<p>There are two types of angular momentum of each planet: orbital angular momentum of the planet around the Sun, and rotational angular momentum of the planet around its rotational axis.</p>
<p>Orbital angular momentum <span class="math-container">$L_{orb}$</span> is typically calculated at perihelion or aphelion as <span class="math-container">$L_{orb}=mvr$</span>, where <span class="math-container">$m$</span> is the mass of the planet, <span class="math-container">$v$</span> is the instantaneous orbital velocity and <span class="math-container">$r$</span> is the distance between the center of the Sun and the center of the planet. For Earth, <span class="math-container">$L_{orb} \approx 2.7 \times 10^{40}$</span> kg m<sup>2</sup>/s.</p>
<p>Rotational angular momentum is <span class="math-container">$L_{rot} = I\omega$</span>, where <span class="math-container">$I$</span> is the moment of inertia and <span class="math-container">$\omega$</span> is the rotation rate. For Earth, <span class="math-container">$L_{rot} \approx 7.1 \times 10^{33}$</span> kg m<sup>2</sup>/s.</p>
<p><strong>As we can see for Earth, the rotational angular momentum is negligible when compared to the orbital angular momentum.  Mars will have a similar result.</strong></p>
<p><span class="math-container">$L_{rot}$</span> and <span class="math-container">$L_{orb}$</span> are instantaneously independent. That is, one can't determine the spin rate of a planet by knowing its orbit or vice-versa. However, angular rotational momentum can be transferred into orbital angular momentum through tidal interactions such as <a href="https://en.wikipedia.org/wiki/Tidal_locking" rel="noreferrer">tidal locking</a>. The total angular momentum of the system (including the rotational momentum of the larger body) is conserved.</p>

CC BY-SA 4.0
2021-07-08T21:31:14.083Z
`+1` My head spins when I think about angular momentum; this is a very grounded and thorough answer! Technically, *I think that* we can calculate (and also conserve) angular momentum about any point in space, but the center of mass of the planet, and the planet-Sun barycenter are the two most useful points to use unless a rogue star enters the solar system... :-0
uhoh
CC BY-SA 4.0
2021-07-09T00:40:51.237Z
"My head spins when I think about angular momentum" - You only have to think about it, not act it out! :)
fluffysheap
CC BY-SA 4.0
2021-07-09T02:25:43.427Z
@uhoh Yes, angular momentum is conserved about any point you care to choose. An easy but unenlightening proof is that physics is unchanged by a rotation about any point, and now Noether’s theorem applies.
Adam Chalcraft
CC BY-SA 4.0
2021-07-09T07:50:41.863Z
@AdamChalcraft thanks for that :-)
uhoh
CC BY-SA 4.0
2021-07-08T18:49:04.300Z
2021-07-08T18:49:04.300Z
Connor Garcia
<p>Is the 'spin' of Earth and the 'spin' of Mars taken into account as a contribution to the conservation of angular momentum? (and a contribution to the increasing/decreasing eccentricity?) Or - is the spin regarded as too small to be measurable?</p>

<orbit><earth><mars><rotation><eccentricity>
Is the 'spin' of Earth and the 'spin' of Mars a significant contribution to their angular momentum?
Rob Clennell
# Is the 'spin' of Earth and the 'spin' of Mars a significant contribution to their angular momentum?

(Asked by: username_0 on 2021-07-08)

Is the 'spin' of Earth and the 'spin' of Mars taken into account as a contribution to the conservation of angular momentum? (and a contribution to the increasing/decreasing eccentricity?) Or - is the spin regarded as too small to be measurable?

**Tags:** orbit, earth, mars, rotation, eccentricity


---

## 1 answers

### Answer by username_1 on 2021-07-08. Score: 6

There are two types of angular momentum of each planet: orbital angular momentum of the planet around the Sun, and rotational angular momentum of the planet around its rotational axis.

Orbital angular momentum $L\_{orb}$ is typically calculated at perihelion or aphelion as $L\_{orb}=mvr$, where $m$ is the mass of the planet, $v$ is the instantaneous orbital velocity and $r$ is the distance between the center of the Sun and the center of the planet. For Earth, $L\_{orb} \approx 2.7 \times 10^{40}$ kg m2/s.

Rotational angular momentum is $L\_{rot} = I\omega$, where $I$ is the moment of inertia and $\omega$ is the rotation rate. For Earth, $L\_{rot} \approx 7.1 \times 10^{33}$ kg m2/s.

**As we can see for Earth, the rotational angular momentum is negligible when compared to the orbital angular momentum. Mars will have a similar result.**

$L\_{rot}$ and $L\_{orb}$ are instantaneously independent. That is, one can't determine the spin rate of a planet by knowing its orbit or vice-versa. However, angular rotational momentum can be transferred into orbital angular momentum through tidal interactions such as [tidal locking](https://en.wikipedia.org/wiki/Tidal_locking). The total angular momentum of the system (including the rotational momentum of the larger body) is conserved.

> **Comments:**
> **username_2**: `+1` My head spins when I think about angular momentum; this is a very grounded and thorough answer! Technically, *I think that* we can calculate (and also conserve) angular momentum about any point in space, but the center of mass of the planet, and the planet-Sun barycenter are the two most useful points to use unless a rogue star enters the solar system... :-0  
> **username_3**: "My head spins when I think about angular momentum" - You only have to think about it, not act it out! :)  
> **username_4**: @username_2 Yes, angular momentum is conserved about any point you care to choose. An easy but unenlightening proof is that physics is unchanged by a rotation about any point, and now Noether’s theorem applies.  
> **username_2**: @username_4 thanks for that :-)  

---

