Celestial Mechanics: Central Forces and Conservation Laws

CM03 reduced Newtonian two-body gravitation to the relative equation

|-------r------------------------| |¨r = − μ-3, μ = G (m1 + m2 ).| --------r-------------------------

(1)

For a spacecraft or other negligible test mass moving about a much larger body, the usual approximation is $μ = GM$ . Equation (1) is already enough to reveal two of the deepest structural properties of Keplerian motion: conservation of angular momentum and conservation of mechanical energy. From those two constants follow planarity, constant areal velocity, the distinction between bound and unbound motion, the circular-orbit and escape-speed scales, and much of the orbit classification used throughout celestial mechanics [1, 2, 3, 4].

This article derives those results directly from Newton’s orbital equation. The purpose is not to memorize orbital formulas, but to see why they exist.

1. Central forces

A force is called central when it always acts along the instantaneous radius vector joining the particle to a fixed force center. The most general central force may be written

F = F (r)^r, r

(2)

where $Fr(r)$ is a signed scalar. An attractive force has $Fr < 0$ . Newtonian gravity has

μm r F = − -r2 ^r = − μm r3.

(3)

The defining geometric fact is therefore

F ∥ r.

(4)

That simple parallelism is what produces angular-momentum conservation.

Figure 1. For a central force, $F$ is parallel or antiparallel to $r$ . The moment arm about the force center is therefore zero, so the torque vanishes identically.

2. Torque and angular momentum

For a particle of mass $m$ , define the angular momentum about the force center as

L = r × p,

(5)

with linear momentum

p = mv.

(6)

Thus

L = mr × v.

(7)

The torque about the origin is

τ = r × F.

(8)

Differentiate $L$ :

	= (r × mv)	(9)
	= r × mv + r × mv.	(10)

Because $˙r = v$ , the first term is

v × mv = 0.

(11)

Newton’s second law gives $m ˙v = F$ , so

dL- = r × F = τ. dt

(12)

This is the angular form of Newton’s second law:

|--------| |dL | |---= τ.| -dt-------

(13)

For a central force,

τ = r × Fr (r)^r = 0,

(14)

because $r$ and $^r$ are parallel. Therefore

|--------| |dL-= 0,| -dt-------

(15)

and hence

|--------------| |L = constant. | ---------------

(16)

This result applies to every central force, not only inverse-square gravity.

3. Specific angular momentum

In celestial mechanics it is convenient to divide angular momentum by the orbiting mass. Define the specific angular momentum

|----------------| | L | |h = m- = r × v.| ------------------

(17)

Because $m$ is constant,

|------| h˙ = 0.| --------

(18)

The SI units of $h$ are

[h] = m2 ∕s.

(19)

In astrodynamics, $km2 ∕s$ is commonly used.

The magnitude of the cross product is

h = rv sin γ,

(20)

where $γ$ is the angle between $r$ and $v$ . If the velocity is decomposed into radial and transverse parts,

v = vr^r + vt^𝜃,

(21)

then only the transverse velocity contributes:

|--------| -h-=-rvt.-

(22)

Purely radial motion has $vt = 0$ , and therefore

h = 0.

(23)

4. Why a central-force orbit is planar

For any instant of motion,

h = r × v.

(24)

A cross product is perpendicular to both of its factors. Therefore

h ⋅ r = 0,

(25)

and

h ⋅ v = 0.

(26)

If $h ⁄= 0$ and is constant, then the radius vector must always remain in the fixed plane through the origin perpendicular to $h$ . Consequently,

|--------------------------------------------| |every nonradial central- force orbit is planar. ---------------------------------------------

(27)

The original three-dimensional differential equation can therefore be reduced once the fixed orbital plane is known. The remaining motion is two-dimensional.

Figure 2. The conserved vector $h = r × v$ is normal to the orbital plane. Because its direction does not change, the plane itself remains fixed in the ideal central-force problem.

The radial case $h = 0$ is degenerate. There is then no unique orbital-plane normal because the motion lies along a single line through the force center.

5. Two-body meaning of the relative angular momentum

For the exact two-body problem, CM03 introduced the center-of-mass coordinate and the relative coordinate

r = r − r . 2 1

(28)

In the center-of-mass frame,

m1r1 + m2r2 = 0.

(29)

Solving for the individual position vectors gives

r1 = − --m2----r, r2 = ---m1----r. m1 + m2 m1 + m2

(30)

The corresponding velocities are

v1 = − --m2----r˙, v2 = ---m1----˙r. m1 + m2 m1 + m2

(31)

The total angular momentum about the center of mass is

L = m r × v + m r × v . total 1 1 1 2 2 2

(32)

Substitution and collection of terms gives

--m1m2--- Ltotal = m1 + m2 (r × ˙r).

(33)

Define the reduced mass

-m1m2---- mred = m1 + m2 .

(34)

Then

|--------------| Ltotal =-mredh.-

(35)

Thus $h$ is a specific angular momentum associated with the relative orbit, while the physical total orbital angular momentum carries the additional reduced-mass factor. In the test-particle limit $m2 ≪ m1$ , the reduced mass approaches $m2$ , recovering the familiar $L ≈ m2h$ .

6. Areal velocity

Angular momentum has a direct geometric interpretation. During a short interval $dt$ , the particle moves through

dr = v dt.

(36)

The radius vector sweeps out an approximately triangular area

1- dA = 2|r × dr|.

(37)

Substitute $dr = v dt$ :

dA	= \|r × v\|dt	(38)
	= dt.	(39)

Divide by $dt$ :

|---------| |dA-= h. | -dt----2--|

(40)

Since $h$ is constant,

|----------------| |dA- | | dt = constant. | -----------------

(41)

This is Kepler’s second law: equal areas are swept out in equal times. Newtonian mechanics therefore explains Kepler’s empirical area law as a direct consequence of zero torque about the force center.

Figure 3. In a short interval $dt$ , the swept area is $dA = 1|r × v|dt 2$ . Conservation of $h$ therefore makes the areal velocity constant.

7. Polar-coordinate form of angular momentum

In the orbital plane,

v = r˙^e + r𝜃˙^e . r 𝜃

(42)

Then

h	= re_r ×	(43)
	= r²𝜃 .	(44)

Thus

|--------| |h = r2 ˙𝜃. ----------

(45)

The areal-velocity relation becomes

dA- 1- 2 ˙ dt = 2 r 𝜃.

(46)

This form will be used in CM05 and CM06 to reduce the two-dimensional central-force problem to a radial equation and then to Binet’s equation.

8. Mechanical energy from the gravitational potential

CM02 derived the Newtonian gravitational potential energy

U(r) = − μm--. r

(47)

The kinetic energy is

1- 2 K = 2 mv .

(48)

Therefore the total mechanical energy is

|------------------| | 1 2 μm | |E = -mv − ----.| ------2--------r---

(49)

Dividing by the particle mass gives the specific mechanical energy

|------------------| | E- v2- μ- | |𝜖 = m = 2 − r .| -------------------

(50)

Its units are

2 2 [𝜖] = m ∕s = J∕kg.

(51)

9. Direct derivation of energy conservation from Newton’s equation

We can prove energy conservation without first assuming the potential-energy formula. Start from Newton’s orbital equation

r- ¨r = − μ r3.

(52)

Take the dot product with the velocity $v = ˙r$ :

v ⋅¨r = − μr-⋅ v-. r3

(53)

The left side is

( ) d v2 v ⋅ ˙v = -- --- . dt 2

(54)

For the gravitational term, note that

( ) d- 1- = − 1-r˙. dt r r2

(55)

Also,

r-⋅ v ˙r = r .

(56)

Therefore

d (1 ) r ⋅ v -- -- = − ----. dt r r3

(57)

Hence

( ) r ⋅ v -d 1- − μ r3 = μdt r .

(58)

Substituting into the dot-product equation gives

d (v2 ) d (1 ) -- --- = μ -- -- . dt 2 dt r

(59)

Bring both derivatives to the same side:

( ) d- v2- μ- dt 2 − r = 0.

(60)

Thus

|-----2------------------| |𝜖 = v--− μ-= constant. | ------2----r-------------|

(61)

This is the specific-energy integral of the Newtonian two-body equation.

10. General central conservative forces

The conservation argument is broader than gravity. Suppose a central force can be derived from a radial potential energy $U (r)$ :

dU- F = − dr ^r.

(62)

Newton’s second law is

m v˙= F.

(63)

Dot with $v$ :

mv ⋅v˙= F ⋅ v.

(64)

The left side is

( ) -d 1- 2 dt 2mv .

(65)

Because a central potential depends only on $r$ ,

dU- dU- dt = dr ˙r.

(66)

Also $F ⋅ v = − (dU∕dr )˙r$ , so

dK dU ----= − ---. dt dt

(67)

Therefore

|----------------| |d- | |dt(K + U ) = 0.| -----------------

(68)

A central force guarantees angular-momentum conservation because its torque is zero. Energy conservation additionally requires that the force be conservative. Newtonian gravity satisfies both conditions.

11. The same result from work and potential energy

The work-energy theorem states

ΔK = W.

(69)

For a conservative gravitational force,

Wg = − ΔU.

(70)

Therefore

ΔK + ΔU = 0,

(71)

|------------------------| -K--+-U-=--E-=-constant.-|

(72)

The force-law derivation and the work-potential derivation are two descriptions of the same conservation law.

12. Why angular momentum and energy are independent pieces of information

The specific energy depends on the speed magnitude:

v2 μ 𝜖 = ---− --. 2 r

(73)

Specific angular momentum depends only on the transverse part of velocity:

h = rvt.

(74)

Two particles at the same $r$ with the same speed $v$ therefore have the same $𝜖$ , but may have different $h$ if their velocity directions differ.

For example, a purely transverse velocity gives

h = rv,

(75)

whereas a purely radial velocity gives

h = 0,

(76)

although the two states have the same kinetic energy if their speeds are equal.

Thus:

|------------------------------------------------------------------------------------| |𝜖 measures the energy class, h encodes rotational geometry and the orbital plane.| --------------------------------------------------------------------------------------

(77)

Later, CM06 will show that together $𝜖$ and $h$ determine the size and shape of the Kepler conic.

13. Energy scale at a given radius

At a fixed radius $r$ , the specific energy is

2 𝜖(v) = v--− μ. 2 r

(78)

Three speed scales are especially useful.

11.1 Zero speed

If $v = 0$ , then

μ- 𝜖 = − r < 0.

(79)

11.2 Circular speed

For circular motion, the inward gravitational acceleration must equal the centripetal acceleration:

2 vc= μ-. r r2

(80)

Therefore

|-----∘----| | μ | |vc = --.| --------r--

(81)

Substitute this into the energy equation:

𝜖_c	= −	(82)
	= −.	(83)

Thus

|----------| |𝜖 = − -μ-.| --c----2r---

(84)

The corresponding angular momentum magnitude is

|-----------√----| -hc =-rvc =---μr.-

(85)

11.3 Escape speed

The minimum escape trajectory has zero speed at infinity and therefore zero total specific energy:

𝜖 = 0.

(86)

At the launch radius,

v2 μ -esc− --= 0, 2 r

(87)

|------∘-----| | 2μ | |vesc = ---.| ----------r---

(88)

Therefore

|------√-----| |vesc = 2vc.| --------------

(89)

14. Orbit classification from specific energy

Because the gravitational potential approaches zero as $r → ∞$ , the sign of $𝜖$ has an immediate physical meaning.

12.1 Negative energy: bound motion

|------| |𝜖 < 0,| -------

(90)

then the particle does not have enough mechanical energy to reach infinity with nonnegative kinetic energy. The motion is gravitationally bound.

For $h > 0$ , the eventual Kepler conic is an ellipse; a circle is the special constant-radius case. CM06 derives that conic result from Binet’s equation. If $h = 0$ , the bound motion is a degenerate radial trajectory rather than a nondegenerate ellipse.

12.2 Zero energy: marginal escape

|------| -𝜖 =-0,|

(91)

then the object is exactly at the escape threshold. It can reach infinity with speed tending to zero.

For $h > 0$ , the corresponding Kepler conic is a parabola. For $h = 0$ , the motion is radial marginal escape.

12.3 Positive energy: unbound motion

|------| -𝜖 >-0,|

(92)

then the motion is unbound. At very large distance the potential term vanishes, so

2 𝜖 = v∞-. 2

(93)

Therefore

|----------| | √ ---| v-∞-=---2𝜖.-

(94)

For $h > 0$ , CM06 shows that the Kepler conic is a hyperbola. For $h = 0$ , the motion is radial unbound motion.

Figure 4. At a fixed radius, the sign of $𝜖 = v2∕2 − μ∕r$ separates bound, marginal-escape, and unbound motion. The circular speed lies below the escape speed by the factor $1∕√2--$ .

15. Compact classification table

The two invariants $𝜖$ and $h$ already organize the major cases:

Energy	Angular momentum	Newtonian motion

$𝜖 < 0$	$h > 0$	bound; ellipse, with circle as a special case
$𝜖 < 0$	$h = 0$	bound radial motion
$𝜖 = 0$	$h > 0$	parabolic escape threshold
$𝜖 = 0$	$h = 0$	radial marginal escape
$𝜖 > 0$	$h > 0$	hyperbolic unbound motion
$𝜖 > 0$	$h = 0$	radial unbound motion

The conic labels in this table are established formally in CM06. At the present stage, the bound-versus-unbound classification follows directly from energy conservation.

A further consequence of angular-momentum conservation can already be obtained without knowing the conic equation. At any radial turning point the radial speed vanishes, so the velocity is purely transverse and

h = rv.

(95)

If a bound orbit has an inner turning radius $rp$ and an outer turning radius $ra$ , then

|----------------| -rpvp =-rava =-h.-

(96)

Therefore

|vp---ra-| |--= --.| -va---rp--

(97)

The body moves fastest at its smaller turning radius and slowest at its larger turning radius. This is the speed counterpart of constant areal velocity.

16. Circular motion as a special simultaneous condition

Negative energy alone does not imply a circular orbit. Circular motion requires both

˙r = 0

(98)

and the exact transverse speed

∘ -- μ- vt = r .

(99)

Equivalently, a circular state at radius $r$ satisfies

|------μ-----------√-----| |𝜖 = − --, h = μr. | -------2r----------------|

(100)

A state at the same radius with the same negative energy but a different angular momentum is not circular.

17. Example: a circular low-Earth orbit

Take

μE = 3.986004418 × 1014 m3 ∕s2,

(101)

and an orbital altitude of $400 km$ above a mean Earth radius

R = 6.371 × 106m. E

(102)

The orbital radius is

r = 6.771 × 106m.

(103)

The circular speed is

∘ ---- μE- vc = r ≈ 7.673 km/s.

(104)

The escape speed at the same radius is

√ -- vesc = 2 vc ≈ 10.851km/s.

(105)

The circular specific angular momentum is

√ ---- 10 2 hc = μEr ≈ 5.196 × 10 m ∕s.

(106)

The specific energy is

μE 7 𝜖c = − ---≈ − 2.943 × 10 J/kg. 2r

(107)

Thus a low-Earth satellite is deeply inside Earth’s gravitational well even though it is continuously falling around the planet.

18. Example: classify a Cartesian initial state

Consider the state

r = (7000,− 1200, 1800)km,

(108)

v = (1.0,7.2,2.0)km/s.

(109)

Using Earth-centered units of km and s,

3 2 μE = 398600.4418 km ∕s .

(110)

The radius and speed magnitudes are approximately

r ≈ 7326.66 km, v ≈ 7.5392km/s.

(111)

The specific energy is

v2 μ 𝜖 = ---− --E ≈ − 25.984 km2 ∕s2. 2 r

(112)

Therefore the motion is bound.

The specific angular momentum is

h = r × v,

(113)

which gives approximately

2 h = (− 15360, − 12200, 51600 )km ∕s.

(114)

Its nonzero magnitude confirms that this is a nonradial planar orbit. The vector $h$ supplies the fixed normal to that orbital plane.

CM06 will use these same invariants to determine the eccentricity and complete conic geometry.

19. Conservation laws as numerical diagnostics

Conservation laws are not only analytical tools. They are also excellent diagnostics for numerical orbit propagation.

For an ideal two-body simulation, evaluate at every numerical time step

v(t)2 -μ-- 𝜖(t) = 2 − r(t) ,

(115)

and

h(t) = r(t) × v(t).

(116)

Ideally,

𝜖(t) = 𝜖(0), h(t) = h(0).

(117)

In an actual numerical integrator, small deviations measure numerical error. A propagator may appear visually plausible while slowly drifting in energy or angular momentum. Monitoring the invariants makes such errors visible.

For perturbed motion, the invariants need not remain constant physically. Their rates of change then contain information about the perturbing force rather than merely numerical error.

20. What changes when the force is not perfectly central?

Suppose

¨r = − μ-r + ap, r3

(118)

where $ap$ is a perturbing acceleration. Then

˙h = r × ap.

(119)

A perturbation with a nonzero moment about the central body can therefore change both the magnitude and direction of $h$ , altering the orbital plane or angular-momentum magnitude.

Likewise,

˙𝜖 = v ⋅ ap

(120)

when $𝜖$ is defined using only the central $− μ∕r$ potential. A perturbation doing positive work increases the Keplerian specific energy; one doing negative work decreases it.

These relations foreshadow the later perturbation and variation-of-elements sections of the CM series.

21. The conservation-law spine of the Kepler problem

The logic developed in CM03 and CM04 can now be summarized compactly:

|----------| | r | |¨r = − μ-3 | --------r--

(121)

implies

|----------| |r × ¨r = 0 | -----------

(122)

and therefore

|----------------------| -h-=-r-×-v-=--constant.-|

(123)

That immediately gives

--------------------- |dA h | |--- = -- = constant | --dt----2------------|

(124)

and a fixed orbital plane.

Taking the dot product with $v$ gives

|-----2------------------| |𝜖 = v--− μ-= constant. | ------2----r-------------|

(125)

The sign of $𝜖$ separates bound, marginal, and unbound motion. The pair $(𝜖,h)$ is therefore the natural bridge from Newton’s differential equation to orbit geometry.

The next steps of the course exploit exactly that bridge:

|------------------------------------------| -h--=⇒---planar--motion-and-polar-reduction--

(126)

followed by

|------------------------------------------------------| -polar-dynamics--=-⇒--Binet-equation--=-⇒--conic-orbit.|

(127)

References

[1] I. Newton, The Principia: Mathematical Principles of Natural Philosophy, translated by I. B. Cohen and A. Whitman, University of California Press, 1999.

[2] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2001.

[3] J. M. A. Danby, Fundamentals of Celestial Mechanics, 2nd ed., Willmann-Bell, 1988.

[4] R. R. Bate, D. D. Mueller, and J. E. White, Fundamentals of Astrodynamics, Dover reprint, 1971.

[5] B. W. Carroll and D. A. Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Pearson, 2007.

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