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:
êr along a line from the origin O to position of a test object P, that is collinear to vector r;
êθ along a line perpendicular to êr.
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·êr
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 asr=√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 {êr, êθ}.
For this, first of all, we express basic {êr, êθ} in terms of basic {i, j} using the fact that vector êr has a unit length and positioned at angle θ to OX-axis, while vector êθ has a unit length and positioned at angle θ+π/2 to OX-axis:
êr = cos(θ)·i + sin(θ)·j
êθ = −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
dêr /dθ = êθ
dêθ /dθ = −êr
Furthermore, the time derivatives of these unit vectors are
êr' = (dêr /dθ)·θ' = θ'·êθ
êθ' = (dêθ /dθ)·θ' = −θ'·êr
Since r = r·êr
v = r' = dr/dt =
= d(r·êr)/dt =
= r'·êr+r·êr' =
= r'·êr+r·θ'·êθ
v = r'·êr+r·θ'·êθ
a = v' = dv/dt =
= r"·êr + r'·êr' +
+ r'·θ'·êθ + r·θ"·êθ + r·θ'·êθ' =
= r"·êr + r'·θ'·êθ +
+ r'·θ'·êθ + r·θ"·êθ − r·θ'²·êr =
= (r"−r·θ'²)·êr +
+ (r·θ"+2r'·θ')·êθ
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 êr are collinear, from which follows that coefficient at êθ must be equal to zero:
r·θ"+2r'·θ' = 0 and
a = (r"−r·θ'²)·êr
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 êθ component in the above expression of acceleration in polar coordinates.
Therefore,
a = (r"−r·θ'²)·êr
a = −r·ω²·êr
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]·êr or
a = [r"−L²/(m²·r³)]·êr
Again, using Newton's Laws, this vector of acceleration should be equal to
a = [r"−r·L²/(m·r²)²]·êr =
= −(G·M/r²)·êr.
Therefore, we can express it as a differential equation
r"−L²/(m²·r³) = −G·M/r²
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.
The area of a triangle formed by two vectors 
