Physics+ Kepler Third Law: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Kepler's Third Law

Kepler's Third Law states that for all objects moving around a fixed source of gravitational field along elliptical orbits the ratio of a square of their period of rotation to a cube of a semi-major axis is the same.

As in the case of the Kepler's First Law, this Third Law has been based on numerous experiments and years of observation.

Based on all the knowledge conveyed in previous lectures on Kepler's Laws, we will derive this Third Law theoretically.

Let's make a simple derivation of Kepler's Third Law in case of a circular orbit.

In this case the velocity vector of an object circulating around a central point is always perpendicular to a position vector from a center to an object.
Since the gravitational force is collinear with a position vector, it is also perpendicular to velocity, which is tangential to a circular orbit. Therefore, gravitational force makes no action along a velocity vector which makes the magnitude of the velocity vector constant.

Let's introduce the following characteristics of motion:
t - absolute time,
r - radius of a circular orbit of a moving object,
F - vector of gravity,
M - mass of the source of gravitational field,
m - mass of object moving in the gravitational field,
r - position vector from the source of gravitational field to a moving object,
r'=v - velocity vector of a moving object,
r"=v'=a - acceleration vector of a moving object,
T - period of circulation,
ω=2π/T - scalar value of angular velocity,
Here bold letters signify vectors, regular letters signify scalars and magnitudes of corresponding vectors, single and double apostrophes signify first and second derivative by time.

Constant magnitude v of velocity vector means constant angular velocity ω and obvious equality v=r·ω.

Magnitude a of an acceleration vector can be simply found by representing a position vector as a pair of Cartesian coordinates (x,y):
x = r·cos(ωt)
y = r·sin(ωt)
x' = −r·ω·sin(ωt)
y' = r·ω·cos(ωt)
x" = −r·ω²·cos(ωt)=−ω²·x
y" = −r·ω²·sin(ωt)=−ω²·y
and, therefore,
a = r" = −ω²·r
(collinear with r and F)
from which follows
a = |a| = |−ω²·r| = ω²·r

According to the Newton's Second Law,
F = m·a
According to the Universal Law of Gravitation,
F = G·M·m/r²
Therefore,
a = ω²·r = G·M/r²
from which follows
ω² = G·M/r³

Since ω=2π/T,
4π²/T² = G·M/r³
T²/r³ = 4π²/(G·M) - constant
End of proof for circular orbit.

Let's prove it in a more complicated general case of any elliptical orbit.
We will use the First and the Second Kepler's Laws as well as the results presented in the previous lecture Planet Orbit Geometry to derive this Third Law.

Recall the Kepler's Second Law (see the lecture Kepler's Second Law in this course).
We have introduced a function A(t) that represents an area of a sector bounded by r(0), r(t) and a trajectory from a planet's position P(0) at time t=0 to its position at any moment of time 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 or, equivalently, that A(t) is a linear function of time t with A(0)=0.

In the same lecture we have proven that
dA(t)/dt = ½|L|/m
where L is an angular momentum of a moving object (constant in a central force field) and m is object's mass.

Therefore,

A(t) = t·½|L|/m

Assume, we want to know how much area is swept by position vector r(t) during a period T of a complete round movement of a planet around the Sun.
Obviously, it's
A(T) = T·½|L|/m
At the same time, A(T) is an area of an elliptical orbit of a planet, and we know that the area of an ellipse along which a planet moves equals to
A(T) = π·a·b
where a is semi-major and b is semi-minor axes (see the lecture More on Ellipse in this course).

Therefore,
π·a·b = T·½|L|/m

Let's assume that at time t=0 a planet is at the furthest from the Sun point (aphelion), it's initial position vector is r₀ and its velocity vector is v₀.
At this initial point on an orbit the position vector r₀, lying along the major axis of an ellipse, and a tangential to an ellipse vector of velocity v₀ are perpendicular to each other.

Therefore, the magnitude of the angular momentum |L| equals to a product of planet's mass, a magnitude of its position vector (that is, a distance of the Sun) and a magnitude of its velocity vector:
|L| = m·|r₀|·|v₀| or L/m = r₀·v₀

Now the above formula for a period T of a planet's rotation around the Sun is
π·a·b = ½T·r₀·v₀

In the previous lecture Planet Orbit Geometry we have derived the expressions for major and minor axes of an elliptical orbit of a planet in terms of its initial position and velocity at aphelion:

a =

r0 2−β

b =	r0√β·(2−β) 2−β

where β=r₀·v₀²/(G·M)

Let's substitute these expression into a formula connecting a period T with an area of an ellipse:
πr₀²√β·(2−β)/(2−β)²=½T·r₀·v₀

Let's square both sides to get rid of a radical:
π²r₀⁴β/(2−β)³=¼T²·r₀²·v₀²

Next is just technicality.
Cancel one r₀ from both sides
π²r₀³β/(2−β)³=¼T²·r₀·v₀²

Replace r₀³/(2−β)³ with a³ (see formula above)
π²a³β=¼T²·r₀·v₀²

Replace β with r₀·v₀²/(G·M) (see formula above)
π²a³r₀·v₀²/(G·M)=¼T²·r₀·v₀²

Cancel r₀·v₀² on both sides
π²a³/(G·M)=¼T²

Final result:

T² a3

=

4π² G·M

The right side is a constant that contains no planet-specific parameters (like initial position and velocity), which means that any planet has the following property (Kepler's Thirt Law).
The ratio of a square of the period of a planet's rotation around the Sun to a cube of a semi-major axis of its elliptical orbit is constant that depends only on a mass of the Sun.

Physics+ Orbit Geometry: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Planet Orbit Geometry

Before studying this material we strongly recommend to study geometrical aspects of ellipse (for example, in Math+ 4 All => Geometry => Ellipse) and physical aspects of movement in the gravitational field in lectures of this course Central Force Field, Planet Orbits, More on Ellipse and Kepler's First Law.

The purpose of this lecture is to determine the geometric characteristics of a planet's orbit based on its initial position and velocity relative to the Sun.

Before addressing these issues let's recall the characteristics of an object of mass m circulating on a circular orbit around a fixed in space central object of mass M.
In particular, we are interested in some relationship between the moving object's linear speed v and a radius r of rotation around a central source of gravitation of mass M in order to stay on a circular orbit.

According to the Newton's Universal Law of Gravitation, the force of gravity F is
F = G·M·m/r²
where G is the gravitational constant.
The Newton's Second Law connects this force to a mass m and centripetal acceleration a of a moving object
F = m·a
Therefore,
G·M·m/r² = m·a
from which follows
a = G·M/r²

We can express the linear speed along an orbit v and centripetal acceleration a in terms of constant radius of uniform rotation r on a circular orbit with constant angular speed of rotation ω using time-dependent Cartesian coordinates {x(t),y(t)} of a moving object as follows.

Position vector:
x(t) = r·cos(ω·t)
y(t) = r·sin(ω·t)

Velocity vector:
x'(t) = −r·ω·sin(ω·t)
y'(t) = r·ω·cos(ω·t)
v = √x'(t)²+y'(t)² = r·ω

Acceleration vector:
x"(t) = −r·ω²·cos(ω·t)
y"(t) = −r·ω²·sin(ω·t)
a = √x"(t)²+y"(t)² = r·ω² = v²/r

Therefore, returning to the Universal Law of Gravitation,
v²/r = G·M/r²
and we conclude that for a circular rotation of an object in a gravitational field

β =

r·v² G·M

= 1

In the above formula we have introduced a symbol β with which we expressed the main condition of object to stay on a circular orbit.
This symbol will be used in our analysis of an elliptical orbit of an object in a central gravitational field.

Let's switch now to a more complicated case of elliptical orbit.

Kepler's First Law states that all planets move around the Sun along elliptical orbits with the Sun in one of the two focal points of their orbits.
Kepler had come up with this law experimentally based on many years of observations.
In the lecture Kepler's First Law we have proven it based on the Newton's Second Law and the Universal Law of Gravitation.

The geometric properties of a planet's elliptical orbit are completely defined by two parameters: its major axis of the length 2a and eccentricity e, from which we can derive a minor axis of the length 2b and focal distance 2c using equations
c = a·e
b²+c²=a² => b=a·√1−e²

Assume, a point-mass M (the Sun) is the source of a gravitational field and is fixed in our space.
Assume further that a point-mass m (a planet) is moving relatively to this point-mass M in its gravitational field with no other forces involved.
Two major characteristics of this motion are the planet's position and velocity.

In the lecture Planet Orbits we have proven that the trajectory of a planet's movement around the Sun is a plane that we called the plane of motion.



This plane goes through the source of gravity and contains two vectors - the vector of initial position r₀=r(0) from the source of gravity to a moving object at some chosen moment in time t=0 (from the Sun to a planet) and the vector of object's initial velocity v₀=v(0) (tangential to a trajectory) at the same time.

In the lecture Central Force Field we have proven that Angular Momentum vector L=m·r(t)⨯v(t) of a planet moving around the Sun is independent of time, is a constant vector perpendicular to a plane of motion.

In this lecture we will examine the geometric properties of an elliptical trajectory of a planet moving around the Sun and see how these geometric properties relate to a planet's initial position and velocity relative to the Sun at some chosen moment in time t=0.

We know that the trajectory of a planet is an ellipse lying within a plane of motion with the Sun at one of this ellipse focal points.
Using this, let's choose the coordinate system and the initial moment in time t=0 to simplify the equation of an orbit.

We will use the polar coordinate systems lying within a plane of motion.
The pole of the polar system will be at the Sun.
The polar axis will coincide with the major axis of an elliptical orbit and will be directed from the Sun towards the furthest from it point on a major axis (aphelion).

As the initial time t=0 we will choose a moment in time when a planet is at this furthest point from the Sun.
At this point of intersection of an ellipse and its major axis the initial velocity, being tangential to an ellipse, is perpendicular to a major axis and, therefore, perpendicular to a position vector.

Therefore, vectors r(0) and v(0) are perpendicular to each other with their magnitudes, correspondingly, r₀ and v₀.

So, at time t=0 a planet is at the polar angle θ=0 and distance from a pole (the Sun) r(0)=r₀.
The velocity vector is perpendicular to a major axis and its magnitude is v(0)=v₀.
Parameters r₀ and v₀ are given. Based on them, we have to determine the characteristics of an elliptical orbit - its semi-major axis a and the eccentricity e.

As described in the Math+ 4 All - Geometry - Ellipse lecture, the canonical equation of our elliptical orbit in these polar coordinates is

r(θ)=

a(1−e²) 1−e·cos(θ)

Setting θ=0 gives the an equation
r(0) = r₀ = a·(1+e)
It's one equation with two variables a and e. We need more equations to find these characteristics of an elliptical orbit.

Setting θ=π in the above equation of an ellipse (that is, considering the opposite point on an ellipse along the major axis called perihelion) gives
r(π) = a·(1−e)
This adds the second equation for a and e but adds another unknown r(π).

While r(π) and v(π) (the magnitudes of the position and the velocity vectors at the perihelion) are unknown, we can use two Laws of Conservation between points θ=0 and θ=π to determine them - the Law of Conservation of Angular Momentum and the Law of Conservation of Energy.

Recall the Conservation of Angular Momentum Law for an object in a central force field
L = m·r⨯v

The velocity and position vectors are perpendicular to each other at point θ=0 as well as at point θ=π.
Therefore, the magnitude of the vector product of these two vectors at both points equals to the product of their magnitudes and is the same because of conservation of angular momentum.
L(0) = m·r₀·v₀ =
= m·r(π)·v(π) = L(π)
From this follows
v(π) = r₀·v₀/r(π)

Next we will use the conservation of energy - the sum of potential energy of an object in the gravitational field and its kinetic energy is constant.

Potential energy of an object of mass m (a planet) in the gravitational field of an object of mass M (the Sun) located at a distance r from it is
U = −G·M·m/r
Kinetic energy of this object, when its speed is v, is
K = ½m·v²
Full energy is E = K + U

Therefore,
E(0) = ½m·v₀² − G·M·m/r₀ =
= ½m·v²(π) − G·M·m/r(π) =
= E(π)

From the equation for conservation of angular momentum we get the value of v²(π) in terms of r(π) and initial parameters r₀ and v₀ as follows
v²(π) = r₀²·v₀²/r²(π)
and put it in the equation for energy conservation getting an equation to determine r(π)
½m·v₀²−G·M·m/r₀ =
= ½m·r₀²·v₀²/r²(π)−G·M·m/r(π)

For brevity, let's temporarily use variable x instead of r(π).
To simplify this equation, let's reduce all members by mass m: ½v₀²−G·M/r₀ =
= ½r₀²·v₀²/x²−G·M/x
and multiply by 2x² gathering all members to the left side of an equation getting
x²·(2G·M/r₀−v₀²) −
−x·(2G·M) + r₀²·v₀² = 0

Recall that earlier in this lecture we analyzed the circular rotation in a central gravitational field and introduced a symbol β=r·v²/(G·M), which was supposed to be equal to 1 in order for an object to stay on a circular orbit.
Multiplying our equation for x by r₀/(G·M) and using this symbol β=r₀·v₀²/(G·M), the equation looks much simpler:
x²·(2−β) −x·(2r₀) + β·r₀² = 0

Canonical quadratic equation
P·x² + Q·x + R = 0
where
P = 2−β
Q = −2r₀
R = β·r₀²
has solutions

x = r(π) =

−Q±√Q²−4P·R 2P

with all components described above known.

The expression under a square root can be simplified.
Q²−4P·R = 4r₀²−4(2−β)βr₀² =
= 4r₀² − 8βr₀² +4β²r₀² =
= 4r₀²·(1−2β+β²) =
= 4r₀²·(1−β)²

Therefore,

x = r(π) =

2r0 ± 2r0·(1−β) 2(2−β)

or

r(π) =

r0·(1±(1−β)) 2−β

If we choose a plus sign in the numerator, the value for r(π) will be equal to r₀, which only possible if our ellipse is a circle with coinciding focal points.
This is a trivial case and we will not consider it here.

In all other cases the solution is
r(π) = r₀·β/(2−β) where β=r₀·v₀²/(G·M), which contains only known values - initial distance from the Sun and speed of a planet at aphelion and other known variables.

Therefore, we have a simple system of two equations with two unknowns a and e:
r(0) = r₀ = a·(1+e)
r(π) = a·(1−e)

This system has solutions
a = ½[r₀+r(π)]
e = 2r₀ / [r₀+r(π)] − 1 =
= [r₀−r(π)] / [r₀+r(π)]
From these we can determine the focal distance
c = a·e = ½[r₀−r(π)]
and semi-minor axis
b = √a²−c² = [r₀·r(π)]^½

Putting the obtained expression for r(π) into above formulas we have the geometric properties in terms of initial distance of a planet from the Sun and its linear speed at aphelion:
a = ½[r₀+r(π)] =
= ½[r₀+r₀·β/(2−β)]
which can be transformed into

a =

r0 2−β

Let's do the same with eccentricity e:
e = [r₀−r(π)] / [r₀+r(π)] =
= [r₀−r₀·β/(2−β)] /
/ [r₀+r₀·β/(2−β)]
which can be transformed into

e = 1 − β

The focal distance:
c = a·e
which can be expressed as

c =	r0(1−β) 2−β

The semi-minor axis:
b² = a² − c² =
= [r₀² − r₀²·(1−β)²] /(2−β)² =
= r₀²·β·(2−β) /(2−β)²
Therefore,

b =	r0√β·(2−β) 2−β

To conclude geometric properties of an elliptical orbit, since we know both semi-axes, we can determine an area of an ellipse A=π·a·b

A=	πr0²√β·(2−β) (2−β)²

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)