Physics+ - Motion in Polar Coordinates: UNIZOR.COM - Physics+ 4 App - La...

Notes to a video lecture on UNIZOR.COM

Motion in Polar Coordinates

The subject of this lecture is to describe characteristics of movement (position, velocity and acceleration) in polar coordinates.
This approach will be useful in analyzing the movement of objects in a central force field (like in a gravitational or an electrostatic fields).
Using of polar coordinates to analyze the movement in a central field seems to result in simpler derivation of important physical results, like the Kepler's Laws.

Consider a model of a space with a fixed position of a point-mass M - the source of a central gravitational field.
Assume, a test object of mass m moves in this gravitational field. As was proven in the lecture Planet Orbits of this course, the trajectory of this test object lies within some plane of motion, and the source of gravitation also lies in this same plane.

Let's associate the origin of some Cartesian coordinate system with the fixed position of the source of gravitation O and XY-plane coinciding with the plane of motion of our test object.
So, the coordinates of the source of gravitation are always {0,0,0} and coordinates of a test object always have Z-coordinate equal to zero, so in most cases we will not even specify it, considering we deal with a two-dimensional XY-space of a plane of motion.

Now we also introduce a polar system of coordinates {r,θ} in a plane of motion with the same origin at the source of gravitation O and the base axis coinciding with the OX-axis.

Consider a position vector r from the source of gravitation O to point P where a test object is located.
If a test object is at Cartesian coordinates {x,y}, its position vector can be represented as
r = x·i + y·j
where i and j are unit vectors along X- and Y-axes correspondingly, forming an orthogonal basis on XY-plane.

Assume, our test object is at polar coordinates {r,θ} related to its Cartesian coordinates as
x = r·cos(θ)
y = r·sin(θ)
To express the same position vector r in some orthogonal basis in the polar system of coordinates, let's introduce two unit vectors:
ê_r along a line from the origin O to position of a test object P, that is collinear to vector r;
ê_θ along a line perpendicular to ê_r.
This orthogonal basis is not fixed in space like unit vectors i and j in the Cartesian system of coordinates, but is moving with a test object.

In this new orthogonal basis the same position vector r can be represented as

r = r·ê_r

where r is the magnitude of vector r - its first coordinate in the polar system {r,θ}, which, in turn, can be expressed in Cartesian system as
r=√x²+y²
As you see, polar representation of a position vector as a vector in some orthogonal basis is simpler than its Cartesian representation - a very important factor for analysis of movement in a central gravitational field.

All the coordinates mentioned above, Cartesian {x,y} and polar {r,θ} are functions of time, as our test object is moving in the gravitational field.

As we know, differentiating a position vector r by time gives the velocity vector v=r', where we use a single apostrophe to indicate a derivative by time.
Vector representation of velocity in Cartesian orthogonal basis, as we know, is
v = x'·i + y'·j

We would like to represent a vector of velocity in polar coordinates as well using the orthogonal basic {ê_r, ê_θ}.
For this, first of all, we express basic {ê_r, ê_θ} in terms of basic {i, j} using the fact that vector ê_r has a unit length and positioned at angle θ to OX-axis, while vector ê_θ has a unit length and positioned at angle θ+π/2 to OX-axis:
ê_r = cos(θ)·i + sin(θ)·j
ê_θ = −sin(θ)·i + cos(θ)·j
where we used trigonometric identities
cos(θ+π/2) = −sin(θ) and
sin(θ+π/2) = cos(θ)

From the above follows an important property of this orthogonal basis
dê_r /dθ = ê_θ
dê_θ /dθ = −ê_r

Furthermore, the time derivatives of these unit vectors are
ê_r' = (dê_r /dθ)·θ' = θ'·ê_θ
ê_θ' = (dê_θ /dθ)·θ' = −θ'·ê_r

Since r = r·ê_r
v = r' = dr/dt =
= d(r·ê_r)/dt =
= r'·ê_r+r·ê_r' =
= r'·ê_r+r·θ'·ê_θ

v = r'·ê_r+r·θ'·ê_θ

Let's extend these calculations and get an acceleration vector represented in the same basis of {ê_r , ê_θ}.

a = v' = dv/dt =
= r"·ê_r + r'·ê_r' +
+ r'·θ'·ê_θ + r·θ"·ê_θ + r·θ'·ê_θ' =
= r"·ê_r + r'·θ'·ê_θ +
+ r'·θ'·ê_θ + r·θ"·ê_θ − r·θ'²·ê_r =
= (r"−r·θ'²)·ê_r +
+ (r·θ"+2r'·θ')·ê_θ

As we know, according to Newton's Second Law, vector of acceleration is collinear to a vector of force.
According to Newton's Universal Law of Gravitation, vector of gravitational force is collinear to a position vector.
Therefore, vectors a and r are collinear.
Consequently, a and ê_r are collinear, from which follows that coefficient at ê_θ must be equal to zero:
r·θ"+2r'·θ' = 0 and
a = (r"−r·θ'²)·ê_r

We can come up with the same result using the Law of Conservation of Angular Momentum.
Recall that the vector of Angular Momentum of an object in a central force field is constant because a central force has no rotational action, no torque.
It is directed along the Z-axis perpendicularly to a plane of motion.
This vector of Angular Momentum is defined as L=m·r⨯v.
We can express the constant magnitude of this vector in terms of {r,θ} as follows.
r = {r·cos(θ), r·sin(θ), 0}.
v = {r'·cos(θ)−r·sin(θ)·θ', r'·sin(θ)+r·cos(θ)·θ', 0}.

According to the rules of vector product in three-dimensional Cartesian coordinates, their vector product L/m = r⨯v is a vector with X- and Y-coordinates equal to zero and its Z-coordinate equal to
r·cos(θ)·[r'·sin(θ)+r·cos(θ)·θ']−
−r·sin(θ)·[r'·cos(θ)−r·sin(θ)·θ']
The above expression evaluates to
r²·[cos²(θ)+sin²(θ)]·θ' = r²·θ'
Therefore,
|L|/m = L/m = r²·θ'
and is a constant of motion in a central field.

Since L/m is a constant of motion, its derivative by time is zero.
Therefore,
d(L/m)/dt = r²·θ" + 2r·r'·θ' = 0
Canceling r as a non-zero multiplier results in
r·θ" + 2r'·θ' = 0

This nullifies the ê_θ component in the above expression of acceleration in polar coordinates.

Therefore,

a = (r"−r·θ'²)·ê_r

Just as a check point, if the motion is circular (r is constant) and uniform (θ'=ω is constant), this formula looks like
a = −r·ω²·ê_r
which is fully in agreement with kinematics and dynamics of a uniform rotation (see lectures in UNIZOR.COM - Physics 4 Teens - Mechanics - Rotational Kinematics and Rotational Dynamics).

Using the established equality L/m=r²·θ' we can substitute θ' in the above equation in a vector of acceleration with L/(mr²) getting
a = [r"−r·L²/(m·r²)²]·ê_r or
a = [r"−L²/(m²·r³)]·ê_r

Again, using Newton's Laws, this vector of acceleration should be equal to
a = [r"−r·L²/(m·r²)²]·ê_r =
= −(G·M/r²)·ê_r.
Therefore, we can express it as a differential equation
r"−L²/(m²·r³) = −G·M/r²

Physics+ More on Ellipse: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

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
A_ellipse = 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

Physics+ Kepler's Second Law: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Kepler's Second Law

This lecture continues studying movement of objects in a central field. The familiarity with material presented in the lectures Central Force Field and Kepler's First Law is essential for understanding this educational material.

Kepler's Second Law states that a segment, connecting our Sun with any planet moving around a Sun, sweeps out equal areas during equal intervals of time.
As in the case of the Kepler's First Law, this Second Law has been based on numerous experiments and years of observation.

This Law can be formulated more mathematically.

Imagine a three-dimensional space with a single source of gravitation - a point-mass M at point O.
Some object comes into this field - a point-mass m. Its position at time t is at point P(t) and its velocity is v(t).

At time t=0 the position of our object is P(0) and velocity vector v(0).

As we know from previous lectures of Laws of Newton part of this course, the trajectory of our object will lie in the plane defined by vectors of initial position OP(0)=r(0) and initial velocity v(0) at time t=0.
Therefore, we can restrict our analysis to a two-dimensional case of trajectory lying completely within the plane defined by r(0) and v(0).

Let's choose a system of polar coordinates in this plane with an origin at point O, where the source of gravity is located, and a base axis defined by direction from point O to a position of our object at time t=0 - point P(0).

If our object during a time interval from t₁ to t₂ moved from point P(t₁) to P(t₂), its position vector r(t) swept up a sector bounded by r(t₁), r(t₂) and a trajectory from P(t₁) to P(t₂).

Let's introduce a function A(t) that represents an area of a sector bounded by r(0), r(t) and a trajectory from P(0) to P(t).
Then the area swept by position vector r(t) during the object's motion from time t₁ to t₂ equals
ΔA_[t1,t2] = A(t₂) − A(t₁)

Using the above symbols, the Kepler's Second Law can be formulated as
If t₂−t₁ = t₄−t₃ then
A(t₂)−A(t₁) = A(t₄)−A(t₃)
The above condition is equivalent to a statement that
dA(t)/dt is constant.

Indeed, let t₁ and t₃ be any two moments of time and t₂=t₁+Δt and t₄=t₃+Δt.
Then t₂−t₁=Δt and t₄−t₃=Δt
Therefore,
A(t₁+Δt)−A(t₁) =
= A(t₃+Δt)−A(t₃)
Dividing both sides by Δt, we get
[A(t₁+Δt)−A(t₁)]/Δt =
= [A(t₃+Δt)−A(t₃)]/Δt
Taking this to the limit, when Δt→0, we get
dA(t)/dt|_t=t1 = dA(t)/dt|_t=t3
Since values t₁ and t₃ are chosen freely, it means that the derivative of a function A(t) is constant.
Hence, we conclude, A(t) is a linear function of time t.

In reverse, if we assume that the first derivative of function A(t) is constant and, therefore, A(t) is a linear function of time t, we can easily prove that
if t₂−t₁ = t₄−t₃ then
A(t₂)−A(t₁) = A(t₄)−A(t₃)

The constant first derivative of function A(t), that represents an area swiped by a position vector during the time from t=0 to some value t, is just a mathematical way of stating the Kepler's Second Law.
Proving this characteristic of function A(t) is a proof of the Kepler's Second Law.

Let's prove it then.
Assume that an object position vector during time interval Δt moved from r(t) to r(t+Δt).
These two vectors form a triangle whose area approximately equal to an area A(t+Δt)−A(t) of a sector swiped up by vector r(t) during the time Δt. The approximation will be better, as interval of time Δt tends to zero.

The third side of this triangle is a vector connecting the end points of position vectors.
An approximation of this vector's magnitude is a magnitude of velocity vector at time t multiplied by time interval Δt:
r(t+Δt) − r(t) ≅ v(t)·Δt
The area of a triangle formed by two vectors a and b equals to the half of a magnitude of a vector product of these vectors because
(i) the area of a triangle with two sides a and b with an angle ∠φ between them is equal to
½·a·h_a = ½·a·b·sin(φ)
where h_a is an altitude onto side a
(ii) from the definition of a vector product
|a⨯b| = |a|·|b|·sin(φ)

Using this, we can say that an area of a triangle formed by r(t), r(t+Δt) and v(t)·Δt equals to ½|r(t)⨯v(t)|·Δt.

Recall that Angular Momentum of an object of mass m moving in some field, having a position vector r(t) and velocity v(t), is defined as
L(t) = m·r(t)⨯v(t).
But if the field is central, the Angular Momentum is a constant because a central force has no torque.
Therefore,
|r(t)⨯v(t)| = |L|/m is a constant.

An immediate consequence from this is that the area of a triangle formed by r(t), r(t+Δt) and v(t)·Δt is ½|L|/m·Δt.

When Δt→0, the area of our infinitesimal triangle tends to the area of a sector swiped up by position vector r(t) during infinitesimal time interval Δt ΔA(t)=A(t+Δt)−A(t).

Therefore,
lim_Δt→0ΔA(t)/Δt = ½|L|/m
which is a constant.
The limit above is a derivative of A(t) by time. Since it is a constant, A(t) is a linear function of time.
That proves the Kepler's Second Law.

Physics+ Kepler's 1st Law: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Kepler's First Law

Kepler's First Law states that all planets move around the Sun on elliptical orbits with the Sun in one of the two focus points of their orbits.
Kepler had come up with this law experimentally based on many years of observations.

Based on our knowledge of Physics, we will prove analytically a more general theorem about objects moving in the central gravitational field:
The trajectory of an object moving in the central gravitational field is a flat curve of the second order (ellipse, parabola or hyperbola) with the source of the gravitational field in the focal point of this curve.

The previous lecture Planet Orbits of this section of this course was about why the trajectory of an object in a central gravitational field is a flat curve with the center of a gravitational field lying within this plane.
So, we assume this topic is covered.

Also covered were the properties of ellipse, hyperbola and parabola, including their canonical equations in Cartesian and polar coordinates (see UNIZOR.COM > Math+ 4 All > Geometry).

Assume, a point-mass M is the source of a gravitational field and is located at point O in space.

Assume further that a point-mass m is moving in the gravitational field of point-mass M with no other forces involved, and at time t it's located at point P(t). Then its position relative to the source of gravity is vector OP(t)=r(t).

The velocity vector of an object is v(t)=r'(t) - a derivative of the position vector by time (here we will use a single apostrophe to signify the first derivative of a function by time and a double apostrophe - to signify its second derivative by time).

Let's choose some moment of time as the beginning of motion and consider it as an initial time t=0.
The initial position of our object moving in a central gravitational field will be r(0) and initial velocity will be v(0).

As we have proven in the previous lecture, an entire trajectory of the motion of point-mass m will be within the plane of motion going through the center of gravity O and vectors r(0) and v(0).
Consequently, at any moment of time t vectors of position r(t) and velocity v(t) will lie in the same plane.

Let's introduce a system of coordinates OXY on this plane with an origin O at the center of gravity and axis OX directed to a position of our object r(0) at time t=0.
Let's fix the directions of axes OX and OY within the plane of motion for now, but keep in mind that we might rotate axes in the future to simplify the final equations of a trajectory.

In such a system of XY-coordinates with point O as an origin the vector OP(t)=r(t) can be represented as a pair
r(t) = {x(t),y(t)}.

In these XY-coordinates the velocity vector v(t) can be represented as
v(t) = r'(t) = {x'(t),y'(t)}.

Our task is to analytically describe the trajectory of an object in a gravitation field as a curve inside a plane OXY.

The first foundation for theoretical derivation of the Kepler's First Law is the Second Law of Newton.

Provided the force F (a vector) acts on an object of mass m, this Law states that F = m·a
where a is a vector of object's acceleration, that is the first derivative from the object's velocity vector v or, equivalently, the second derivative of object's position vector r
a(t) = v'(t) = r"(t)

The second component is the Newton's Universal Law of Gravitation.

It assumes that an object of mass M is fixed in space at the origin of coordinates and is the only source of gravitation.

It further assumes that an object of mass m is moving in the gravitational field produced by an object of mass M and at time t is at position defined by a position vector r(t) that stretches along the line between the source of gravitation (the origin of coordinates) and the object's location.

Then the Law of Gravitation states that
(a) the vector of gravitational force F(t) produced by object of mass M acting on an object of mass m at any time t is collinear with the position vector r(t) and directed towards the object of mass M and
(b) the magnitude F(t) of the gravitational force F(t) is
F(t) = G·M·m/r²(t)
where
G is the universal gravitational constant whose value is 6.67430·10⁻¹¹ N·m²/kg² and
r(t) is the magnitude of the position vector r(t) - the distance of an object from the origin of coordinates.

Combining conditions (a) and (b) together, we can express the force of gravity in vector form as
F(t) = −G·M·m·r(t)/r³(t)

The Newton's Second Law, which connects a force and an acceleration, and the Universal Law of Gravitation, which defines the value of a gravitational force, allow to establish the differential equation that defines the motion of our object in a gravitational field
F(t) = m·a(t) = m·r"(t) =
= −G·M·m·r(t)/r³(t)
or, canceling mass m,
r"(t) = −G·M·r(t)/r³(t)

The fact that mass m can be canceled is quite remarkable. It says that different objects would follow the same trajectory in the gravitational field, if at some moment their positions and velocities are the same.
This is not supposed to be a surprise. Recall Galileo's experiments with different objects dropped from the Tower of Pisa. They would spend the same time to hit the ground regardless of their mass.
If, in addition to dropping them down, you give them the same horizontal speed, they would spend the same time going horizontally and fall on the same distance from a vertical, thus going along the same trajectory, regardless of mass.

The above equation is a vector differential equation that can be transformed into a system of two differential equations for Cartesian coordinates {x(t),y(t)} of vector r(t) with
r"(t) = {x"(t),y"(t)} and
r(t) = √x²(t)+y²(t)

In coordinate form this system of two differential equations with two unknown functions x(t) and y(t) looks like
x"(t)=−G·M·x(t)/r³
y"(t)=−G·M·y(t)/r³

Directly solving this system of differential equations is not an easy task to do.
But let's keep in mind that its solution will give us more than we asked for. We wanted to prove that objects in a gravitational field move along ellipses, hyperbolas or parabolas. We just needed a shape of trajectories, not a time-dependent position at every moment in time.

Having this smaller task in mind, we do not have to derive explicit functions x(t) and y(t) that are solutions to the above system of equations, we just have to concentrate on geometrical characteristics of trajectories.

Consider a position vector r from a fixed center of gravity (point O in our system of coordinates) to a position of an object at any moment of time.
Apparently, having a dependency of its length r=|r| on its angle θ from OX-axis is sufficient to determine geometric characteristics of its trajectory.
Basically, we can derive the geometric characteristics of a trajectory by analyzing its equation in polar coordinates {r,θ}. Deriving a function r=r(θ) is a much easier task than solving the above system of differential equations.
Let's concentrate on this smaller task.

Consider an angle ∠θ(t) from the positive direction of X-axis to position vector r(t).
Using this, coordinates of position are
x(t) = r(t)·cos(θ(t))
y(t) = r(t)·cos(θ(t))

At this point let's concentrate not on time-dependency of coordinates, but on their dependency on the angle θ. That will allow to derive the shape of a trajectory without knowing exactly the coordinates of an object moving along this trajectory at any moment of time.
Now the coordinates of an object's position, as functions of θ, are
x(θ(t)) = r(θ(t))·cos(θ(t))
y(θ(t)) = r(θ(t))·cos(θ(t))
Now we are not interested in function θ(t) and the above equations can be viewed as x(θ) = r(θ)·cos(θ)
y(θ) = r(θ)·cos(θ)

Using this representation, the original system of differential equations is
x"(t) = −G·M·cos(θ)/r²(θ)
y"(t) = −G·M·sin(θ)/r²(θ)

Recall from the lecture Planet Orbits of this part of the course that vector of Angular Momentum L=m·r·v of an object moving in a central gravitational field is a constant.
This vector L is perpendicular to both r and v and, therefore, is directed along the third dimension OZ-axis that is perpendicular to a plane of object motion.

The length L=|L| (a constant) of an Angular Momentum vector equals to L=m·r²·θ'.

Proof

In three dimensional XYZ space the position vector r and velocity vector v=r' have Z-coordinate equal to zero:
r = {r·cos(θ), r·sin(θ), 0}.
v = {r'·cos(θ)−r·sin(θ)·θ', r'·sin(θ)+r·cos(θ)·θ', 0}.

According to the rules of vector product in three-dimensional Cartesian coordinates, their vector product is
L/m = r⨯v = {0, 0, r·cos(θ)·[r'·sin(θ)+r·cos(θ)·θ'] −
− r·sin(θ)·[r'·cos(θ)−r·sin(θ)·θ']} =
= {0, 0, r²·[cos²(θ)+sin²(θ)]·θ'} =
= {0, 0, r²·θ'}
Therefore,
L = m·r²·θ'

Now we can express r² in terms of θ' and constants
r² = L/(m·θ')
Therefore, the system of our differential equations can be expressed only in terms of θ(t) as follows
x"(t) = −θ'·cos(θ)·(G·M·m/L)
y"(t) = −θ'·sin(θ)·(G·M·m/L)
In these equations θ(t) is an unknown function of time, while x"(t) and y"(t) are components of an acceleration vector and are the first derivatives of the components of a velocity vector {x'(t),y'(t)}.

Notice that expressions on the right side of both differential equations are full derivative by time of simple functions of θ, which is easy to integrate.
Therefore, integrating by time left and right sides of equations, we obtain
x'(t) = −sin(θ)·(G·M·m/L) + c₁
y'(t) = cos(θ)·(G·M·m/L) + c₂
or, substituting for simplicity a constant K for G·M·m/L,
x'(t) = −sin(θ)·K + c₁
y'(t) = cos(θ)·K + c₂
where constants c₁ and c₂ are defined by initial conditions at time t=0, when θ(0)=0:
x'(0) = −sin(0)·K + c₁
y'(0) = cos(0)·K + c₂
c₁ = x'(0)
c₂ = y'(0)−K
The initial components of velocity at x'(0) and y'(0) are given and K=G·M·m/L is a known constant.

The remaining task is to find the length of a position vector r(θ), as a function of angle θ from the above expressions for components of the velocity.

The components of a position vector, as functions of θ, are
x = r·cos(θ)
y = r·sin(θ)
differentiating these by time, we obtain
x'(t) = r'·cos(θ) − r·sin(θ)·θ'
y'(t) = r'·sin(θ) + r·cos(θ)·θ'

Comparing these with the expressions for components of a velocity vector above, give
r'·cos(θ) − r·sin(θ)·θ' =
= −sin(θ)·K + c₁
r'·sin(θ) + r·cos(θ)·θ' =
= cos(θ)·K + c₂

The next step is to get rid of r' and θ'.
The way to do it is to multiply the first equation by sin(θ), the second - by cos(θ) and subtract the first from the second.
r·cos²(θ)·θ' + r·sin²(θ)·θ' =
= cos²(θ)·K + c₂·cos(θ) +
+ sin²(θ)·K − c₁·sin(θ)
Now we can use trigonometric identity sin²(θ)+cos²(θ)=1 getting
r·θ' = K + c₂·cos(θ) − c₁·sin(θ)

Also, recall that L=m·r²·θ' and, therefore, θ'=L/(m·r²).
Using this result in the following
L/(m·r)=K+c₂·cos(θ)−c₁·sin(θ)

Resolving for r results in the expression of r as a function of θ - exactly what we are looking for
r = 1/[A+B·cos(θ)−C·sin(θ)]
where all coefficients are known constants:
A = K·m/L
B = c₂·m/L = [y'(0)−K]·m/L
C = c₁·m/L = x'(0)·m/L
K = G·M·m/L

The above looks like a complete solution r(θ), but we can improve it a little.
The expression B·cos(θ)−C·sin(θ) can be transformed into D·cos(θ+φ) as follows
B·cos(θ)−C·sin(θ) =
= √B²+C²·[cos(θ)·B/√B²+C² −
− sin(θ)·C/√B²+C²]
We can always find an angle φ such that
B/√B²+C² = cos(φ)
C/√B²+C² = sin(φ)
which results in
B·cos(θ)−C·sin(θ) =
= √B²+C²·[cos(θ)·cos(φ)−
+sin(θ)·sin(φ)] =
= √B²+C²·cos(θ+φ)

Therefore,
r = 1/[A+√B²+C²·cos(θ+φ)]
Let's do a small correction in the direction of the X-axis.
Originally, we placed the beginning of coordinates point O at the point where the source of gravitation is located. Then we chose OX axis directed to the initial position of our object at time t=0.
Now, knowing angle φ we can change the direction of the X-axis by turning it by an angle φ.
As a result, in this new coordinate system the equation of the trajectory looks even simplier:
r = 1/[A+√B²+C²·cos(θ)]
which, depending on the values of all constants involved, is an equation in polar coordinates of either an ellipse or a hyperbola, or a parabola, as explained in the previous lecture dedicated to 2-nd order curves.

End of proof.

Physics+ 2nd Order Curves: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

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 value 2b, where b²=a²−c², is called the length of minor axis.
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).
The equation of an ellipse in Cartesian coordinates (x,y) with X-axis coinciding with the perpendicular line from a focus to a directrix and origin of coordinates is a midpoint between focus and directrix
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:
Physical trajectory described by this equation would be when an object moves towards but slightly off the source of gravity. The gravitational force will turn it around the origin, but it will not be sufficient to keep it on an orbit around a center, and an object will move away from the center of gravitation to infinity.

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
which, from the physical standpoint, would be an orbit of an object moving around a central point-mass.

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 The physical meaning of this trajectory corresponds to an object moving straight towards the center of gravity or directly from it.

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

u v

=

x y

⨯

cos(β) sin(β) −sin(β) cos(β)

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

x y

=

u v

⨯

cos(β) −sin(β) sin(β) cos(β)

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)

Math+ Parabola Optics: UNIZOR.COM - Math+ 4 All - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ Parabola Optics

Though the word "optics" sounds very much like a topic of Physics, we will consider strictly mathematical aspect of it.
One of the properties of parabola is that if its contour is reflective, a ray of light emitted from its focus will be reflected parallel to its axis of symmetry regardless of the direction it was sent.

Reflection of a curve occurs exactly as if, instead of a curve at the point of incidence, there was a tangential line to a curve, and reflection was of that tangential straight line.

Consider a parabola with focus F(c,0), directrix d, point P(x,y) on this parabola and tangential BC to a parabola at point P(x,y).
The ray of light is emitted from a focus F(c,0), goes to point P(x,y) and reflects along PS.
Because of the laws of reflection, angles ∠FPB and ∠SPC are equal, they are marked as α.
If we prove that these angles α are also equal to angle ∠ψ the tangential at point P line BC makes with the axis of symmetry of parabola BX, it will prove the parallelism of reflected ray PS with line BC, regardless of position of point P on a parabola.

Our plan to prove it is to prove that tan(α)=tan(ψ), from which the equality α=ψ follows because function tan() is monotonic for these angles.

Since sum of angles of a triangle ΔBPF equals to π,
α + ψ + (π−φ) = π
Therefore,
α = φ − ψ
We can calculate tan(α) using a formula for tangent of difference between angles, we can express tan(α) in terms of tan(φ) and tan(ψ).
tan(α) =
= [tan(φ)−tan(ψ)]/[1+tan(φ)·tan(ψ)]

Angle ∠φ=∠XFP.
Knowing coordinates of points P(x,y) and F(c,0), it's easy to calculate
tan(φ) = y/(x−c)

As we know, a tangent of an angle between a tangential line to function f(x) at some point and X-axis is a function's derivative f'(x) at that point.
Therefore, tan(ψ)=y'(x)
Since an equation of a parabola is y²=4c·x, differentiating this equation we get
2y·y' = 4c
Hence, y' = 2c/y = tan(ψ).

Now we have all the components to calculate tan(α):
tan(α) =
= [y/(x−c)−2c/y]/[1+(y/(x−c))·(2c/y)]
Let's simplify this expression.
Its numerator equals to
(y²−2c·x+2c²)/(x·y−c·y)
But for each point of a parabola y²=4c·x.
Use it in the formula above, getting the same numerator as
(4c·x−2c·x+2c²)/(x·y−c·y) =
= 2c·(x+c)/(x·y−c·y)
The denominator in the formula above can be simplified as
1 + (y/(x−c))·(2c/y) =
= 1 + 2c·y/((x−c)·y) =
= (x·y−c·y+2c·y)/((x−c)·y) =
= y·(x+c)/(x·y−c·y)

Dividing the numerator
2c·(x+c)/(x·y−c·y)
by denominator
y·(x+c)/(x·y−c·y)
we get the value of tan(α) as
tan(α) = 2c/y

But this is the same value as tan(ψ), which proves that angles ∠α and ∠ψ are equal, which, in turn, proves that lines PS (reflected ray of light) and FX (axis of symmetry of a parabola) are parallel regardless of the position of point P on a parabola.

Therefore, all the rays from the parabola's focus directed in any direction will be reflected in one direction - parallel to the axis of symmetry of this parabola.

Math+ Parabola: UNIZOR.COM - Math+ 4 All - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ Parabola

Parabola is a class of curves on a plane with the following defining properties.

For each curve of this class (that is, for each parabola) there is one specific point called its focus and a specific straight line called its directrix not going through a focus, such that this parabola consists of all points on a plane (or is a locus of all points on a plane) equidistant from its focus and directrix.

As we see, the position of a focus point F and a directrix d (should not go through focus F) uniquely identifies a parabola.

Obviously, a midpoint of a perpendicular from focus F onto directrix d belongs to a parabola defined by them.
From this point we can draw a parabola point by point maintaining equality of the distance from each point to both a focus and a directrix.

Choosing an X-axis perpendicular to a directrix with an origin of coordinates 0 at midpoint between a focus and a directrix, we will have the following picture of a parabola on a coordinate plane.
We can derive an equation that defines this parabola using the main characteristic property of every point on this parabola to be equidistant from focus and a directrix.

The distance from any point P(x,y) on a parabola to a focus F that has coordinates (c,0) can be calculated using the known formula of a distance between two points.
The distance from point P to a directrix is calculated along a perpendicular from point P to a directrix that is parallel to X-axis.

If the distance between a focus and a directrix is 2c, as on a drawing above, the equality of the distances to a focus and a directrix is
√(x−c)²+(y−0)² = x−(−c)

We can simplify this as follows
(x−c)²+(y−0)² = (x+c)²
y² = (x+c)²−(x−c)²
y² = 4c·x

The only parameter that defines the shape of a parabola is c (half of a distance between a focus and a directrix), which is called the focal distance (or focal length) of a parabola.

The perpendicular from a focus to a directrix is an axis of symmetry of a parabola.
The midpoint of this perpendicular is called a vertex of a parabola.

Another item of interest is parabola's focal width. This is the length of a segment drawn through a focus parallel to a directrix with its endpoints being its intersections with a parabola.
Using a formula of a parabola y²=4c·x we determine that if x=c then y²=4c² and y=2c, which is a half of a segment described above.
Therefore, parabola's focal width is 4c.

Let's derive a formula r=r(θ) for a parabola in polar coordinates with an origin at its focus and base axis perpendicular to a directrix.
The distance from point (r,θ) to a focus is, obviously, r.
The distance from this point to a directrix is
AB = AF + FB = 2c+r·cos(θ).
From this the equation of a parabola in polar coordinates is

r = 2c/[1−cos(θ)]