Classical-Mechanics · Unit 24 · Video 4 · Interactive Practice

Orbit Plus Spin: Angular Momentum Splits at the Center of Mass, and the Earth Carries Four Million Times More in Its Orbit

IKey Formulas

Formula	Name	What it says
$\vec{\mathbf{L}}_S^{\text{total}} = \vec{\mathbf{L}}_S^{\text{orbital}} + \vec{\mathbf{L}}_{cm}^{\text{spin}}$	Orbit–spin split	Holds for any system of particles and any point $S$
$\vec{\mathbf{L}}_S^{\text{orbital}} = \vec{\mathbf{r}}_{S,cm}\times\vec{\mathbf{p}}_{\text{sys}}$ , $\ \vec{\mathbf{p}}_{\text{sys}} = m_T\vec{\mathbf{V}}_{cm}$	Orbital angular momentum	The system as one particle of mass $m_T$ at the center of mass; depends on $S$
$\vec{\mathbf{L}}_{cm}^{\text{spin}} = \sum_i \vec{\mathbf{r}}_{cm,i}\times m_i\vec{\mathbf{v}}_{cm,i} = I_{cm}\,\vec{\boldsymbol{\omega}}_{cm}$	Spin angular momentum	Motion relative to the center of mass; no $S$ in it (second form: rigid body about a symmetry axis)
$\dfrac{L_S^{\text{orbital}}}{L_{cm}^{\text{spin}}} = \dfrac{5}{2}\,\dfrac{r_{s,e}^2}{R_e^2}\,\dfrac{T_{\text{spin}}}{T_{\text{orbit}}} \approx 3.77\times10^{6}$	Earth: orbit vs spin	Uniform sphere, circular orbit; the squared distance ratio dominates

Key Insight: The two cross terms vanish because, measured from the center of mass, the mass-weighted positions and velocities both sum to zero: $\sum_i m_i\vec{\mathbf{r}}_{cm,i} = \vec{\mathbf{0}}$ and $\sum_i m_i\vec{\mathbf{v}}_{cm,i} = \vec{\mathbf{0}}$ .

IIVisualization 1 — Orbital Depends on S, Spin Does Not

For a spinning body sliding in a straight line, the particle-by-particle sum always equals orbital plus spin.

Center of mass Time t 2.0 s

IIIVisualization 2 — Why the Orbit Wins by Millions

With the mass cancelled, the ratio is a product of three factors; on a log scale they add.

Body

💡 All three bodies are modelled as uniform spheres, $I_{cm} = \tfrac{2}{5}mR^2$ ; Jupiter is centrally condensed ( $I_{cm} \approx 0.25\,mR^2$ ), so its true ratio is about $1.6$ times larger.

IVVisualization 3 — Sidereal Day and Solar Day

Because the Earth also moves along its orbit, facing the Sun again takes longer than one turn against the stars.

Year length Time 0 h

VQuiz Questions

Problem 1 · Orbit Plus Spin for a Disk

Given: A uniform disk of mass $2\ \text{kg}$ and radius $0.5\ \text{m}$ spins at $4\ \text{rad/s}$ about its symmetry axis (vertical, through its center). Its center moves at $3\ \text{m/s}$ in a straight horizontal line that passes $2\ \text{m}$ from a point $S$ in the same plane. Seen from above, both the spin and the motion of the center about $S$ are counterclockwise. Find the magnitude of $\vec{\mathbf{L}}_S^{\text{total}}$ .

✅ Correct! Orbital

12

plus spin

1

, both along the same vertical direction, gives

13\ \text{kg m}^2/\text{s}

❌ Not quite.

12

is only the orbital piece

d\,m_T V_{cm}

; the disk's rotation about its own center adds spin.

❌ Check the moment of inertia. A uniform disk about its symmetry axis has

I_{cm} = \tfrac{1}{2}MR^2

, not

MR^2

❌ Check the directions. Both motions are counterclockwise seen from above, so both vectors point up and add.

❌ Not quite. Compute

L^{\text{orbital}} = d\,m_T V_{cm}

and

L^{\text{spin}} = I_{cm}\omega

separately, then add.

Show solution

Orbital: the center moves on a straight line at perpendicular distance $d = 2\ \text{m}$ from $S$ , so

L_S^{\text{orbital}} = d\,m_T V_{cm} = (2)(2)(3) = 12\ \text{kg m}^2/\text{s}.

Spin: for a uniform disk about its symmetry axis,

L_{cm}^{\text{spin}} = I_{cm}\,\omega = \tfrac{1}{2}(2)(0.5)^2(4) = 1\ \text{kg m}^2/\text{s}.

Both are counterclockwise seen from above, so both point up:

L_S^{\text{total}} = 12 + 1 = 13\ \text{kg m}^2/\text{s}.

Problem 2 · A Straight Path Still Carries Orbital Angular Momentum

Given: A puck, a uniform disk of mass $3\ \text{kg}$ and radius $0.10\ \text{m}$ , slides along the $x$ -axis with $\vec{\mathbf{V}}_{cm} = 4\,\hat{\mathbf{i}}\ \text{m/s}$ while spinning counterclockwise (seen from $+z$ ) at $20\ \text{rad/s}$ . Find its angular momentum about $S = (0,\,-1.5\ \text{m})$ and about $S' = (5\ \text{m},\,0)$ , which lies on its path.

Angular momentum about S

Angular momentum about S′ (on the path)

✅ Correct! The orbital piece

-18\,\hat{\mathbf{k}}

depends on the point; the spin

+0.3\,\hat{\mathbf{k}}

is the same about both.

❌ Check the direction of the orbital piece.

\vec{\mathbf{r}}_{S,cm}

points up (

+\hat{\mathbf{j}}

) and

\vec{\mathbf{p}}_{\text{sys}}

points along

+\hat{\mathbf{i}}

;

\hat{\mathbf{j}}\times\hat{\mathbf{i}} = -\hat{\mathbf{k}}

❌ Check the spin's direction. Counterclockwise seen from

+z

means the spin is along

+\hat{\mathbf{k}}

, opposite to the orbital piece here.

❌ A straight path still has orbital angular momentum. About a point off the line,

\vec{\mathbf{r}}_{S,cm}\times\vec{\mathbf{p}}_{\text{sys}} \neq \vec{\mathbf{0}}

❌ Only the orbital piece vanishes. On the path

\vec{\mathbf{r}}_{S',cm}

is parallel to

\vec{\mathbf{p}}_{\text{sys}}

, but the spin has no

S

in it and is still there.

❌ The orbital piece depends on the point. Only zero total momentum makes

\vec{\mathbf{L}}

the same about every point; here

\vec{\mathbf{p}}_{\text{sys}} \neq \vec{\mathbf{0}}

❌ Check the value about S. Add the orbital piece

\vec{\mathbf{r}}_{S,cm}\times\vec{\mathbf{p}}_{\text{sys}}

and the spin

I_{cm}\omega\,\hat{\mathbf{k}}

❌ Check the value about S′. On the path,

\vec{\mathbf{r}}_{S',cm}

is parallel to

\vec{\mathbf{p}}_{\text{sys}}

Show solution

Spin (the same about every point): $I_{cm} = \tfrac{1}{2}(3)(0.10)^2 = 0.015\ \text{kg m}^2$ , so

\vec{\mathbf{L}}_{cm}^{\text{spin}} = (0.015)(20)\,\hat{\mathbf{k}} = 0.3\,\hat{\mathbf{k}}\ \text{kg m}^2/\text{s}.

Orbital about $S$ : with the puck at $(x, 0)$ , $\vec{\mathbf{r}}_{S,cm} = x\,\hat{\mathbf{i}} + 1.5\,\hat{\mathbf{j}}$ and $\vec{\mathbf{p}}_{\text{sys}} = 12\,\hat{\mathbf{i}}\ \text{kg m/s}$ :

\vec{\mathbf{L}}_S^{\text{orbital}} = (x\,\hat{\mathbf{i}} + 1.5\,\hat{\mathbf{j}})\times 12\,\hat{\mathbf{i}} = 18\,(\hat{\mathbf{j}}\times\hat{\mathbf{i}}) = -18\,\hat{\mathbf{k}}\ \text{kg m}^2/\text{s},

the same at every $x$ . Total: $\vec{\mathbf{L}}_S = -18\,\hat{\mathbf{k}} + 0.3\,\hat{\mathbf{k}} = -17.7\,\hat{\mathbf{k}}\ \text{kg m}^2/\text{s}$ .

About $S'$ : $\vec{\mathbf{r}}_{S',cm} = (x - 5)\,\hat{\mathbf{i}}$ is parallel to $\vec{\mathbf{p}}_{\text{sys}}$ , so the orbital piece is zero and $\vec{\mathbf{L}}_{S'} = +0.3\,\hat{\mathbf{k}}\ \text{kg m}^2/\text{s}$ .

Problem 3 · Orbit Versus Spin for Mars

Given: Mars orbits the Sun on a (nearly) circular orbit of radius $2.28\times10^{11}\ \text{m}$ with period $687\ \text{d}$ , has mean radius $3.39\times10^{6}\ \text{m}$ , and spins once every $24\ \text{h}\ 37\ \text{min}$ relative to the stars. Model Mars as a uniform sphere. Find $L_S^{\text{orbital}}/L_{cm}^{\text{spin}}$ with $S$ at the Sun.

✅ Correct! About

17

million: Mars is farther out than the Earth and smaller, so its squared distance ratio is even larger.

❌ A factor is missing. The

\tfrac{2}{5}

I_{cm}

sits in the denominator of the ratio, so it contributes

\tfrac{5}{2}

❌ The factor is upside down.

\tfrac{2}{5}

belongs to the spin, which is in the denominator, so the ratio carries

\tfrac{5}{2}

, not

\tfrac{2}{5}

❌ Square the distance ratio. Orbital goes as

m r^2\omega_{\text{orbit}}

and spin as

\tfrac{2}{5}mR^2\omega_{\text{spin}}

, so the ratio has

(r/R)^2

❌ Not quite. Use

\dfrac{L^{\text{orbital}}}{L^{\text{spin}}} = \dfrac{5}{2}\,\dfrac{r^2}{R^2}\,\dfrac{T_{\text{spin}}}{T_{\text{orbit}}}

Show solution

Divide $L^{\text{orbital}} = m r^2\,\dfrac{2\pi}{T_{\text{orbit}}}$ by $L^{\text{spin}} = \tfrac{2}{5}mR^2\,\dfrac{2\pi}{T_{\text{spin}}}$ ; the mass and $2\pi$ cancel:

\frac{L_S^{\text{orbital}}}{L_{cm}^{\text{spin}}} = \frac{5}{2}\,\frac{r^2}{R^2}\,\frac{T_{\text{spin}}}{T_{\text{orbit}}}.

Distance factor: $r/R = 2.28\times10^{11}/3.39\times10^{6} = 6.73\times10^{4}$ , squared $4.52\times10^{9}$ .

Period factor: $T_{\text{spin}} = 24\ \text{h}\ 37\ \text{min} = 88{,}620\ \text{s}$ and $T_{\text{orbit}} = 687 \times 86{,}400\ \text{s} = 5.94\times10^{7}\ \text{s}$ , so $T_{\text{spin}}/T_{\text{orbit}} = 1.49\times10^{-3}$ .

\frac{L_S^{\text{orbital}}}{L_{cm}^{\text{spin}}} = 2.5 \times 4.52\times10^{9} \times 1.49\times10^{-3} \approx 1.69\times10^{7}.

Problem 4 · The Solar Day on Venus

Given: Venus spins once relative to the stars every $243.0$ Earth days, but in the sense opposite to its orbital motion, and it orbits the Sun once every $224.7$ Earth days. Find the length of its solar day (Sun overhead to Sun overhead).

✅ Correct! With the spin opposite to the orbit, the orbital motion brings the Sun round to meet the marker sooner, so the solar day is shorter than the sidereal one.

❌ Not quite. The solar day differs from both periods: the direction to the Sun turns at

2\pi/T_{\text{orbit}}

while the planet turns at

2\pi/T_{\text{sidereal}}

against the stars.

❌ That is the Earth's rule. Subtracting the rates applies when spin and orbit are in the same sense; for Venus they are opposite, so the rates add.

Show solution

Relative to the stars, the planet turns at $\omega_{\text{spin}} = 2\pi/T_{\text{sidereal}}$ , while the direction from the planet to the Sun turns at $\omega_{\text{orbit}} = 2\pi/T_{\text{orbit}}$ in the orbital sense.

For the Earth the two are in the same sense, so a marker gains on the Sun at $\omega_{\text{spin}} - \omega_{\text{orbit}}$ (that is the extra degree per day). For Venus they are opposite, so it gains at $\omega_{\text{spin}} + \omega_{\text{orbit}}$ :

\frac{1}{T_{\text{solar}}} = \frac{1}{T_{\text{sidereal}}} + \frac{1}{T_{\text{orbit}}} = \frac{1}{243.0\ \text{d}} + \frac{1}{224.7\ \text{d}}

T_{\text{solar}} = \frac{(243.0)(224.7)}{243.0 + 224.7}\ \text{d} \approx 116.7\ \text{d}.

Using the Earth's rule, $1/T_{\text{solar}} = 1/T_{\text{sidereal}} - 1/T_{\text{orbit}}$ , would give a negative rate with magnitude $1/2984\ \text{d}^{-1}$ , which is wrong here.

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