Angular Momentum, Areal Velocity, and Kepler's Second Law

(Topic)

Angular Momentum, Areal Velocity, and Kepler’s Second Law

Kepler’s second law states that the line from an orbiting body to the attracting center sweeps out equal areas in equal times.

In modern mechanics, this law follows directly from conservation of angular momentum.

The essential chain is

|-------------------------------------------------------------| | dA- | central force =⇒ τ = 0 =⇒ L = constant = ⇒ dt = constant. | --------------------------------------------------------------

(1)

The result is more general than Newtonian gravity.

Any motion governed by a central force has constant areal velocity about the force center.

1 Central force and zero torque

A central force has the form

|------------| |F = F (r)^er.| --------------

(2)

The force is parallel or antiparallel to the position vector

r.

(3)

The torque about the force center is therefore

τ	= r × F	(4)
	= 0.	(5)

Angular momentum obeys

dL ---= τ. dt

(6)

Hence

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

(7)

and

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

(8)

2 Planarity of central force motion

For a particle,

|------------| L = r × mv. | --------------

(9)

The vector

L

(10)

is perpendicular to both

r

(11)

and

v.

(12)

L

(13)

is constant in direction, the position and velocity remain in one fixed plane perpendicular to

L.

(14)

Thus nonradial central force motion is planar.

Figure 1. For central force motion, the constant angular-momentum vector is perpendicular to the orbital plane. The position and velocity vectors remain in that plane.

3 Angular momentum in polar coordinates

In the orbital plane,

v = ˙r^er + r𝜃˙^e𝜃.

(15)

Then

L	= mr × v	(16)
	= m ×.	(17)

The radial term vanishes because

^er × ^er = 0.

(18)

Therefore

|------------| |L = mr2 𝜃˙^n,| --------------

(19)

where

^n

(20)

is the fixed unit Normal to the orbital plane.

The magnitude is

|----------| |L = mr2 ˙𝜃.| ------------

(21)

Define specific angular momentum

|--------------| | L | |h = -- = r2 ˙𝜃. -----m---------

(22)

4 The infinitesimal swept area

During a small time interval

dt,

(23)

the radius vector changes by

dr.

(24)

The small triangle swept by the radius vector has area

1 dA = -|r × dr|. 2

(25)

Since

dr = v dt,

(26)

|r × v|dt.

(27)

Thus

|----------------| |dA 1 | |--- = -|r × v |.| --dt----2--------

(28)

Because

L = m |r × v|,

(29)

we obtain

|----------| |dA- -L--| |dt = 2m .| ------------

(30)

Equivalently,

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

(31)

Figure 2. In time dt, the radius vector sweeps an infinitesimal triangular area dA = 1
2 |r × v|dt.

5 Areal velocity

The quantity

|--------| |˙ dA-| |A = dt | ----------

(32)

is called the areal velocity.

For central force motion,

L

(33)

is constant, so

|-------------------------| A˙= -L--= h-= constant. | -----2m-----2--------------

(34)

The units are

|------| |m2 ∕s.| -------

(35)

This is the modern mechanical statement of Kepler’s second law.

6 Areal velocity in polar coordinates

A small sector of radius

r

(36)

and angle

d 𝜃

(37)

has area

|-------------| | 1 | dA = --r2d𝜃. | ------2-------

(38)

Divide by

dt :

(39)

|------------| |dA- = 1-r2 ˙𝜃. -dt----2-----|

(40)

Since

h = r2 ˙𝜃,

(41)

this again gives

|--------| | h- | |A˙= 2 .| ---------

(42)

7 Vector form of areal velocity

A signed area vector can be defined perpendicular to the orbital plane:

|--------------| |dA-- 1- | |dt = 2 r × v.| ----------------

(43)

Using

L = mr × v,

(44)

|-----------| dA-- -L-- | |dt = 2m . | -------------

(45)

Thus both the magnitude and direction of the areal-velocity vector are constant in a central force.

8 Kepler’s second law

Because

A˙

(46)

is constant,

ΔA = A˙Δt.

(47)

Therefore equal time intervals produce equal swept areas:

|--------------------------------| |Δt1 = Δt2 =⇒ ΔA1 = ΔA2. | ----------------------------------

(48)

This is Kepler’s second law:

|------------------------------------------------------------------------------------| |A line joining the orbiting body to the force center sweeps equal areas in equal times. --------------------------------------------------------------------------------------

(49)

Figure 3. Equal time intervals correspond to equal swept areas. Near periapsis the body traverses a longer arc in the same time because the radius is smaller and the angular rate is larger.

9 Why the body moves faster near periapsis

Specific angular momentum is

h = r2 ˙𝜃.

(50)

Therefore

|--------| |˙𝜃 = -h .| -----r2--|

(51)

When

r

(52)

is small,

˙𝜃

(53)

is large.

When

r

(54)

is large,

˙𝜃

(55)

is smaller.

Thus equal area does not mean equal angular displacement.

It also does not mean equal arc length.

10 Tangential speed and angular momentum

The transverse component of velocity is

|--------| |v𝜃 = r˙𝜃.| ----------

(56)

Using

h = r2 ˙𝜃,

(57)

|--------| | h | |v𝜃 = r-.| ---------

(58)

At smaller radius, the transverse speed required by fixed angular momentum is larger.

This is the same angular-momentum effect that produced the small-radius barrier in M07-06.

11 Periapsis and apoapsis

At a radial turning point,

˙r = 0.

(59)

Therefore the velocity is purely transverse:

v = v𝜃.

(60)

At periapsis,

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

(61)

At apoapsis,

|---------| h = r v . | -----a-a---

(62)

Hence

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

(63)

Thus

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

(64)

Figure 4. At periapsis and apoapsis the radial velocity vanishes, so angular-momentum conservation gives r_pv_p = r_av_a.

12 Example 1: apsidal speed ratio

Suppose

ra = 2rp.

(65)

Then

rpvp = 2rpva.

(66)

Therefore

|---------| vp = 2va. | -----------

(67)

The body moves twice as fast at periapsis as at apoapsis.

This statement follows from angular momentum alone.

13 Example 2: areal velocity in a circular Earth orbit

Consider a circular orbit at

r = 7.00 × 106 m

(68)

around Earth.

Using

μE = 3.9860 × 1014m3 ∕s2,

(69)

the circular speed is

∘ ---- μE- 3 vcirc = r ≈ 7.546 × 10 m ∕s.

(70)

The specific angular momentum is

h	= rv_circ	(71)
	≈ 5.282 × 10¹⁰ m²∕s.	(72)

Therefore the areal velocity is

|----------------------------| |A˙= h-≈ 2.641 × 1010 m2 ∕s.| ------2----------------------|

(73)

14 Circular-orbit check using total area

For a circle,

A = πr2. orbit

(74)

Since

A˙ = h∕2,

(75)

the orbital period is

T	=	(76)
	= .	(77)

For a circular orbit,

h = rvcirc,

(78)

|----------| | 2πr | |T = ----.| ------vcirc-

(79)

Thus the area-law description reproduces the familiar circumference-over-speed result.

15 Angular rate versus radius

Since

˙𝜃 = h∕r2,

(80)

the angular rate scales as

|------| |˙ 1-| 𝜃 ∝ r2| --------

(81)

along one orbit with fixed

h.

(82)

If the radius at one point is twice the radius at another,

|------------| |˙ 1˙ | 𝜃far = 4𝜃near.| --------------

(83)

Figure 5. For fixed specific angular momentum, the instantaneous angular rate varies as 1∕r². A larger orbital radius therefore corresponds to slower angular sweep.

16 Equal areas do not require an ellipse

The derivation used only

τ = 0.

(84)

It did not use the inverse-square form

F ∝ 1∕r2.

(85)

Therefore

|----------------------------------------------------------| |constant areal velocity is a property of every central force.| -----------------------------------------------------------

(86)

The inverse-square gravitational law is needed for the specific conic shapes of Keplerian trajectories, not for the area law itself.

This distinction is important:

central force gives Kepler’s second-law structure,
inverse-square central force gives the Keplerian conic orbit.

17 Historical interpretation

Kepler inferred the equal-areas law from astronomical observations before Newtonian mechanics existed.

Newton’s mechanics later explained why the law follows from a central gravitational force.

In modern language:

|----------------------------------------------------------------------------| |Kepler’s second law is angular-momentum conservation written geometrically.| ------------------------------------------------------------------------------

(87)

18 Finite-area integration

Because

dA = 1r2d𝜃, 2

(88)

the area swept as the orbit moves from

𝜃1

(89)

𝜃2

(90)

----------------------- | ∫ 𝜃2 | A12 = 1- r2(𝜃)d𝜃. | | 2 𝜃1 | -----------------------

(91)

The elapsed time is

|------------| | 2A | |Δt = ---12-.| --------h----

(92)

Once

r(𝜃)

(93)

is known, the area law can therefore be used to relate orbital position and elapsed time.

The orbit equation will be developed in M07-08.

19 Period from the area law

Suppose a bound orbit is known to be an ellipse with semimajor axis

a

(94)

and semiminor axis

b.

(95)

Its total area is

|--------------| -Aellipse-=-πab.-|

(96)

One full orbital period sweeps the entire ellipse, so

h πab = --T. 2

(97)

Therefore

|----------| T = 2πab-.| -------h----

(98)

This is a purely area-law relation.

Additional gravitational dynamics are needed to express

h

(99)

in terms of

a

(100)

and the central mass.

Figure 6. Over one full period an elliptical orbit sweeps its entire area πab. Constant areal velocity therefore gives T = 2πab∕h.

20 Connection to Kepler’s third law

The area law by itself gives

T = 2πab ∕h.

(101)

For Newtonian inverse-square gravity, the conic-orbit dynamics will later give a relation among

h, a, b, μ.

(102)

Combining those results leads to Kepler’s third law:

2 3 T ∝ a .

(103)

We postpone that derivation until the orbital conic has been established.

21 Exact two-body relative motion

For two finite masses

m 1

(104)

and

m2,

(105)

define the relative position

|------------| |r = r2 − r1.| -------------

(106)

The relative equation of motion is central:

|---------------------| | -r | ¨r = − G (m1 + m2 )r3. | -----------------------

(107)

The relative angular momentum is

|--------------| -Lrel-=-μrr-×-˙r,-

(108)

where the reduced mass is

|---------------| | m1m2 | μr = --------. | ------m1-+-m2----

(109)

Since the relative force is central,

|----------------| |Lrel = constant.| -----------------

(110)

Figure 7. In the exact two-body problem, the relative position vector obeys a central equation. Its constant relative angular momentum produces constant relative areal velocity.

22 Relative areal velocity

Define the specific relative angular momentum

|------------------| | Lrel | |h = ----= |r × r˙|.| ------μr------------

(111)

Then the area swept by the relative position vector obeys

|------------------| |dArel Lrel h- | | dt = 2μr = 2 .| -------------------

(112)

Thus Kepler’s second law applies exactly to the relative orbit of the two-body problem.

The approximation of a small body orbiting a fixed massive body is obtained when one mass is much larger than the other.

23 Each body about the Center of Mass

In the center of mass frame,

m1r1 + m2r2 = 0.

(113)

The positions of the two bodies are proportional to the relative vector:

|------------------| |r1 = − ---m2----r,| --------m1-+--m2---|

(114)

and

|----------------| | ---m1---- | |r2 = m1 + m2 r.| -----------------

(115)

Each body therefore sweeps area at a constant rate about the Center of Mass as well.

The two rates differ because the orbital sizes differ.

24 Example 3: continuing the M07-06 Earth orbit

M07-06 considered a turning point at

r = 7.00 × 106m p

(116)

with purely tangential speed

vp = 8.00 × 103 m ∕s.

(117)

The specific angular momentum is

h	= r_pv_p	(118)
	= 5.60 × 10¹⁰ m²∕s.	(119)

Therefore

|----------------------| |A˙= 2.80 × 1010 m2 ∕s.| -----------------------

(120)

The other turning radius found in M07-06 was

ra ≈ 8.9805 × 106m.

(121)

Angular-momentum conservation gives the apoapsis speed:

v_a	=	(122)
	≈ 6.236 × 10³ m∕s.	(123)

Figure 8. For the M07-06 Earth example, the radius is larger and speed is smaller at apoapsis, but the product rv_𝜃 = h and therefore the areal velocity remain constant.

25 Example 3 continued: orbital area and period

For an ellipse with apsidal radii

rp

(124)

and

ra,

(125)

the semimajor axis is

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

(126)

For the numerical example,

a ≈ 7.9903 × 106 m.

(127)

Using the ellipse geometry

b = √r--r-, p a

(128)

we obtain

b ≈ 7.9287 × 106 m.

(129)

Then

2πab- T = h

(130)

gives

|------------------------------| |T ≈ 7.108 × 103 s ≈ 118.5min. | --------------------------------

(131)

This is a useful geometric preview of the orbital-period relations developed later.

26 Equal area versus equal distance

Suppose two equal time intervals occur at different radii.

Kepler’s second law requires

ΔA1 = ΔA2.

(132)

It does not require

Δs1 = Δs2.

(133)

Near periapsis:

radius is smaller,
angular rate is larger,
tangential speed is larger.

Near apoapsis:

radius is larger,
angular rate is smaller,
tangential speed is smaller.

The changing speed is precisely what preserves equal swept area.

27 Radial motion as a limiting case

L = 0,

(134)

then

h = 0.

(135)

The areal velocity is

|------| A˙-=-0.-

(136)

The position vector does not sweep area because the motion is entirely along one radial line.

This is consistent with Kepler’s second law in its angular-momentum form.

28 Common mistakes

Assuming Kepler’s second law requires an elliptical orbit.
Forgetting that the law follows from zero torque, not specifically from inverse-square gravity.
Confusing angular momentum L with specific angular momentum h.
Missing the factor 1∕2 in dA = r²d𝜃.
Writing Ȧ = h instead of Ȧ = h∕2.
Treating equal swept areas as equal arc lengths.
Treating equal swept areas as equal angular displacements.
Forgetting that 𝜃 = h∕r² varies around a noncircular orbit.
Assuming the orbiting body stops at periapsis or apoapsis.
Forgetting that r_pv_p = r_av_a applies because the velocity is transverse at the apsides.
Applying the apsidal-speed relation at arbitrary points with nonzero radial velocity.
Assuming planarity without connecting it to the fixed direction of L.
Confusing the force center with the geometric center of an ellipse.
Using the test-particle form when exact two-body relative motion is required.
Forgetting that zero angular momentum gives zero areal velocity.

29 Practice exercises

Show that a central force produces zero torque about the force center.
Derive L = mr²𝜃n from the polar-coordinate velocity.
Derive dA = |r × dr|.
Derive Ȧ = L∕(2m) = h∕2.
Show that equal time intervals produce equal swept areas.
Explain why the proof applies to every central force.
Derive 𝜃 = h∕r² and explain its orbital meaning.
Derive r_pv_p = r_av_a.
For r_a = 3r_p, find v_p∕v_a.
Reproduce the circular-Earth-orbit areal velocity example.
Derive T = 2πr²∕h for a circular orbit.
Derive T = 2πab∕h for a known elliptical orbit.
Derive the exact relative areal velocity for the two-body problem.
Reproduce the apoapsis speed in the M07-06 Earth example.
Explain why purely radial motion has zero areal velocity.

30 Summary

For any central force,

|------| -τ-=-0--

(137)

and therefore

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

(138)

In planar polar coordinates,

|------------------------| |L = mr2 ˙𝜃, h = r2𝜃˙. | -------------------------

(139)

The swept area satisfies

|------------| | 1 2 | dA = -r d𝜃.| -------2------

(140)

Therefore the areal velocity is

|-----------------------| dA- 1-2 ˙ h- -L-- | -dt-=--2r-𝜃-=-2--=-2m-.--

(141)

Since angular momentum is conserved,

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

(142)

This is Kepler’s second law.

At periapsis and apoapsis,

|r-v-=--r-v-.| --p-p----a-a-|

(143)

For a known ellipse,

|----------| | 2πab | T = -----.| -------h----

(144)

Kepler’s second law is therefore the geometric expression of angular-momentum conservation in central force motion.

References

[1] PhysicsLibrary, M07-06, Effective Potential and Radial Motion in Central Gravity.

[2] PhysicsLibrary, M04-08, Conservation of Angular Momentum and Central Force Motion.

[3] PhysicsLibrary, M01-10, Polar-Coordinate Particle Kinematics.

[4] J. Kepler, Astronomia Nova, 1609.

[5] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

"Angular Momentum, Areal Velocity, and Kepler's Second Law" is owned by bloftin.
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Angular Momentum, Areal Velocity, and Kepler's Second Law