Laws of Newton -
2nd-order Curves
We talk here about second-order curves because they describe the trajectories that objects move along in a central gravitational field.
These curves were discussed in the UNIZOR.COM course Math+ 4 All in the part Geometry, where you can find lectures on ellipse, hyperbola and parabola, their defining characteristics and equations in Cartesian and polar coordinates.
Here is a quick recap of this material.
Ellipse
Ellipse is a locus of points on a plane that satisfy the following condition.
The sum of distances from each such point P to two given points F1 and F2 (called foci) is equal to a given positive real number called its length of major axis (2a on a graph below), that is supposed to be greater than the distance between its two foci (2c on a graph below).
The equation of an ellipse in Cartesian coordinates (x,y) with X-axis coinciding with the line between the foci and origin of coordinates is a midpoint between foci is
x²/a² + y²/b² = 1
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 the ratio e=c/a is called eccentricity of an ellipse and it's always less than 1.
Hyperbola
Hyperbola is a locus of points on a plane that satisfy the following condition.
The difference of distances from each such point P from two given points F1 and F2 (called foci) is equal by absolute value to a given positive real number called its length of major axis (2a on a graph below), that is supposed to be less than the distance between its two foci (2c on a graph below).
The value 2b, where b²=c²−a², is called the length of minor axis.
The equation of a hyperbola in Cartesian coordinates (x,y) with X-axis coinciding with the line between the foci and origin of coordinates is a midpoint between foci is
x²/a² − y²/b² = 1
The equation in polar coordinates (r,θ) with an origin at one of the hyperbola' foci and a base axis coinciding with the line between the foci is
r= a·(e²−1)/[1−e·cos(θ)]
where the ratio e=c/a is called eccentricity of a hyperbola and it's always greater than 1.
Parabola
Parabola is a locus of points on a plane that satisfy the following condition.
Each such point P is equidistant from a given point F (called focus) and a given straight line d (called directrix) with the distance between a focus and a directrix being a given positive number (2c on a graph below).
y² = 4c·x
The equation in polar coordinates (r,θ) with an origin at parabola' focus and a base axis coinciding with the perpendicular line from a focus to a directrix is
r= 2c/[1−cos(θ)]
Generally speaking, a second-order curve is defined in Cartesian coordinates as a locus of (x,y) points on a plane that satisfy a general equation of a second order for two variables
Ax²+Bxy+Cy²+Dx+Ey+F=0
We can show that any second-order curve belongs to one of the three types presented above, it's either ellipse or hyperbola, or parabola.
If coefficient B at xy is zero, the expression above can be easily converted into one of the three canonical equations by shifting and stretching coordinates (which shifts and stretches the graph without changing its type, an ellipse will remain an ellipse etc.) and/or exchanging places of x and y (which changes the orientation of a graph from horizontal to vertical).
If coefficient B is not zero, we can turn the system of coordinates in such a way that the equation of the same curve in the new system will not contain a non-zero coefficient at xy.
Details of the transformation of a general equation into one of three canonical forms are at the end of these notes.
The main purpose of the above recap of the properties of second-order curves is to analyze the motion of a point-mass object in the central gravitational field of another point-mass object fixed at the origin of an inertial reference frame.
More precisely, we will prove in the next lecture that a trajectory of an object in a central gravitational field is some second-order curve (ellipse, hyperbola or parabola) depending on its position and velocity relative to the center of gravitational field at initial moment of time t=0.
To prove it, we will, primarily, rely on representation of second-order curves in polar coordinates.
The reason for this is that all polar coordinate representations of three types of curves (ellipse, hyperbola or parabola) are very similar and differ only in certain parameters.
If we derive the equation of a trajectory in polar coordinates as
r(θ) = A/[B+C·cos(θ)],
where A, B and C depend only on position and velocity of our object relative to a center of the gravitational field at the beginning of motion, we can say that a trajectory would be either ellipse or hyperbola, or parabola.
Moreover, analyzing the values of position and velocity of our object relative to a center of the gravitational field at the beginning of motion, we will be able to predict the type of trajectory it will take.
Transformation of General Second-Order Equation into One of Three Canonical Forms
The general form of a second-order curve on an XY-plane as an equation in Cartesian coordinates is
A·x²+B·x·y+C·y²+D·x+E·y+F=0
It's called 'second-order' because X- and Y-coordinates participate in this equation as polynomial with combined exponent 2.
Our task is to analyze the shape of a curve defined by such an equation depending on coefficients A, B. C etc.
Let's start with a simpler case of B=0.
Then our equation looks like
A·x²+C·y²+D·x+E·y+F=0
In this form we will analyze the shape of a curve in two different cases:
1. Either A or C is zero.
2. Both A and C are not equal to zero.
Both of them cannot be equal to zero because then the equation would not be of a second order.
1. CASE Either A or C is zero
We will consider two separate subcases.
Subcase A≠0 but C=0
In this case we can easily transform the original equation into a form
A·(x+½D/A)²+E·y=K
where
K = D²/(4A) − F
Let's shift our coordinate system parallel to itself moving the origin to point (½D/A,0).
This transformation will not change the shape of a curve, but will allow to express it in a simple form in new coordinates
A·x² + E·y = K
or
E·y = −A·x² + K
which is a parabola for any E≠0.
For example, for E=1, A=−1, K=−4 this parabola looks like this:
Subcase A=0 but C≠0
In this case we can easily transform the original equation into a form
C·(y+½E/C)²+D·x=K
where
K = E²/(4C) − F
Let's shift our coordinate system parallel to itself moving the origin to point (0,½E/C).
This transformation will not change the shape of a curve, but will allow to express it in a simple form in new coordinates
C·y²+D·x=K
or
D·x= −C·y² + K
which is a parabola for any D≠0.
For example, for D=1, C=−1, K=−4 this parabola looks like this:
2. CASE A≠0 and C≠0
In this case we can easily transform the original equation into a form
A·(x+½D/A)²+C·(y+½E/C)²=K
where
K = D²/(4A) + E²/(4C) − F
Let's shift our coordinate system parallel to itself moving the origin to point (½D/A,½E/C).
This transformation will not change the shape of a curve, but will allow to express it in a simple form in new coordinates
A·x²+C·y²=K
Subcase A and C are
both positive or both negative
If K is of the same sign as A and C, it's an ellipse.
For example, if A=4, C=9 and K=36, this ellipse looks like this
If K=0, the only point (0,0) fits the equation. You can consider it as a degenerate ellipse with zero dimension in any direction.
Physical meaning of this is that this is a trajectory of an object in central gravitational field with its initial position at its center and initial velocity zero.
So, it's stuck at the center where the source of gravity is located and will not escape without any initial velocity, which is meaningless for our analysis.
Finally, if K is of the opposite sign to A and C, there are no points that satisfy the equation.
In a physical sense it means that there is no object in the gravitational field.
Subcase A and C are
of opposite signs
The curve will be a hyperbola
(if signs of A and K are the same)
or
(if signs of C and K are the same)
From the physical standpoint the trajectory like that will be of an object moving fast enough to overcome the force of gravitation and, after its trajectory has been curved by this force, still manage to fly away from the source of gravitation force to infinity.
If K=0 we will have two straight lines defined by equation
A·x² = C·y²
which can be simplified to
y=±√(A/C)x
with the graph looking like this
All the above cases assumed that in the original equation of a second-order curve
A·x²+B·x·y+C·y²+D·x+E·y+F=0
the coefficient B=0.
What if it's not?
We will show that with a proper turning of our system of Cartesian coordinates (which would not change the shape of a curve, only its representation as a second-order equation in Cartesian coordinates) we can change the equation into a form with B=0.
That would reduce a general task of analysis of the shape of any second-order curve to a case analyzed above, when B=0.
Assume, we can turn our Cartesian system of coordinates by angle β.
Recall that the new system will have coordinates u and v that are related to old coordinates x and y in the original system as
u = x·cos(β) + y·sin(β)
v = −x·sin(β) + y·cos(β)
or in vector form
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Old coordinates x and y can be expressed in terms of new ones u and v by turning the new system of coordinates by angle −β:
x = u·cos(β) − v·sin(β)
y = u·sin(β) + v·cos(β)
or in vector form
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Now we can substitute new coordinates u and v into original second-order equation for a curve getting another second order equation for the same curve in terms of new coordinates.
Can we find such an angle β that all mixed terms that contain a product u·v will cancel each other?
If yes, that would prove that in some coordinate system the equation of a second-order curve contains no terms with product of different coordinates, like x·y.
In terms of u and v the new equation will contain terms with u², u·v, v², u, v and a constant.
Let's collect only terms with a product of different coordinates u·v that we want to nullify.
These are:
A·x² → −2·A·u·v·cos(β)·sin(β)
B·x·y → B·u·v·(cos²(β)−sin²(β))
C·y² → 2·C·u·v·sin(β)·cos(β)
We would like the sum of them to be zero.
After obvious usage of the formulas for sin and cos of double angle, it means that we have to find such an angle β that
A·sin(2β)=B·cos(2β)+C·sin(2β)
The above trigonometric equation can be easily solved by dividing by cos(2β) getting
tan(2β) = B/(A−C)
β = ½arctan[B/(A−C)]
It works only for A≠C. In case A=C the equation for β looks like
B·cos(2β) = 0
from which follows
β=π/4
Therefore, by rotating the system of Cartesian coordinates we can find the one where the second-order equation of our curve will have no terms with a product of different coordinates x·y.
After that we can use the technique presented above to shift the origin of our system of coordinates to simplify this equation even further to a form
A·x² + E·y = K (parabola open along Y-axis) or
C·y² + D·x = K (parabola open along X-axis) or
A·x² + C·y² = K (ellipse if all multipliers are positive or hyperbola if A and C are of opposite signs)