal to half of the major axis plus half of a focal distance, that is a+c.
Indeed, if we substitute θ=0 into an equation of an ellipse in polar coordinates, we obtain
r(0) = a·(1−e²)/[1−e·cos(0)] =
= a·(1−e²)/[1−e] =
= a·(1+e) = a + a·c/a = a + c
To reach the opposite end of an ellipse (the shortest distance from an origin) we have assign θ=π, which should result in r=a−c.
Let's check it by substituting θ=π in our equation of an ellipse.
r(π) = a·(1−e²)/[1−e·cos(π)] =
= a·(1−e²)/[1+e] =
= a·(1−e) = a − a·c/a = a − c
Since
r(0) = a + c and
r(π) = a − c
we can derive the half of the major axis
a = (1/2)·[r(0) + r(π)]
and the half of the focal distance
c = (1/2)·[r(0) − r(π)]
As we know, the half of minor axis b equals
b = √a²−c²
Short calculations show that in terms of r(θ) it will be
b = √r(0)·r(π)
Ellipse Area
The equation of an ellipse in Cartesian coordinates (x,y) with X-axis coinciding with the line between the foci and the origin of coordinates being at a midpoint between foci is
x²/a² + y²/b² = 1
Here a is a half of a major axis and b is a half of a minor axis of an ellipse.
If we consider only top half of an ellipse, this equation can be resolved for y to represent it as a function y(x)
y²/b² = 1 − x²/a²
y² = b²·(1−x²/a²)
y² = (b²/a²)·(a²−x²)
y = √(b²/a²)·(a²−x²)
y = (b/a)·√(a²−x²)
Let's compare this function with a function describing the top half of a circle of radius a and equation
x² + y² = a²
from which follows
y = √(a²−x²)
Graphically the functions describing an ellipse and a circle look like this
As you see, for any abscissa x the ordinate of an ellipse is smaller than the ordinate of a circle by the same factor b/a.If you take a look at any vertical bar from the X-axis up, it's height to an intersection with an ellipse is smaller than to an intersection with a circle by a factor b/a.
That means, the area of a portion of that bar below the ellipse is smaller that the area of a bar below a circle by the same factor b/a.
The area of a circle and the area of an ellipse can be comprised from an infinite number of such bars of infinitesimal width (integration!), which means that the total area of an ellipse is smaller than the one of a circle by the same factor b/a.
Since the area of a circle of a radius a is πa², the area of an ellipse is
Aellipse = πa²·(b/a) = πab
Area and a Period
Consider an object moving along an elliptical trajectory.
Let's introduce a system of polar coordinates with an origin at one of the ellipse foci and base axis coinciding with a line between foci.
Let the semi-axes of an elliptical trajectory be a and b.
Let r be a position vector of a moving object - a vector from the ellipse' focus chosen as an origin of polar coordinates to an object's position at any time.
Let θ be an angle between the base axis and vector r. This angle, obviously changes with time as an object moves along its trajectory.
An object's movement in this system of coordinates along its elliptical trajectory is described in polar coordinates as r(θ) with angle θ being, in turn, a function of time t.
Assume farther that an object moves on an elliptical trajectory with certain periodicity T, that is, it returns to the same position at each interval of time T
θ(t+T)=θ(t) + 2π.
Consider a function A(θ) equal to an area of an ellipse swept by position vector r(θ) from its position at θ=0 to a position at angle θ.
Since an angle θ, in turn, depends on time, area A(θ) can be considered as a function of time A(t) as well.
By the time from t=0 to t=T an angle θ will make a full turn by 2π and vector r will swipe an entire area of an ellipse.
Therefore,
A(T) = A(θ(T)) = πab
If we know the function A(t), we can determine the period of rotation T based on geometrical characteristics of a trajectory.
Actually, when we will discuss the Kepler's Laws of planetary movements, we will prove that
A(t) = k·t
where k - some constant of motion.
That allows to calculate the period T:
A(T) = k·T = πab
Therefore,
T = πab/k
No comments:
Post a Comment