3
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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. The total angular momentum of the system (including the rotational momentum of the larger body) is conserved.

$\endgroup$
4
  • 1
    $\begingroup$ +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 $\endgroup$
    – uhoh
    Jul 8 at 21:31
  • 2
    $\begingroup$ "My head spins when I think about angular momentum" - You only have to think about it, not act it out! :) $\endgroup$ Jul 9 at 0:40
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 9 at 2:25
  • $\begingroup$ @AdamChalcraft thanks for that :-) $\endgroup$
    – uhoh
    Jul 9 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.