Periodicity is a characteristic feature of Elliptic Functions. Elliptic Functions may be either doubly-periodic, singly-periodic, or (trivially) non-periodic. The Exponential Function is the singly-periodic Elliptic Function. Thus, the Exponential Function possesses the modulus of periodicity of 2 pi i
exp(x + 2 pi i) = exp(x) for any x in Complex.
And, by saying that 2 pi i is the "modulus of periodicity" we mean that 2 pi i is the smallest, in magnitude, constant for which the foregoing identity is true. Incidentally, we thus have defined pi. An approximate value of pi is 3.14159, a Real number. Each of the trigonometric functions will inherit this modulus of periodicity, except that the tangent and cotangent functions have a modulus of periodicity only half the size.
Possession of an addition theorem is another characteristic feature of Elliptic Functions. The addition theorem for the Exponential Function is
exp(x + y) = exp(x) exp(y) for any (x, y) in the Cartesian product of Complex by Complex.
The third characteristic property of the Elliptic Functions is that they satisfy a certain differential equation. Thus, the exponential function may be characterized by the differential equation
d exp(x) / dx - exp(x) = 0, with exp(0) = 1,
for any x in Complex.
The MacLaurin's series for the exponential function is
exp(x) = 1 + x + x^2 / 2! + x^3 / 3! + ... + x^n / n! + ....
For an x in Complex, we define the nine Hyperbolic Trigonometric functions as follows:
That the two definitions of the haversed hyperbolic sine are equivalent follows from a sneak preview of the double-angle formula for the cosine. The first definition is the one that motivates the name; but it entails excessive round-off error.
It is obvious that the cosine is an even function, while the sine is an odd function. It follows that the secant, versed sine, and haversed sine are even, while the tangent, cotangent, and cosecant are odd.
For any x in the Cartesian product of Complex by Complex, from the foregoing, the following identities are obvious:
For any (x, y) in the Cartesian product of Complex by Complex, from the definitions of the Hyperbolic Trigonometric function, the following real addition theorems are obvious:
By substitution of the addition formulae into the right-hand side of the following, one may obtain these real product theorems:
Let u = x + y and v = x - y. Substitution in the foregoing three equations yields the Sums or Differences
For any (x, y) in the Cartesian product of Complex by Complex, from the foregoing real addition theorems, with the replacement of y by i y and the use of the circular trigonometric functions, the following complex addition theorems are obvious
The special cases where x is zero are as follows
On the other hand, we may invert the first three of the foregoing complex addition theorems. Set the right-hand side equal to u + i v. Then collect the real and imaginary parts on one side of the equation; each part has to be zero. Then by employing the identities, we obtain
In the foregoing real addition theorems, take y equal to x, to yield the following double-angle formulae:
By the use of the first of the identities, the double angle formula for the hyperbolic cosine may be written in either of the additional two forms:
The half-angle formulae are as follows:
By substitution of i x for x in the multiple-angle formulae of the circular-trigonometric functions, we obtain the multiple-angle formulae for the hyperbolic-trigonometric functions.
Of course, these series terminate before the exponents become negative.
Let y = Arcsinh(x) and solve for x to obtain
x = sinh(y) = (exp(y) - exp(- y)) / 2 = (exp(2 y) - 1) / (2 exp(y)).
Solution of this quadratic equation yields the first of the following formulae of the inverse hyperbolic trigonometric functions in
terms of the logarithmic function. The derivation of the others is similar.
However, the formulae for the Acoth(x), Asech(x), and Acsch(x) may be obtained from those for the Atanh(x), Acosh(x), and Asinh(x), respectively,
as well.
Considered as complex-functions, these functions have an unrestricted domain; i.e., all complex numbers. However, considered as
real-functions, they have the indicated domains.
| Here is a detailed derivation of the first of the foregoing formulae. We define the Arcsinh(x) function as the principal inverse of
the hyperbolic sine function. Let
y = Asinh(x). By this definition, it is the inverse of x = sinh(y). Substitute the definition of the hyperbolic sine (in terms of the exponential function), to obtain x = (exp(y) - exp(-y)) / 2. Multiply through by (2 exp(y)), and re-arrange, to obtain a quadratic equation in exp(y): (exp(y))^2 - 2 x exp(y) - 1 = 0. Anytime that one multiplies by the unknown, one might -- and indeed, did -- introduce an extraneous solution. Solve by the quadratic-formula, to obtain exp(y) = (- (- 2 x) +- sqrt((- 2 x)^2 - 4 (1) (-1))) / 2. Simplify the right-hand side, to obtain exp(y) = x +- sqrt(x^2 + 1). Take logarithms, to obtain y = ln(x +- sqrt(x^2 + 1)). Substitute into the original equation, to obtain Asinh(x) = ln(x + sqrt(x^2 + 1)). The negative sign is extraneous. Check: Take the sinh of each side, to obtain x =? sinh(ln(x + sqrt(x^2 + 1))). Employ the definition of the sinh in terms of the exp, to obtain = (exp(ln(x + sqrt(x^2 + 1)) - exp(- ln(x + sqrt(x^2 + 1)))) / 2 Since the logarithm is the inverse of the exponential function, we obtain = ((x + sqrt(x^2 + 1)) - 1 / (x + sqrt(x^2 + 1))) / 2, where the second-term is in the denominator; because, of the negative sign inside of the exponential. Multiply the numerator and denominator of the second-term by the conjugate -- ( x - sqrt(x^2 + 1)) -- of its denominator, to obtain = ((x + sqrt(x^2 + 1)) - (x - sqrt(x^2 + 1)) / (x^2 - (x^2 + 1))) / 2. Since the denominator of the second-term is (- 1), this expression simplifies to = x. QED. |
As a reminder, we provide the
Definition: A function f(x) is said to be odd iff (= if and only if) f(x) + f(-x) = 0.
Theorem: The inverse hyperbolic sine is an odd function.
Proof: Substitute the foregoing logarithmic expression for the inverse hyperbolic sine function into the definition of an odd function, to
obtain
ln(x + sqrt(x^2 + 1)) + ln((-x) + sqrt((-x)^2 + 1)) = ln((x + sqrt(x^2 + 1) (- x + sqrt(x^2 + 1)) = ln(- x^2 + (x^2 + 1))
= ln(1) = 0.
QED.
Theorem: The inverse hyperbolic cosecant is an odd function.
Proof: Substitute the foregoing logarithmic expression for the inverse hyperbolic cosecant function into the definition of an odd function,
to obtain
ln((1 + sqrt(1 + x^2)) / x) + ln((1 - sqrt(1 + (-x)^2)) / (-x)) = ln(((1 + sqrt(1 + x^2)) / x) (- (1 - sqrt(1 + x^2) / x))
= ln(- (1 - (1 + x^2)) / x^2) = ln(1) = 0.
QED.
Sometimes, one is called upon to find the inverse hyperbolic cosine for an angle in [-1, 1]. Of course, one may evaluate the foregoing
formula as a complex function, and let the chips fall where they may. However, since the result simplifies considerably, we provide the
Theorem: The inverse hyperbolic cosine is
| Proof: Step one -- assume that x is in [-1, 1]. Employ the formula from the previous table, namely (We define y here so that
we may use it later.)
y = Arccosh(x) = ln(x + sqrt(x^2 - 1)) = ln(x + i sqrt(1 - x^2)) = (1 / 2) ln(x^2 + (1 - x^2)) + i Atan((sqrt(1 - x^2)) / x), where we have employed the formula for the complex logarithm. Since the logarithm of one is zero, the first term vanishes. We obtain = i Atan(sqrt(1 - x^2) / x). Take the tangent of each side, to obtain tan(y / i) = sqrt(1 - x^2) / x. Square it, to obtain (tan(y / i))^2 = (1 - x^2) / x^2. Anytime that one multiplies (like in squaring) by the unknown, one might -- and indeed, did -- introduce an extraneous solution. Add one, to obtain (sec(y / i))^2 = (tan(y / i))^2 + 1 = (1 - x^2) / x^2 + 1 = 1 / x^2. Take its reciprocal, to obtain (cos(y / i))^2 = x^2. Take its square-root, to obtain cos(y / i) = x, where the negative square-root is extraneous. Now, take the inverse cosine, to obtain Arccosh(x) = y = i Acos(x) x in [-1, 1]. Step two -- for any complex x. The foregoing equation obviously is true for any complex-value of x.. QED. |
Corollary: The inverse circular cosine is
We usually associate the negative sign with negative x and the positive sign with positive x. However, actually, the sign is ambiguous. This formula has applications in tunneling and in the logarithmic-attenuator which employs a wave-guide beyond its cutoff frequency.
A parametric equation of a hyperbola, in the Cartesian product of Complex by Complex, is given
by
(x, y) = (a cosh(t), b sinh(t)) for any t in Complex and any constant (a, b), called the semi-axes, in the Cartesian product of Complex by Complex. In the case of real t, this formula yields only one branch of the hyperbola. To obtain the other branch, change the sign of a.
Passive (that is with time being unknown) navigation employs hyperbolas and hyperboloids of revolution of two-sheets.
Historically, these functions have been called hyperbolic because of this parameterization of a hyperbola.
This discussion applies to any conic section: circle, ellipse, parabola, or hyperbola.
Let the x-axis be the directrix -- usually indicated by the Greek letter lambda.
Let the focal point be at (0, f), with a given f > 0.
Let the eccentricity e >= 0 be given.
Then the conic section is the locus of points P = P(x, y), such that
the distance y from the directrix to P times the eccentricity e equals
the distance sqrt(x^2 + (y - f)^2) from P to the focus. Written as an equation, it is
y e = sqrt(x^2 + (y - f)^2).
Square this equation; then collect terms, to obtain
x^2 + (1 - e^2) y^2 - 2 f y + f^2 = 0.
Definition: The latus rectum is the chord, parallel to the directrix, which passes through the
focus. Hence, the ordinate of each end of the latrus rectum is y = f. Substitute into the first equation above and solve for x, to
obtain the abscissa as x = +- f e. Thus, the length of the latus rectum becomes 2 f e.
Now, you have three cases
Case #1: ellipse e is in [0, 1).
Case #2: parabola e = 1.
Case #3: hyperbola e is in (1, oo).
Case #1. ellipse e is in [0, 1).
Complete the square in y, to obtain
x^2 + (1 - e^2) (y - f / (1 - e^2))^2 = f^2 (1 / (1 - e^2) - 1) = f^2 e^2 / (1 - e^2).
Divide through by the right-hand side, to obtain
(x / (f e / sqrt(1 - e^2)))^2 + ((y - f / (1 - e^2)) / (f e / (1 - e^2)))^2 = 1.
This is an ellipse, centered at (0, f / (1 - e^2)), with semi-axes of
b = f e / sqrt(1 - e^2) in the x-direction
a = f e / (1 - e^2) in the y-direction,
where we have introduced the constants a and b for brevity. Now, we may re-write the equation of the ellipse in its usual standard form
(x / b)^2 + (y / a)^2 = 1.
Let us define another constant c as
c = sqrt(a^2 - b^2) = (f e / (1 - e^2)) e.
Then, we see that the eccentricity e is
e = c / a = sqrt(1 - (b / a)^2).
We may solve this equation for b, to obtain
b = a sqrt(1 - e^2).
Substitute back into the equation of the ellipse, to obtain
x^2 + y^2 (1 - e^2) = a^2 (1 - e^2).
Substitute the eccentricity e into the equation for the major semi-axis a, and solve for the focal-length, to obtain
f = a (1 - e^2) / e = a (b / a)^2 / sqrt(1 - (b / a)^2).
Case #2. parabola e = 1.
Substitute e = 1, to obtain
x^2 - 2 f y + f^2 = 0.
Solve for y, to obtain
y = (1 / (2 f)) (x^2 + f^2).
This is a parabola, concave upwards, with the vertex at (0, f / 2).
The distance between the vertex and the focus is f / 2.
At x = +- f, y = f.
Case #3. hyperbola e is in (1, oo).
First, observe that
(1 - e^2) = - (e^2 - 1).
To keep all signs positive, we will make this substitution. Thus, when we complete the square in y, we obtain
x^2 - (e^2 - 1) (y + f / (e^2 - 1))^2 = - f^2 e^2 / (e^2 - 1).
Divide through by the right-hand side, to obtain
- (x / (f e / sqrt(e^2 - 1)))^2 + ((y + f / (e^2 - 1)) / (f e / (e^2 - 1)))^2 = 1.
This is a hyperbola, centered at (0, - f / (e^2 - 1)), with semi-axes of
b = f e / sqrt(e^2 - 1) in the x-direction
a = f e / (e^2 - 1) in the y-direction,
where we have introduced the constants a and b for brevity. Now, we may re-write the equation of the hyperbola in its usual standard form
- (x / b)^2 + (y / a)^2 = 1.
Let us define another constant c as
c = sqrt(a^2 + b^2) = (f e / (e^2 - 1)) e.
Then, we see that the eccentricity e is
e = c / a = sqrt(1 + (b / a)^2).
We may solve this equation for b, to obtain
b = a sqrt(e^2 - 1).
Substitute back into the equation of the hyperbola, to obtain
- x^2 + y^2 (e^2 - 1) = a^2 (e^2 - 1).
Substitute the eccentricity e into the equation for the a, and solve for the focal-length f, to obtain
f = a ( e^2 - 1) / e = a (b / a)^2 / sqrt(1 + (b / a)^2).
These conic sections may be translated, rotated, or dilitated. With Linear Algebra, it is easy to prove that any quadratic in x and y
a x^2 + 2 b x y + c y^2 + d x + e y + f = 0
is a conic section. These constant coefficients a, b, c, d, e, and f are not related to the previous constants a, b, c, and e. Furthermore, it is an ellipse, parabola, or hyperbola depending upon the sign of its discriminant
a c - b^2
being strictly positive, zero, or strictly negative, respectively. However, to make this presentation self-contained, we provide a brief summary in the following.
Special cases
As already stated in the foregoing, the quadratic is a parabola iff (= if and only if) a c - b^2 = 0.
Since a circle is degenerate, both conditions are required. Thus, the quadratic is a circle iff a - c = b = 0.
Since the trace a + c is invariant under rotation, the quadratic is a rectangular hyperbola iff a + c = 0,
regardless of the value of b.
Eigenvalues
A coordinate system is said to be principal iff in it the value of the coefficient of each cross-product term is zero; that is,
iff b' = 0.
In two-dimensional space, the angle of rotation required to transform a given quadratic to the principal coordinates is trivial. At least, as
compared to doing so in any higher dimensional space. :-) We leave it as an exercise for the reader to find this angle. Hint:
start with phi / 2.
The characteristic polynomial of the quadratic is
x^2 - (a + c) x + (a c - b^2) = 0.
Its roots, called the eigenvalues, are given by the quadratic formula, as
x = ((a + c) +- (a - c) sqrt(1 + (2 b / (a - c))^2)) / 2.
After the suitable rotation -- that which carries the quadratic into the principal coordinate system --, these eigenvalues become the new coefficients of the quadratic terms, namely
a' = ((a + c) +- (a - c) sqrt(1 + (2 b / (a - c))^2)) / 2
b' = 0
c' = ((a + c) -+ (a - c) sqrt(1 + (2 b / (a - c))^2)) / 2.
Since the two roots for x are conjugate, they could be interchanged; but, this first choice is preferable. Because, then, if the original b were zero, the new a' = a and c' = c are equal to the respective old values.
Invariance
The class -- ellipse, parabola, or hyperbola -- of a conic section is invariant under translation, rotation, or dilitation. The specific
shape -- circle of ellipse or rectangular-hyperbola -- is invariant under translation, rotation, or isomorphic dilitation.
Theorem: An ellipse is the locus of points, the combined distance to each of the focal-points is a constant 2a.
Proof:
The distance from a focal-point to a point (x, y) on ellipse
x^2 + y^2 (1 - e^2) = a^2 (1 - e^2)
is
sqrt(x^2 + (y - (+- a e))^2).
Expand it, to obtain
sqrt(a^2 (1 - e^2) - y^2 (1 - e^2) + (y^2 -+ 2 y a e + a^2 e^2)).
Collect terms, to obtain
sqrt(a^2 + y^2 e^2 -+ 2 y a e) = a -+ y e,
which is non-negative; because |y e| <= a. Now, show that their sum is identically 2 a
(a + y e) + (a - y e) ?= 2 a..
All of the terms cancel-out, leaving us with
0 ?= 0.
QED.
Theorem: A hyperbola is the locus of points, the difference of the distances to each of the focal-points is a
constant 2a.
Proof follows the same steps as the foregoing.
The equation of the hyperbola is the same as that of the ellipse; but, the distance is
y e -+ a,
which is non-negative; because |y e| >= a. Now, show that their difference is identically 2 a
(y e + a) - (y e - a) ?= 2 a.
Again, all of the terms cancel-out, leaving us with
0 ?= 0.
QED.
center (0, f / (1 - e^2)).
focal-point (0, f), with f > 0.
a = f e / (1 - e^2).
From which it follows that:
center - focal-point = (f / (1 - e^2)) - (f) = f (1 - (1 - e^2)) / (1 - e^2) = f e^2 / (1 - e^2) = a e.
vertex-1 = center - a = (f / (1 - e^2)) - (f e / (1 - e^2)) = f (1 - e) / (1 - e^2) = f / (1 + e).
vertex-2 = center + a = (f / (1 - e^2)) + (f e / (1 - e^2)) = f (1 + e) / (1 - e^2) = f / (1 - e).
focal-point - vertex-1 = (f) - (f / (1 + e)) = f e / (1 + e). By Kepler's Second Law, the speed proportional to (1 + e).
vertex-2 - focal-point = (f / (1 - e)) - (f) = f e / (1 - e). By Kepler's Second Law, the speed proportional to (1 - e).
speed-1 / distance-2 = (1 + e) / (f e / (1 - e)) = (1 - e^2) / (f e).
speed-2 / distance-1 = (1 - e) / (f e / (1 + e)) = (1 - e^2) / (f e);
distance between the two focal-points = 2 a e, by symmetry from the first item.
This summary reflects the classical Greek view of the planetary motion. The Newton view is developed in the next section.
Based upon the extensive astronomical observations by Tycho Brache (1546-1601), Johannes Kepler (1571-1630) promulgated three laws of planetary motion. They also are applicable to the motion of a satellite (including a moon) around a planet.
The orbit is a conic section -- ellipse, parabola, or hyperbola --, with the parent-body at a focal-point. By implication, the orbit is planar. Newton showed that it is the center of mass of the two bodies, taken together, which is at the focal-point.
The vector from the focal-point to the planet sweeps out a constant area per unit time. Again, so does the vector to the parent-body. This law is just the conservation of angular momentum.
The square of the period is proportional to the cube of the "mean radius" -- the major semi-axis. This law, obviously, is applicable only to the elliptical orbits; because only they are periodic.
First Law.
Now, place the origin at the focal point. Consequently, the distance from the focal point to the point on the ellipse becomes r. Measure the angle theta counter-clockwise from the vertex -- the perihelion. Then, y becomes y = a e + r cos(theta). Substitution into the distance equation yields
r = a - (a e + r cos(theta)) e
Collecting terms, we obtain
r (1 + e cos(theta)) = a (1 - e^2)
r = a (1 - e^2) / (1 + e cos(theta))
p = (1 + e cos(theta)) / (a (1 - e^2))
We define the variable p as the reciprocal of the distance r. It is apparent that such a change of variables will make life easier; because, the theta then will be exclusively in the numerator. The specific symbol 'p' is conventional for the equation of an ellipse. This equation provides the distance r = 1 / p of a planet, as a function of its angle from the perihelion, subject to Kepler's first law. In particular, at perihelion or aphelion, we have, respectively
p(0 or pi) = (1 +- e) / (a (1 - e^2)) = 1 / (a (1 -+ e))
In anticipation of Newton, differentiate this last equation, to obtain
dp / d-theta = - e sin(theta) / (a (1 - e^2))
Now, let us eliminate the angle theta between the two equations. Square each equation and add, employing the trigonometric identity, to obtain the first-order ordinary differential-equation.
(dp / d-theta)^2 + (p - 1 / (a (1 - e^2)))^2 - (e / (a (1 - e^2)))^2 = 0
Second Law. In classical Greece already, it was discovered that the Second Law could be approximated by a projection onto a circle, centered at the other focal point of the ellipse, travels at a constant angular speed. This approximation, obviously, is applicable only to the elliptical orbits; because only they are periodic.
Hence, an object on a constant-speed circular orbit, centered on one focal-point will be seen from the other focal-point as moving at the correct speed, as it passes each vertex of the ellipse.
Third Law. Newton showed that this proportionality constant, in the third law, is (2 pi / sqrt(G M)), where G is the gravitational constant and M = ms^3 / (mp + ms)^2, where ms is the mass of the sun and mp is the mass of the planet. Namely, the period T is given by
T^2 = 4 pi^2 a^3 / (G M).
We may solve for M, to obtain
M = 4 pi^2 a^3 / (G T^2).
This is how astronomers weigh things.
Sir Isaac Newton (1642-1727). Newton's law of motion states that the mass m times the acceleration A of a point object is equal to the force acting upon it. We write it as the equation
m A = F
Acceleration is the second derivative of the radius vector from the fixed central point body of mass M. By Newton's law of gravitation, the force is proportional to the inverse-square of the distance r between the two objects. Thus, the equation becomes
m d^2 R / dt^2 = - G m M R / r^3
The R is the radius-vector from the central body to the orbiting object, G is Newton's gravitational constant, and the t is time. Divide through by m and transpose, to obtain
d^2 R / dt^2 + G M R / r^3 = 0
Now, let us employ the cylindrical coordinate system. Here m and n are the cylindrical unit-vectors. Substitute, to obtain
(m (d^2 r / dt^2 - r (d-theta / dt)^2) + n (r d^2 theta / dt^2 + 2 dr/dt d-theta / dt) + k d^2 z / dt^2) + G M (m r + k z) / r^3 = 0
In the context of Celestial Mechanics, this angle theta is known as the true anomaly. Collect terms in the unit vectors, to obtain
m (d^2 r / dt^2 - r (d-theta / dt)^2 + G M / r^ 2) + n (r d^2 theta / dt^2 + 2 dr/dt d-theta / dt) + k d^2 z / dt^2 = 0
Since each component individually must be zero for their sum to be zero, we have a set of three simultaneous ordinary differential equations
d^2 r / dt^2 - r (d-theta / dt)^2 + G M / r^ 2 = 0
r d^2 theta / dt^2 + 2 dr/dt d-theta / dt = 0
d^2 z / dt^2 = 0
Let us consider these equations in reverse order. With the choice of coordinates so that the z-axis is normal to the plane of the R and dR/dt at the initial-position, the third equation integrates twice, to give us
z = 0
identically, for all time. That is, the orbit is planar. This condition was so obvious that Kepler did not devote a law to it. Let us call it Kepler's zero-th law. The second equation integrates to
r^2 d-theta / dt = c
I omega = r^2 d-theta / dt = (integral, from 0 to r, of (r dr)) d-theta / dt = r^2 omega = c
where c is a constant. Considering the Jacobian of the cylindrical coordinate system, this equation gives us Kepler's second law. The moment of inertia I is the mass times the square of the radius. The derivative of the angle theta wrt (= with respect to) time t is called the angular-speed lower-case omega, which is equal to (2 pi / T), where T is the period. Angular momentum is the product of the moment of inertia and the angular speed; hence, we have shown the conservation of angular momentum. Once Kepler's first law is proven, the radius r will become a known function of the angle theta. Then, this will become a first-order ordinary differential-equation, whose solution will be the equation of time. For the special case of constant r -- that is, a circular orbit --, by substitution into the first equation, we obtain
- r (2 pi / T)^2 + G M / r^2 = 0
which may be solved for the period T, to obtain
T^2 = (4 pi^2 / G M) r^3
This is first half of Kepler's third law. It remains to show that the period is independent of the eccentricity of the orbit. Now, that we have disposed of the circular special-case, which has r a constant, let us define a new variable u as
u = dr / dt
Then
d^2 r / dt^2 = du/dt = (du/dr) (dr/dt) = u du/dr
This technique serves to decrement the order of any differential equation, which does not have explicitly the independent variable, t in our case. Substitute this and the integral of the second equation into the first equation, to obtain
u du/dr - r (c / r^2)^2 + G M / r^2 = 0
It integrates to
(1 / 2) u^2 + (1 / 2) c^2 / r^2 - G M / r = constant
Back-substitute the integral of the second equation, to obtain the energy-integral
(1 / 2) u^2 + (1 / 2) r^2 omega^2 + (- G M / r) = constant
where the first term is the kinetic-energy corresponding to the radial-speed, the second term is the angular kinetic-energy, and the third term is the gravitational potential-energy. That the third term indeed is the potential energy may be verified by taking its gradient. It becomes the inverse-square law of gravitation.
del (- G M / r) = - G M del (1 / r) = - G M (- 1 / r^2) del r = - G M (- 1 / r^2) (R / r) = G M R / r^3
Thus, it satisfies the definition of potential energy as the integral of a force. Since this equation shows that the sum of the kinetic and potential energies is a constant, we have shown the conservation of energy. In passing, we observe that since the variable u had been defined as dr/dt, we may eliminate the dt from the integral of the second equation, namely
u = dr / dt = dr / (r^2 d-theta / c) = (c / r^2) dr / d-theta
Substitute this new expression for u into the integral of the first equation, to obtain
(1 / 2) ((c / r^2) dr / d-theta)^2 + (1 / 2) c^2 / r^2 - G M / r = constant
Multiply through by (2 r^4 / c^2), to obtain
(dr / d-theta)^2 + r^2 - (2 G M / c^2) r^3 - (2 constant / c^2) r^4 = 0
We define the variable p as the reciprocal of r. Then, we have
r = 1 / p
dr / d-theta = - (1 / p^2) dp / d-theta
Substitute, to obtain
(- (1 / p^2) dp / d-theta)^2 + 1 / p^2 - (2 G M / c^2) 1 / p^3 - (2 constant / c^2) 1 / p^4 = 0
(dp / d-theta)^2 + p^2 - (2 G M / c^2) p - (2 constant / c^2) = 0
(dp / d-theta)^2 + (p - G M / c^2)^2 - ((G M / c^2)^2 + 2 constant / c^2) = 0
This is a first-order ordinary differential-equation in the radius r = 1 / p as a function of the angle theta. By comparing with the differential equation of an ellipse, we see that
c^2 = G M a (1 - e^2)
constant = - G M / (2 a)
And we have Kepler's first law. Substituting the value of the constant into the energy equation, we obtain the conservation of energy, as
(1 / 2) u^2 + (1 / 2) r^2 omega^2 + (- G M / r) + G M / (2 a) = 0
radial-KE + angular-KE + PE + BE = 0
The terms on the left-hand side are the radial component of kinetic energy, the angular component of kinetic energy, the gravitation potential energy, and the binding energy. It checks for the special case of a circular orbit: Since the radius is a constant, the variable u is identically zero. Since the centrifugal force omega^2 r is equal to the gravitational force G M / r^2, the angular speed is constant and equal to sqrt(G M / r^3). Finally, the radius is equal to a. Substitution of these values into the conservation of energy equation yields zero, as it should. Alternative usage is to define the potential energy as a positive value, then place a negative sign in front of it. Likewise, the binding energy may be defined as a negative value. Then place a negative sing in front of it, as well.From the conservation of energy, we see that the gravitational force is conservative. Actually, any central force is conservative.
Let us substitute the value of c into the integral of the second equation, to obtain the conservation of angular momentum, as
I omega = r^2 d-theta / dt = sqrt(G M a (1 - e^2))
where I is the moment of inertia and omega is the angular speed. Now, that we have established Kepler's first law, we may substitute the r of the ellipse, to obtain
(a (1 - e^2) / (1 + e cos(theta)))^2 d-theta / dt = sqrt(G M a (1 - e^2))
(1 - e^2)^(3 / 2) / (1 + e cos(theta)))^2 d-theta / dt = sqrt(G M / a ^3)
(1 - e^2)^(3 / 2) / (1 + e cos(theta)))^2 - 1) d-theta / dt + d-theta / dt = sqrt(G M / a^3)
Integrate with respect to time t, to obtain the indefinite integral
- 2 Atan(e sin(theta) / (1 + sqrt(1 - e^2) + e cos(theta))) - e sqrt(1 - e^2) sin(theta) / (1 + e cos(theta)) + theta = sqrt(G M / a^3) t
The right-hand side is known as the mean anomaly. The derivation follows a well-known algorithm; but it is very tedious. However, I cheated. I employed the symbolic mathematics program Derive (version 4) to evaluate this quadrature. For one revolution of the planet about its sun, the angle theta is 2 pi. Thus, we obtain
0 + 2 pi = sqrt(G M / a^3) T
T^2 = (4 pi^2 / (G M)) a^3
This is Kepler's third law. The period T , indeed, is independent of the eccentricity of the ellipse, as Kepler had stated. The first two terms of the indefinite integral depend exclusively upon the eccentricity e of the orbit.
- 2 Atan(e sin(theta) / (1 + sqrt(1 - e^2) + e cos(theta))) - e sqrt(1 - e^2) sin(theta) / (1 + e cos(theta)) [exact] -->
-2 Atan(e sin(theta) / (2 + e cos(theta)) - e sin(theta) / (1 + e cos(theta)) [e^2 neglected] -->
--> - 2 e sin(theta) [e neglected]
as e^2 or even e is neglected. Thus, we see that it is approximately sinusoidal. This same expression may be interpreted as providing the angular position theta -- the true anomaly --, measured from the perihelion, of the planet in its orbit as a function of the time t -- proportional to the mean anomaly.
There is a saying, "If you want something done right, do it yourself!" The foregoing is good; but, we can do better.
Consider an ellipse, centered at the origin, with its major axis along the x-axis. In rectangular Cartesian coordinates (x, y), in standard form, its equation is
(x/a)^2 + (y/b)^2 = 1.
Consider an angle E, at the origin, as usual, measured counter-clockwise from the x-axis. Let
x = a cos(E)
y = b sin(E).
Then substitution yields
(a cos(E) / a)^2 + (b sin(E) / b)^2 = 1.
In view of the identity, the foregoing is an identity in E. Hence, the aforementioned pair of equations provide a parametric definition of the ellipse. Recall that the focal-point is at (a e, 0). The sine and cosine of theta are
r sin(theta) = y = b sin(E) = a sqrt(1 - e^2) sin(E)
r cos(theta) = x - a e = a cos(E) - a e = a (cos(E) - e).
To find the radius r, square and add, to obtain
r^2 =
= a^2 ((1 - e^2) (1 - (cos(E))^2) + (cos(E) - e)^2) =
= a^2 ((1 - e^2 - (cos(E))^2 + e^2 (cos(E))^2) + ((cos(E))^2 - 2 e cos(E) + e^2)) =
= a^2 (1 - 2 e cos(E) + e^2 (cos(E))^2) =
= a^2 (1 - e cos(E))^2.
Now, take square-roots, to obtain the radius r in terms of the eccentric anomaly E, as
r = a (1 - e cos(E)).
Substitute back into the equations for the sine and cosine of theta, to obtain
(1 - e cos(E)) sin(theta) = sqrt(1 - e^2) sin(E)
(1 - e cos(E)) cos(theta) = cos(E) - e.
Finally, let us compute the angle theta, in terms of the angle E. Employ the half-angle formula for the tangent function
tan(theta / 2) =
= sin(theta) / (1 + cos(theta)) =
= sqrt(1 - e^2) sin(E) / ((1 - e cos(E)) + (cos(E) - e)) =
= sqrt(1 - e^2) sin(E) / ((1 - e) (1 + cos(E))) =
= sqrt((1 + e) / (1 - e)) sin(E) / (1 + cos(E)) =
= sqrt((1 + e) / (1 - e)) tan(E / 2),
where the angles theta/2 and E/2 always are in the same quadrant. Square this equation, to obtain
(tan(theta / 2))^2 = ((1 + e) / (1 - e)) (tan(E / 2))^2.
Employing the identity, add one to each side of the equation, to obtain
(sec(theta/ 2))^2 =
= 1 + ((1 + e) / (1 - e)) (tan(E / 2))^2 =
= ((1 - e) + (1 + e) (tan(E / 2))^2) / (1 - e).
Now, find the derivative of the tangent of (theta / 2), as
(sec(theta / 2))^2 d-theta / dE = sqrt((1 + e) / (1 - e)) (sec(E / 2))^2.
Divide by the previous equation, and multiply the numerator and denominator by (cos(E / 2))^2, to obtain
d-theta / dE = sqrt(1 - e^2) / ((1 - e) (cos(E / 2))^2 + (1 + e) (sin(E / 2))^2) =
= sqrt(1 - e^2) / (1 - e cos(E)),
where we have employed an identity and a double-angle formula. At long last, we want to integrate the expression
r^2 d-theta =
= (a (1 - e cos(E)))^2 (sqrt(1 - e^2) / (1 - e cos(E))) dE =
= a^2 (sqrt(1 - e^2) (1 - e cos(E)) dE.
Its integral is
a^2 (sqrt(1 - e^2) (E - e sin(E)) plus a constant.
In passing, we note that in the field of Celestial Mechanics, the angle theta is called "true anomaly" and the angle E is called the "eccentric anomaly". Hence, the integral of the conservation of angular momentum becomes
a^2 (sqrt(1 - e^2) (E - e sin(E)) = sqrt(G M a (1 - e^2))
E - e sin(E) = sqrt(G M / a^3) t = mean anomaly.
This last equation is known as Kepler's equation. The constant of integration is zero; because, by convention, we measure time -- and, thus, the mean anomaly -- from the perihelion. This equation in the integral counterpart to the usual differential statement of the Second Law of Kepler. A historical question: "Did Kepler actually perform this integration, or was the equation so named to honor Kepler?" Perhaps, Kepler just discerned this equation from the observations of Tycho Brache. For one revolution of the planet about its sun, the angle theta is 2 pi. Thus, we obtain
2 pi - 0 = sqrt(G M / a^3) T
T^2 = (4 pi^2 / (G M)) a^3
This is Kepler's third law. The period T , indeed, is independent of the eccentricity of the ellipse, as Kepler had stated.
We may express the sine of the eccentric anomaly E in terms of the true anomaly theta, as
sin(E) =
= 2 sin(E/2) cos(E/2) = 2 tan(E/2) / (sec(E/2))^2 = 2 tan(E/2) / (1 + (tan(E/2))^2) =
= 2 sqrt((1-e)/(1+e)) tan(theta/2) / (1 + ((1-e)/(1+e)) (tan(theta/2))^2) =
= 2 sqrt(1-e^2) sin(theta/2) cos(theta/2) / ((1+e) (cos(theta/2))^2 + (1-e) (sin(theta/2))^2) =
= sqrt(1-e^2) sin(theta) / (1 + e cos(theta))
We have made use of the half-angle formula for the tangent function and the identity. Now, we may substitute into the Kepler's equation, to obtain
2 atan(sqrt((1-e)/(1+e)) tan(theta/2)) - e sqrt(1-e^2) sin(theta) / (1 + e cos(theta)) = sqrt(G M / a^3) t = mean anomaly
The left-hand side is equivalent to that which was provided by the program Derive.
Kepler's equation is
tan(theta / 2) = sqrt((1 + e) / (1 - e)) tan(E / 2)
sqrt(G M / (a^3) t = mean anomaly = E - e sin(E)
E = (mean anomaly) + 2 sum, i running from 1 to oo, of ((J-sub-i (i e) / i) sin(i (mean anomaly)))
tan( (projected anomaly) - (true anomaly of the vernal equinox)) = cos( (obliquity of the ecliptic)) tan( (true anomaly) - (true anomaly of the vernal equinox))
equation of time = ((mean anomaly) - (true anomaly)) + ((true anomaly) - (projected anomaly)) = (mean anomaly) - (projected anomaly) - constant
where E is the eccentric anomaly and half of it is to be taken in the same quadrant at half of the true anomaly theta. The eccentric anomaly may be expressed as a Fourier series, in terms of the mean anomaly. Only the sine terms are present. Their coefficients are the Bessel functions of the first-kind, divided by the subscript. For the purpose of time-keeping, we project the true anomaly from the ecliptic to the equator, employing a Mollweide formula. Again, the projected anomaly is to be taken in the same quadrant as the true anomaly. Observe that the projected anomaly has twice the frequency of the true anomaly. Also, observe that the projected anomaly is measured from the vernal equinox, while each of the other anomalies is measured from the perihelion. Then, the equation of time is the sum of the Keplerian trajectory and the spherical projection. The constant of integration is chosen to make the equation of time be zero at a specified epoch, usually the first of January of some convenient year. By convention, the equation of time is to be subtracted from the apparent solar time -- that shown by a sun-dial or astrolabe -- to obtain the mean solar time, from which the time-zone correction has to be subtracted and daylight saving time has to be added to obtain the civil time. There are two equations which involve the inverse tangent, with the specification that its value has to be in the same quadrant. It may be implemented as
f(a, x) = atan(a * tan(x)) + (x - atan(sign(a) * tan(x))
where the last two terms place the function into the same quadrant as the argument x.
I have implemented this computation as a Microsoft Excel 2000 spread-sheet. (it is only 62 KBytes.) It accepts the orbital elements of a planet as input: the eccentricity of the orbit, the obliquity -- the angle (in radians) between the rotation-axis of the planet and the orbital revolution-axis, the angle (in radians) between the vernal equinox and the perihelion, the angle (in radians) between 1 January and the perihelion, and the constant of integration (also in radians). And, for comparison, a table of the official daily equation of time. This program computes the daily equation of time. Here is a spread-sheet -- regrettably dead -- of the results, including the two graphs. Chart 1 shows the precision of the six-harmonic Fourier analysis. Chart 2 shows the computed values -- slightly larger amplitude -- and the official values. Also, the error -- the small-amplitude curve. The standard deviation of the error is computed as 0.548 minutes of time = 32.9 seconds of time. The computation should be good to less than a second; but, I have been unsuccessful in finding a consistent set of orbital parameters and a a table the equation of time. A detailed description of this spread-sheet follows. This is a description of Sheet1. Sheet2 and Sheet3 are blank; but if you delete either of them, the whole thing vanishes. Excel must have some hidden internal use for these extra sheets.
input data. The orbital elements change; because the Earth is not a rigid spherically-symmetrical body and because both our Moon and Jupiter make it more than a two-body problem. With only minor changes, this spread-sheet could be applied to any planet about any star.
B1 is the step-size. It determines how many harmonics it will be possible to compute. The relationship is (step-size (in degrees)) * (highest harmonic) = 90 degrees.
B2 is the eccentricity of the orbit of the planet. It is almost a constant.
B3 is the obliquity of the orbit. This is the angle (in radians) between the rotation-axis of the planet and the orbital revolution-axis. It varies only very slowly.
B4 is the angle (in radians) between the vernal equinox and the perihelion. It varies on an approximately 20000 year cycle, known as the precession of the equinoxes.
B5 is the constant of integration (also in radians). It depends upon the epoch with respect to which the orbital elements are stated.
B6 is the angle (in radians) between 1 January and the perihelion. It varies on the four-thousand-year leap-year cycle.
F6 is a copy of the standard-deviation of the error (in minutes of time) in the final computation of the equation of time, from cell AC8. It is displayed here to provide a visual feedback of the effect of any changes to the foregoing input parameters.
Newton-Kepler computation of the equation of time.
column A This column provides the vertical-axis for the columns B through W, inclusive.
A10 is the initial angle (in degrees). It has to begin at least one step-size before zero; because the subsequent Fourier analysis employs a step-size from the previous to the following rows. We actually go much farther back to be able to verify the inverse tangent computations in columns C and E.
A11 has the master incremental computation.
all subsequent rows are a copy of A11.
column B is the true-anomaly (in radians). B10 is the master computation. All subsequent rows are a copy of B10.
column C is the eccentric-anomaly (in radians). C10 is the master computation. All subsequent rows are a copy of C10.
column D is the mean-anomaly (in radians). D10 is the master computation. All subsequent rows are a copy of D10.
column E is the projected-anomaly (in radians). E10 is the master computation. All subsequent rows are a copy of E10.
column F is the equation of time (in radians). F10 is the master computation. All subsequent rows are a copy of F10.
column G adjusts the equation of time (in radians) by the constant of integration in cell B5. G10 is the master computation. All subsequent rows are a copy of G10.
column H converts the equation of time to time in minutes. H10 is the master computation. All subsequent rows are a copy of H10.
Fourier analysis and check. Columns I through U, inclusive follow the same pattern. They will be described in logical order.
The harmonics.
I9 is zero. J9 is the master incremental computation. Cells K9 through O9, inclusive, are a copy of K9.
P9 is one. Q9 is the master incremental computation. Cells R9 through U9, inclusive, are a copy of Q9. They are analogous to the preceding entry.
I11 is the master computation of the weighted integrand. It is copied horizontally through P11, which is modified to have the sine -- instead of the cosine -- function. Then copied horizontally through U11. The row I11 through U11 is copied downward to one less than column A.
I8 is the master computation of the sum of these columns, over the semi-closed interval [0, 360) of column A. It is copied (and slightly modified) to J8, which is copied through U8.
I7 through O7 are the computation of the amplitude of each of the harmonics. I7 is unique. J7 is the master, which is copied through O7.
P7 through U7 are the computation of the phase-angle of each of the harmonics. P7 is the master, which is copied through U7.
The check is performed in columns V and W. The sine and cosine functions are orthonormal.
V10 is the master computation of the Fourier synthesis. It is copied downward to the bottom.
W10 is the master computation of the error (multiplied by 1000, for better visual scaling) of column V with respect to column H. It is copied downward to the bottom.
W7 is the computation of the standard-deviation of these errors. It is only an approximate computation; because, we did not weight it. But, it is adequate as an internal check.
The Chart1 is a graph of columns H and W versus D. It is a visual demonstration of the accuracy and precision of our Fourier analysis.
Fourier synthesis and check.
column X is days for one year. X10 is one. X11 is the master incremental computation. All subsequent rows are a copy of X11.
column Y is adjusted for the value in B6 of 1 January (in days) from the perihelion. Y10 is the master computation. All subsequent rows are a copy of Y10.
column Z converts Y to radians. Z10 is the master computation. All subsequent rows are a copy of Z10.
column AA is the Fourier synthesis, employing the amplitudes in I7 through O7 and the phase-angles in P7 through U7, with the independent variable -- the angle -- in column Z. AA10 is the master computation. All subsequent rows are a copy of AA10.
column AD is the month. These are entered manually.
column AE is the day of the month. AE10 is one. AE11 is the master incremental computation. Most subsequent rows are a copy of AE11. The exceptions are each first day of a month, which has a value of one.
Official equation of time.
column AF is the manually-entered official equation of time -- minutes, with a minus sign for "slow".
column AG is the manually-entered official equation of time -- seconds, with (if the minutes in AF are zero) a minus sign for "slow".
column AB is the official equation of time, in minutes. AB10 is a master, which is copied down into most cells. However, the cells corresponding to those rows with AF having a value of zero employ their own master.
AC10 is the master computation of the error of column AA with respect to column AB. It is copied downward to the bottom.
AC8 is the computation of the standard-deviation of these errors (in minutes). This computation is exactly accurate; because, it does not require any weighting.
The Chart2 is a graph of columns AA, AB, and AC versus column X. It is a visual demonstration of the precision of our Fourier synthesis of the equation of time.
Reduced mass. By Newton's laws of motion, it is obvious that it is the center of mass -- rather than the sun -- which remains stationary. Hence, that it is the center of mass which is at the focus of the ellipse of the orbit. Can we find some effective mass -- called the reduced mass -- which will satisfy the orbital problem? Given the
mp = mass of the planet
ms = mass of the sun
r1 + r2 = the dynamic distance between them
f = G mp ms / (r1 + r2)^2 = force acting upon each of them
Let
mo = reduced mass at the focus
m1 = reduced mass of the planet; i.e., the central-mass, as seen by the planet
m2 = reduced mass of the sun; i.e., the central-mass, as seen by the sun
Then we have the relationships
mp r1 = ms r2, by definition of center of mass
f = G m1 mo / r1^2 = force acting upon the planet
f = G m2 mo / r2^2 = force acting upon the sun
Solve each force equation for the distance, to obtain
r1 + r2 = sqrt(G mp ms / f)
r1 = sqrt(G m1 mo / f)
r2 = sqrt(G m2 mo / f)
Substitution gives us
mp sqrt(G m1 mo / f) = ms sqrt(G m2 mo / f)
sqrt(G m1 mo / f) + sqrt(G m2 mo / f) = sqrt(G mp ms / f)
Multiply through by sqrt(f / G), to obtain
mp sqrt(m1 mo) = ms sqrt(m2 mo)
sqrt(m1 mo) + sqrt(m2 mo) = sqrt(mp ms)
Now, since we have eliminated both the force f and the two distances r1 and r2, we observe that there are no time-dependent variables remaining. Hence, we see that some fixed values for m0, m1, and m2 are possible. Square the first of these equations and divide through by mo, to obtain
mp^2 m1 = ms^2 m2
Solve the second equation for mo, to obtain
mo = mp ms / (sqrt(m1) + sqrt(m2))^2
Solve the first equation for m2 and substitute into the second, to obtain
mo = mp ms / (sqrt(m1) + (mp / ms) sqrt(m1))^2 = mp ms / (m1 * (1 + mp / ms)^2) = (mp ms / (mp + ms)^2) (ms^2 / m1)
Solve for m1, to obtain
m1 = (mp ms / (mp + ms)^2) (ms^2 / mo)
m2 = (mp ms / (mp + ms)^2) (mp^2 / mo)
The second of these equations either is by analogy or by following the same steps for m2 as we have for m1. We may take
mo = mp ms / (mp + ms)
m1 = ms^2 / (mp + ms) = (ms / mp) mo
m2 = mp^2 / (mp + ms) = (mp / ms) mo
as the reduced masses. This is not a unique choice: The right-hand side of the first equation could have been multiplied by an arbitrary constant k. Then, the right-hand sides of each of the other two equations divided by the same constant. That is
mo = mp ms / (mp + ms) k
m1 = ms^2 / (mp + ms) = (ms / mp) mo / k
m2 = mp^2 / (mp + ms) = (mp / ms) mo / k
Observe that each of the products mo m1 and mo m2 would remain independent of this constant. Substitution into the force equations checks that indeed, each force is the same force f as that which we had acting between the planet and the sun. The differential equations for the two bodies, then, become
mp d^2 R1 / dt^2 = G m1 mo R1 / r1^3
ms d^2 R2 / dt^2 = G m2 mo R2 / r2^3
Divide by the mass of the object and substitute the reduced masses, to obtain
d^2 R1 / dt^2 = G Ms R1 / r1^3, where Ms = ms^3 / (ms + mp)^2
d^2 R2 / dt^2 = G Mp R2 / r2^3, where Mp = mp^3 / (ms + mp)^2
Each of these equations is the same as that which we have solved in the foregoing.
Point mass. By brute-force integration, Newton showed that the gravitational force seen outside of a spherically-symmetrical body is that of a point-mass, whose value is that of the body, at its center. At any point inside of the body, the mass outside of the sphere which passes through that point has no force. These are triple-integrals -- sometimes assigned as an extra-credit problem for Calculus students. Carl Friedrich Gauss (1727-1855) derived the Gauss theorem and from it the flux-divergence theorem. The foregoing two results are an obvious consequence of the flux-divergence theorem.
Converse. We have shown that the Newton's laws of motion and gravitation, as embodied in the differential equation, imply the Kepler's three laws of planetary motion. The question is, "is the converse true, as well?" The short answer is, "No." However, the question deserves a longer discussion. Given Newton's laws of motion, we already have the law of conservation of angular momentum. Then, Kepler's second law may be reinterpreted as implying a central force. The orbits of many of the planets may be approximated as being circular. Then, Kepler's third law implies an inverse-square law of force. And, as we have shown in the foregoing, Newton's law of gravitation does imply Kepler's three laws. However, Kepler's third law does not say anything about the force, except at the instantaneous location of a planet. Conceivably, the force could be time-dependent or have any value between the orbits of adjacent planets. Then, by invoking Occam's razor, we may say that the Kepler's three laws only suggest Newton's law of gravitation.
A summary of astronomy. We have shown that
Newton's laws of motion and gravitation: d^2 R / dt^2 + G M R / r^3 = 0
imply
Kepler's three laws of planetary motion.
the planetary orbit is planar: z = 0
the orbit is a conic section, with the center of mass of the sum-planet system at the focus: 1 / r = p = (1 + e cos(theta)) / (a (1 - e^2))
the conservation of angular momentum. I omega = r^2 d-theta / dt = sqrt(G M a (1 - e^2))
the period T is independent of the eccentricity and is T^2 = (4 pi^2 / (G M)) a^3
for a being the semi-major axis of the orbit of the planet, Ms = ms^3 / (ms + mp)^2
for a being the semi-major axis of the orbit of the sun, Mp = mp^3 / (ms + mp)^2
the conservation of energy. radial-KE + angular-KE + PE + BE = (1 / 2) u^2 + (1 / 2) r^2 omega^2 + (- G M / r) + G M / (2 a) = 0
the equation of time depends exclusively upon the eccentricity of the orbit and is approximately sinusoidal:
sqrt(G M / (a^3) t - theta = ??? see note in the main text. The following three values are wrong.
= - 2 Atan(e sin(theta) / (1 + sqrt(1 - e^2) + e cos(theta))) - e sqrt(1 - e^2) sin(theta) / (1 + e cos(theta)) [exact] -->
--> -2 Atan(e sin(theta) / (2 + e cos(theta)) - e sin(theta) / (1 + e cos(theta)) [e^2 neglected] -->
--> - 2 e sin(theta) [e neglected]
the reduced masses are
mo = mp ms / (mp + ms) [the reduced mass at the focus]
m1 = ms^2 / (mp + ms) = (ms / mp) mo [the central-mass, as seen by the planet]
m2 = mp^2 / (mp + ms) = (mp / ms) mo [the central-mass, as seen by the sun]
the gravitational force seen outside of a spherically-symmetrical body is that of a point-mass at its center; zero inside
And, conversely, the Kepler's three laws only suggest Newton's law of gravitation.
These derivative formulae are obvious from the definition of the functions. For any x in Complex we have the following derivative formulae:
Let x = sinh(y) and differentiate it to obtain dx / dy = cosh(y). Employ the appropriate identity to obtain dx / dy = sqrt(1 + (sinh(y))^2). Then dy / dx = 1 / sqrt(1 + (sinh(y))^2). Thus, we have obtained the first of the derivative formulae of the inverse hyperbolic trigonometric functions
Their primary utility is as antiderivatives.
The Riemann Integral is an ant-derivative, thus each of these integral formulae may be verified by differentiation. For any x in Complex we have the integral formulae:
The MacLaurin's series for the hyperbolic sine and cosine may be obtained from their definitions and the series for the exponential function.
Take the infinite Geometric series
1 / (1 - x) = 1 + x + x^2 + x^3 + ... + x^n + ....
for any x, in Complex, whose the absolute value is less than one. Replace x by x^2 and integrate to obtain the MacLaurin's series for the hyperbolic arctangent.
The hyperbolic infinite products are not interesting. They have to be obtained from the corresponding circular infinite products, employing the purely imaginary relationship between the corresponding functions.
The purely imaginary counterpart of the Hyperbolic Trigonometric functions is called the Circular Trigonometric functions.
Copyright © 1997-8, 9, 2000, 2, 3, 4, 6 R. I. 'Scibor-Marchocki last modified on Thursday 13-th April 2006. Typographical correction Su 05-th November 2006.