Orbit Equation, Conic Sections, and Kepler's First Law

(Topic)

Orbit Equation, Conic Sections, and Kepler’s First Law

M07-05 classified gravitational motion using orbital energy.

M07-06 introduced the effective potential and radial turning points.

M07-07 showed that angular momentum conservation gives constant areal velocity.

We now derive the actual shape of the orbit in the orbital plane.

For Newtonian inverse-square gravity, the result is

|------------------------| |r(𝜃) = -------p--------.| | 1 + ecos(𝜃 − 𝜃0) | --------------------------

(1)

This is the polar equation of a conic section with the attracting body at a focus.

The parameters are:

p, the semilatus rectum,
e, the eccentricity,
𝜃₀, the direction of periapsis.

The value of

e

(2)

determines the orbit type:

|--------------------------| | e = 0 : circle, | | | | 0 < e < 1 : ellipse, | | e = 1 : parabola, | | e > 1 : hyperbola. | ----------------------------

(3)

Figure 1. Orbital geometry in the plane. The polar angle is measured from the periapsis direction, and the attracting body lies at the focus of the conic.

1 Start from the radial equation

M07-06 gave the radial equation for Newtonian central gravity:

|------------| | 2 | ¨r = h--− μ-,| -----r3---r2--

(4)

where

μ = GM

(5)

in the fixed-central-body approximation and

h = r2𝜃˙

(6)

is the conserved specific angular momentum.

Our goal is not

r (t).

(7)

Instead, we seek

r(𝜃),

(8)

the geometric orbit.

2 Reciprocal-radius substitution

Introduce

|-------1--| |u(𝜃) = --.| --------r--|

(9)

Then

r = 1. u

(10)

Since

h = r2 ˙𝜃,

(11)

we have

|--------| |˙𝜃 = hu2.| ----------

(12)

Use the chain rule:

dr-˙ r˙= d𝜃𝜃.

(13)

Because

−1 r = u ,

(14)

dr 1 du ---= − -2---. d𝜃 u d𝜃

(15)

Therefore

ṙ	= −hu²	(16)
	= −h.	(17)

Figure 2. Binet’s substitution replaces the radial coordinate r by the reciprocal radius u = 1∕r and uses angular momentum to exchange time derivatives for derivatives with respect to orbital angle.

3 Second radial derivative

Differentiate

˙r = − h du d𝜃

(18)

with respect to time.

Since

h

(19)

is constant,

r	= −h	(20)
	= −h𝜃.	(21)

Using

˙𝜃 = hu2,

(22)

we obtain

|---------------| | 2 2d2u | ¨r = − h u --2. | -----------d𝜃---

(23)

4 Substitute into Newton’s radial equation

The radial equation is

h2 μ ¨r = -3-− -2. r r

(24)

Since

r = 1∕u,

(25)

1 3 1 2 r3 = u , r2 = u .

(26)

Therefore

2 − h2u2 d-u-= h2u3 − μu2. d𝜃2

(27)

For

u ⁄= 0,

(28)

divide by

2 2 − h u :

(29)

|--------------| |d2u μ | |--2-+ u = -2.| -d𝜃---------h---

(30)

This is the Binet equation specialized to Newtonian inverse-square gravity.

5 General Binet equation

For a general central radial acceleration

f (r ),

(31)

defined by

¨r − r𝜃˙2 = f (r),

(32)

the corresponding orbit equation can be written

|--------------------| |d2u- f-(1∕u) | -d𝜃2-+-u-=-−--h2u2--.-

(33)

For Newtonian gravity,

f(r) = − μ-, r2

(34)

f(1∕u ) = − μu2,

(35)

which gives the constant right-hand side

μ∕h2.

(36)

This special simplification is what makes inverse-square gravity produce conic sections.

6 Solve the orbit differential equation

We must solve

d2u μ --2-+ u = -2. d𝜃 h

(37)

A particular solution is

μ-- up = h2.

(38)

The homogeneous equation

′′ uh + uh = 0

(39)

has solution

uh = A cos 𝜃 + B sin 𝜃.

(40)

Therefore

-μ- u = h2 + A cos 𝜃 + B sin 𝜃.

(41)

The two sinusoidal constants can be combined into one amplitude and one phase:

|-----μ--------------------| |u = --2 [1 + ecos(𝜃 − 𝜃0)]. -----h---------------------|

(42)

7 The gravitational orbit equation

Since

u = 1∕r,

(43)

|--------------2---------| |r(𝜃) = -----h--∕μ------.| --------1 +-ecos(𝜃-−-𝜃0)--

(44)

Define the semilatus rectum

|--------| | h2- | |p = μ .| ---------

(45)

Then

|--------------p---------| |r(𝜃) = ----------------.| --------1 +-ecos(𝜃-−-𝜃0)--

(46)

Choose the angular origin at periapsis:

𝜃0 = 0.

(47)

Then the standard form is

|------------------| | ----p----- | |r(𝜃) = 1 + ecos 𝜃.| --------------------

(48)

8 Semilatus rectum and angular momentum

The conic parameter

p

(49)

has a direct dynamical interpretation:

|--------| | h2- | |p = μ .| ---------

(50)

Thus angular momentum sets the orbital length scale.

For fixed

μ,

(51)

larger

h

(52)

means larger

p.

(53)

For a circle,

e = 0,

(54)

and

r = p.

(55)

Therefore

p = h2 ∕μ

(56)

reduces to the circular radius found in M07-06.

9 True anomaly

The angle measured from periapsis is conventionally called the true anomaly.

Denote it by

ν.

(57)

Then

|-----------p------| |r(ν) = ----------.| --------1-+-ecos-ν--

(58)

In the planar derivation above,

ν = 𝜃 − 𝜃0.

(59)

Using

ν

(60)

makes the orbit equation independent of the arbitrary orientation of the coordinate axes.

10 Conic-section classification

The polar equation

r = ----p----- 1 + ecosν

(61)

is the focus-centered equation of a conic section.

The eccentricity determines the type:

|--------------------------| | e = 0 : circle, | | 0 < e < 1 : ellipse, | | e = 1 : parabola, | | e > 1 : hyperbola. | ----------------------------

(62)

Figure 3. The same orbit equation produces the conic family. Eccentricity e controls the transition from circle to ellipse, parabola, and hyperbola.

11 Periapsis radius

At periapsis,

ν = 0.

(63)

Since

cos0 = 1,

(64)

|-----------| rp = --p--. | -----1-+-e---

(65)

This formula applies to ellipse, parabola, and hyperbola.

12 Apoapsis radius for an ellipse

For an ellipse,

0 ≤ e < 1.

(66)

Apoapsis occurs at

ν = π.

(67)

Since

cosπ = − 1,

(68)

|------------| | --p-- | |ra = 1 − e. | -------------

(69)

The finite apoapsis exists because

1 − e > 0.

(70)

For

e ≥ 1,

(71)

there is no finite apoapsis.

13 Ellipse geometry

For an ellipse with semimajor axis

a

(72)

and eccentricity

e,

(73)

the focal distance is

|-------| c-=-ae.--

(74)

The periapsis and apoapsis radii measured from the occupied focus are

|--------------| -rp =-a(1-−-e),|

(75)

and

|--------------| |r = a(1 + e).| --a------------

(76)

Therefore

|------------| | rp + ra | |a = -------.| --------2-----

(77)

Also,

|------------| | ra −-rp | |e = ra + rp.| --------------

(78)

Figure 4. Ellipse geometry with the attracting body at one focus. The periapsis and apoapsis radii are a(1 − e) and a(1 + e).

14 Semilatus rectum of an ellipse

Use

--p-- rp = 1 + e = a(1 − e).

(79)

Multiply by

1 + e :

(80)

p = a(1 − e)(1 + e).

(81)

Thus

|--------------| | 2 | -p-=-a(1-−-e-).

(82)

Since the semiminor axis satisfies

2 2 2 b = a (1 − e ),

(83)

we also have

|--------| | b2 | |p = -- .| ------a--

(84)

15 Kepler’s first law

For a gravitationally bound orbit,

𝜀 < 0.

(85)

We will shortly show that this implies

0 ≤ e < 1.

(86)

Therefore every noncollision bound Newtonian two-body orbit is an ellipse, with the circle as the

e = 0

(87)

special case.

This is Kepler’s first law:

|------------------------------------------------------------------------------| |A bound gravitational orbit is an ellipse with the attracting body at one focus. | -------------------------------------------------------------------------------

(88)

For planetary motion in the two-body approximation, the Sun occupies one focus of the planetary ellipse.

16 Connect eccentricity to energy

M07-06 gave the radial energy equation

1- 2 h2-- μ- 𝜀 = 2 ˙r + 2r2 − r .

(89)

Using

u = 1∕r

(90)

and

r˙= − h du, d𝜃

(91)

we obtain

[( ) ] h2 du 2 2 𝜀 = --- --- + u − μu. 2 d𝜃

(92)

For

μ u = -2(1 + ecosν ), h

(93)

du-= − μe-sin ν. d𝜃 h2

(94)

Substitution and simplification give

|----------------| | 2 | |𝜀 = μ--(e2 − 1).| -----2h2----------

(95)

Therefore

|--------------| | 2 2𝜀h2 | e = 1 + --2-.| -----------μ----

(96)

17 Energy and conic type

The relation

2 2𝜀h2- e = 1 + μ2

(97)

connects M07-05’s energy classification directly to the conic geometry.

For

h ⁄= 0 :

(98)

𝜀 < 0 gives e < 1: ellipse.
𝜀 = 0 gives e = 1: parabola.
𝜀 > 0 gives e > 1: hyperbola.

Thus

|--------------------------| | 𝜀 < 0 ⇐ ⇒ 0 ≤ e < 1, | | 𝜀 = 0 ⇐ ⇒ e = 1, | | | --𝜀->-0--⇐-⇒---e->-1.------|

(99)

Figure 5. Orbital energy and eccentricity describe the same bound-unbound transition. Negative energy corresponds to ellipses, zero energy to the parabola, and positive energy to hyperbolas.

18 Circular orbit as the minimum-eccentricity case

For

e = 0,

(100)

the orbit equation becomes

r = p.

(101)

Thus the radius is constant.

The energy relation becomes

2 𝜀 = − μ--. 2h2

(102)

This is exactly the minimum effective-potential energy from M07-06.

Therefore the circular orbit is the

e = 0

(103)

limit of the ellipse family and the minimum-energy orbit at fixed

h.

(104)

19 Turning radii from the conic equation

For an ellipse,

--p-- --p-- rp = 1 + e , ra = 1 − e.

(105)

Using

2 p = h ∕μ,

(106)

|---------2----| |rp = ---h----,| | μ(1 + e) | ----------------

(107)

and

|--------------| | 2 | |ra = ---h----.| ------μ(1-−-e)--

(108)

These are the same two positive radial turning points obtained from the effective-potential quadratic in M07-06.

Figure 6. The ellipse reaches its minimum radius at true anomaly ν = 0 and maximum radius at ν = π. These are the periapsis and apoapsis turning radii.

20 Recover the effective-potential roots

M07-06 gave the turning-point equation

2𝜀r2 + 2 μr − h2 = 0.

(109)

Using

2 2𝜀h2 e = 1 + --2-, μ

(110)

the square root in the quadratic solution becomes

∘ ---------- μ2 + 2𝜀h2 = μe.

(111)

Therefore

r = −-μ-±-μe-. t 2𝜀

(112)

After using the energy-eccentricity relation, these roots reduce to

|--------------------------| | --p-- -p---| |rp = 1 + e, ra = 1 − e| ----------------------------

(113)

for

e < 1.

(114)

Thus the geometric orbit equation and the radial-energy method are completely consistent.

21 Parabolic orbit

For

e = 1,

(115)

------------------ | p | r(ν) = --------. | --------1 +-cos-ν-|

(116)

The periapsis radius is

|--------| |rp = p-.| ------2--|

(117)

ν → π,

(118)

the denominator approaches zero and

r → ∞.

(119)

This is the zero-energy escape trajectory.

22 Hyperbolic orbit

For

e > 1,

(120)

|------------------| |r(ν) = ----p-----.| --------1-+-ecos-ν--

(121)

The denominator vanishes at the asymptotic true anomaly

ν∞

(122)

satisfying

1 + e cosν∞ = 0.

(123)

Therefore

|--------------| | 1- | |cosν ∞ = − e .| ---------------

(124)

As the trajectory approaches this angle,

r → ∞.

(125)

The orbit has one finite periapsis and two asymptotic branches.

Figure 7. Parabolic and hyperbolic gravitational trajectories are unbounded conics with the attracting body at a focus. The hyperbola approaches finite asymptote angles determined by cosν_∞ = −1∕e.

23 Hyperbolic excess speed and eccentricity

For an unbound orbit,

v2∞- 𝜀 = 2 .

(126)

Insert this into

2 2𝜀h2- e = 1 + μ2 .

(127)

Then

|---------2--2--| e2 = 1 + v∞h--. | -----------μ2----

(128)

Thus both asymptotic speed and angular momentum determine the hyperbola’s eccentricity.

24 Example 1: determine an ellipse from turning radii

Suppose

rp = 7000km, ra = 9000 km.

(129)

The semimajor axis is

a	=	(130)
	= 8000 km.	(131)

The eccentricity is

e	=	(132)
	= 0.125.	(133)

The semilatus rectum is

p	= a(1 − e²)	(134)
	= 7875 km.	(135)

The orbit equation is therefore

|----------------------| | ---7875-km---- | |r(ν ) = 1 + 0.125 cos ν.| ------------------------

(136)

25 Example 2: orbital radius at a given true anomaly

For the previous orbit, evaluate the radius at

∘ ν = 90 .

(137)

Since

∘ cos90 = 0,

(138)

∘ r(90 ) = p.

(139)

Therefore

|----∘-------------| -r(90-) =-7875-km.-|

(140)

The semilatus rectum is literally the orbital radius at true anomaly

∘ 90

(141)

and

∘ 270

(142)

for an ellipse.

26 Example 3: continuing the M07-06 Earth orbit

M07-06 used

6 rp = 7.00 × 10 m

(143)

and found

6 ra ≈ 8.9805 × 10 m.

(144)

Therefore

a	=	(145)
	≈ 7.9903 × 10⁶ m.	(146)

The eccentricity is

e	=	(147)
	≈ 0.12394.	(148)

Using

h = 5.60 × 1010m2 ∕s

(149)

and

μE = 3.9860 × 1014m3 ∕s2,

(150)

the semilatus rectum is

p	=	(151)
	≈ 7.8675 × 10⁶ m.	(152)

These values satisfy

|--------------| |p = a(1 − e2).| ---------------

(153)

Figure 8. The continuing Earth example is a modest-eccentricity ellipse. The focus is at Earth, with periapsis near 7000 km and apoapsis near 8980.5 km.

27 Example 3 energy check

The M07-06 specific energy was approximately

𝜀 ≈ − 2.494 × 107 J∕kg.

(154)

Use

2 e2 = 1 + 2𝜀h-. μ2E

(155)

Substituting the numerical values gives

|------------| -e ≈-0.12394,--

(156)

in agreement with the eccentricity found from the turning radii.

This provides an independent consistency check between:

energy,
angular momentum,
turning radii,
conic geometry.

28 Focus versus geometric center

For an ellipse, the attracting body is not generally at the geometric center.

The distance from the ellipse center to either focus is

c-=-ae.-| ---------

(157)

The two foci lie symmetrically about the ellipse center.

In the gravitational two-body orbit equation, the origin of polar coordinates is placed at the occupied focus.

This is why the radius

r

(158)

is measured from the attracting body rather than from the ellipse center.

29 Why the inverse-square law matters

The area law followed from any central force.

The conic orbit equation is more specific.

Newtonian gravity gave

u ′′ + u = μ-, h2

(159)

whose right-hand side is constant.

That produces a sinusoidal reciprocal radius:

u = μ-(1 + ecosν ). h2

(160)

Other central force laws generally produce different differential equations and different orbit shapes.

Thus:

|----------------------------------------| -central force =-⇒-constant-areal velocity,

(161)

while

|------------------------------------------| inverse-square-central force-=⇒-conic-orbit.-

(162)

30 Exact two-body interpretation

For two finite masses,

m 1

(163)

and

m , 2

(164)

the relative coordinate obeys

r- ¨r = − μ r3,

(165)

with

|------------------| -μ-=-G-(m1-+--m2-).-|

(166)

Therefore the exact relative orbit is

|------------------| | ---h2∕μ--- | |r(ν) = 1 + ecos ν.| --------------------

(167)

The same conic equation applies, but the gravitational parameter contains the sum of the two masses.

When one mass is much smaller than the other,

μ ≈ GM.

(168)

31 Relation to the Laplace-Runge-Lenz direction

The phase angle

𝜃0

(169)

fixes the orientation of the conic in the orbital plane.

For inverse-square gravity there is an additional conserved vector, the Laplace-Runge-Lenz vector, that points toward periapsis and encodes both:

the eccentricity,
the periapsis direction.

We do not need that vector to derive Kepler’s first law, so it is deferred to a later advanced article.

32 Common mistakes

Confusing u = 1∕r with u = r.
Forgetting to use angular-momentum conservation when converting time derivatives to angular derivatives.
Dropping the minus sign in ṙ = −hdu∕d𝜃.
Writing the Newtonian Binet equation with a nonconstant right-hand side.
Confusing semilatus rectum p with semimajor axis a.
Forgetting that p = h²∕μ.
Treating eccentricity as having units.
Assuming all conics are bound.
Forgetting that e = 1 is the parabolic escape threshold.
Forgetting that e > 1 corresponds to positive orbital energy.
Placing the attracting body at the geometric center of an ellipse.
Using r_a = p∕(1 − e) for a parabola or hyperbola.
Confusing true anomaly with an angle measured from the ellipse center.
Forgetting that the circular orbit is the e = 0 ellipse.
Assuming Kepler’s first law follows from every central force rather than specifically the inverse-square law.

33 Practice exercises

Starting from h = r²𝜃, derive ṙ = −hdu∕d𝜃 for u = 1∕r.
Derive r = −h²u²d²u∕d𝜃².
Derive the Newtonian Binet equation u′′ + u = μ∕h².
Solve the Binet equation and obtain the conic orbit equation.
Show that p = h²∕μ.
Derive the periapsis radius r_p = p∕(1 + e).
For an ellipse, derive r_a = p∕(1 − e).
Show that p = a(1 − e²).
Derive e = (r_a − r_p)∕(r_a + r_p).
Derive e² = 1 + 2𝜀h²∕μ² from the energy equation.
Use the energy-eccentricity relation to classify ellipse, parabola, and hyperbola.
For a parabolic orbit, show that r_p = p∕2.
For a hyperbola, derive cos ν_∞ = −1∕e.
Reproduce the 7000 km by 9000 km ellipse example.
Reproduce the eccentricity and semilatus rectum of the continuing M07-06 Earth example.