Planetary Orbits
----------------
We will derive some of the basic formulas for planetary orbits from
first principles. Let r be the distance of the satellite from the
primary and @ measure the angle relative to the primary. Assume that
the satellite has negligible mass compared to the primary.
Derivatives in Polar Coordinates
--------------------------------
Let us start out with the formula for x in polar coordinates and take
a couple of derivatives with respect to time:
x = r cos(@)
x' = r' cos(@) - r sin(@) @'
2
x'' = r'' cos(@) - 2 r' sin(@) @' - r cos(@) @' - r sin(@) @''
if @ = 0, that is, the coordinates are rotated so that the y component
of the position is 0, then x'' represents the radial acceleration of the
satellite:
-GM 2
--- = r'' - r @' [1]
r^2
if @ = pi/2, that is, the system is rotated so that the x component of
the position is 0, then x'' is the acceleration perpindicular to the
radius:
0 = 2 r' @' + r @'' [2]
Kepler's Second Law (Equal Area in Equal Time)
----------------------------------------------
Multiply the right hand side of [2] by r/2 and you have the derivative
of 1/2 r^2 @', which is the area swept out by the satellite per unit
time. Equation [2] says that 1/2 r^2 @' is a constant
1 2
- r @' = k [3]
2 1
Equation [3] is precisely Kepler's Second Law of Planetary Motion; a
planet sweeps out equal areas in equal times.
Conservation of Energy
----------------------
Let us look at the square of the velocity, which is proportional to the
kinetic energy of the satellite
2 2 2 2
v = r' + r @' [4]
Take half the time derivative of [4], then apply [2] and then [1]:
2 2
v v' = r' r'' + r r' @' + r @' @''
2
= r' r'' - r r' @'
-GM
= r' --- [5]
r^2
If we integrate [5] and rearrange, we get
1 2 GM
- v - -- = k [6]
2 r 2
When [6] is multiplied by the mass of the satellite, it becomes the law
of conservation of energy. 1/2 mv^2 is the kinetic energy and -GMm/r is
the potential energy of the satellite.
If k_2 is positive, the satellite has enough kinetic energy to escape
the primary's gravity. If k_2 is negative, the satellite will stay in
orbit around the primary.
Using Initial Information
-------------------------
So now we have two constants. Both can be computed from the initial
position, p, and velocity, v, of the satellite relative to the primary:
1
k = - |p x v| [3a]
1 2
1 2 GM
k = - |v| - --- [6a]
2 2 |p|
Kepler's First Law (Elliptical Orbits)
--------------------------------------
Dividing [4] by @'^2 and rearranging, then using [3] and [6] to supply
@' and v in terms of r, gives:
r' 2 v 2 2
( - ) = ( - ) - r
@' @'
GM 2 k_1 2 2
= 2(k + --)/( ----- ) - r
2 r r^2
k_2 4 GM 3 2
= ------- r + ------- r - r [7]
2 k_1^2 2 k_1^2
If we divide [7] by r^4 and change variables to s = 1/r, we get
s' 2 2 GM k_2
( - ) = -s + ------- s + -------
@' 2 k_1^2 2 k_1^2
GM 2 GM 2 k_2
= (-------) - (s - -------) + -------
4 k_1^2 4 k_1^2 2 k_1^2
2 2
= c - (s-b) [8]
where b = GM/(4 k_1^2) and c^2 = b^2 + k_2/(2 k_1^2). Thus, rearranging
the terms in [8] and taking square roots, we get
(s-b)'
@' = -------------------
sqrt(c^2 - (s-b)^2)
u'
= ----------- [9]
sqrt(1-u^2)
where u = (s-b)/c. The right hand side of [9] is the derivative of
acos(u). Thus, if we integrate [9] and solve for r, we get
1
r = ---------------- [10]
b + c cos(@-k_3)
Equation [10] is the polar equation for an ellipse with a focus at the
origin. k_3 is the angle of periapsis, usually taken to be 0. Since
we know that the minimum distance is a(1-e) and the maximum is a(1+e),
we get that
a(1-e^2)
r = ------------ [10a]
1 + e cos(@)
where a(1-e^2) = 1/b and e = c/b. Equation [10a] is Kepler's First Law
of Planetary Motion; the orbit of a planet is an ellipse with the Sun at
one focus.
In terms of the basic orbital constants from [3a] and [6a],
GM
a = - ----- [10b]
2 k_2
and
k_1 2
e = sqrt( 1 + 8 k_2 ( --- ) ) [10c]
GM
Solving for @ yields
1 4 k_1^2
cos(@) = - ( ------- - 1 ) [10d]
e GM r
Vis-Viva Equation
-----------------
Equation [4] says that when r' = 0, v = r @'. Equation [3] says that
r @' = 2 k_1/r. Thus, when r' = 0,
k_1
v = 2 --- [11]
r
Combining [11] with [6], we get that when r' = 0,
k_1^2 GM
2 ----- - -- = k [12]
r^2 r 2
multiplying [12] by r^2 and rearranging, we get that when r is maximum
or minimum,
2 2
k r + GMr - 2 k = 0 [13]
2 1
The sum of the roots of [13] is -GM/k_2. Thus, the semimajor axis, the
average of the minimum and maximum distance of an elliptical orbit, is
a = -GM/(2k_2). Therefore, we get
GM
k = - -- [14]
2 2a
Using [14] in [6], we get the Vis-Viva equation
2 2 1
v = GM ( - - - ) [15]
r a
Kepler's Third Law (Period-Distance Relation)
---------------------------------------------
The product of the roots of [13] is -2k_1^2/k_2 while the product of the
minimum and maximum distance of an elliptical orbit is a^2(1-e^2). With
[14] we get that
2 k_2 2 2 GM 2 2 GMa 2
k = - --- a (1-e ) = -- a (1-e ) = --- (1-e ) [16]
1 2 4a 4
We also know that k_1 is the constant rate that area is swept out along
the elliptical orbit. Over one whole period P, the area swept out is
pi a^2 sqrt(1-e^2), thus
2 2
k P = pi a sqrt(1-e ) [17]
1
Thus, squaring [17], using [16], and cancelling, we get
2 2 3
GM P = 4 pi a [18]
Equation [18] verifies Kepler's Third Law of Planetary Motion; the
square of the period of a planet is proportional to the cube of the
semimajor axis of its orbit.
Time Dependence
---------------
Combining [10d] and [18] and using the results from Kepler's Equation,
we can compute the position in orbit at any time.
Rob Johnson