I have problem understanding total angular momentum of earth about center of sun. Consider that earth has an orbital angular velocity of $\omega_0 \hat z$ and a spin angular velocity of $\omega_s \hat s$, distance between the centers of sun and earth as $R$ and radius of earth as $r$ and that earth has mass $m$. My understanding is that total angular momentum of center of mass of earth + angular momentum about CM of earth is $$L = m R^2 \omega_0 \hat z + \frac{2}{5} m r^2 \omega_s \hat s $$ however a it's given in my text book as $$L = m R^2 \omega_0 \hat z + \frac{2}{5} m r^2(\omega_s \hat s + \omega_0 \hat z) $$

Why does the the $\omega_0$ appear when calculating angular momentum about CM of earth?


1 Answer 1


You have to be a little careful what you mean by "spin" in a system like the one you're talking about. Ideally, we'd measure everything in an inertial reference frame (one that is fixed with respect to the "fixed" stars at infinity), but often that is not actually how we talk about things in an everyday sense. When you talk about the spin of the earth, for example, would you say that it rotates once every twenty-four hours? Because, actually, 24 hours is the time it takes for the earth to rotate once with respect to the direction of the sun (or, viewed from earth's surface, the interval of time between when the sun is directly north or south in the sky, and the next time it is directly north or south). But because of the earth's revolution around the sun, the direction to the sun changes with respect to the fixed stars changes over time. So, 24 hours (the length of a so-called stellar day) is NOT the rotation period of the earth with respect to an inertial frame. The latter happens to be shorter by about four minutes (24 hours / 365) and is called the sidereal day.

All this may seem a little unrelated to your question, but bear with me. If we want to separately consider the spin angular momentum and orbital angular momentum in a system such as yours, the mathematics (which I won't derive in detail here) allows us to do so if either:

1) The spinning orbiter is so small that it can be considered a point-mass compared to the overall system (note in your case, that the book's first term swamps the $2/5 m r^2 w_o$ term regardless of the relative sizes of $w_o$ and $w_s$).


2) For the orbital part, we consider the system rotating as a whole as a rigid body. This means, in particular, that for the orbital part, we must assume that the SAME side of the earth faces the sun at all times. If instead, we let the earth remain facing the same direction with respect to the fixed stars, the corresponding earth-sun system would not be rigid. (To picture all this, it may help to imagine drawing the sun as a circle at the center of an old-style audio record. Then, draw the earth as a circle at the edge of the record and shade the side that points towards the "sun". Clearly, as the record rotates rigidly, the shaded side of the earth points in all directions around the room. If we tried to keep it facing, say, North, at all times while the record rotated, then the record could not be rigid.)

So here then is the overall answer to your question. We want to break the total angular momentum of the earth into an orbital piece and a spin piece. Due to the discussion above, the orbital piece must include a contribution from the fact that the earth would rotate once every 365 days if it was rigidly attached to the sun. The last term that the book adds is exactly this piece... even though it is grouped with the spin contribution, the appearance of $w_o$ shows that it is actually "orbital" in origin. The book is then considering $w_s$ to be the spin of the earth relative to the sun (that is relative to if the earth actually was rigidly tied to the sun), rather than to the fixed stars. Your formula is correct if $w_s$ was instead the spin relative to the fixed stars (which would be harder to measure from here on earth, but would not be invalid if done that way). It comes back to the difference between a stellar and a sidereal day, and how one wants to break things up. Both can be correct, but one must be careful to define one's terms and use symbols consistently.

Hope that helps.

  • $\begingroup$ Was just about to start typing more or less the same thing :) $\endgroup$
    – Kyle Oman
    Commented Sep 10, 2015 at 22:57
  • $\begingroup$ Spot on with the difference between sidereal and solar day, and the likely interpretation of the symbols as used in the question. Note also that in a practical sense, the difference is tiny, given that the first term (with $R^2$) will completely dominate the result. $\endgroup$
    – Floris
    Commented Sep 10, 2015 at 23:02
  • 1
    $\begingroup$ Re the parenthetical remark "(which would be harder to measure from here on earth ...)" -- The sidereal day is easier to measure from here on earth rather than harder. The sidereal day is what is measured in precision astronomy. Multiple quasars give a much better reference than does that one big bright thing that is only visible during daytime. Measuring sidereal time means the equation of time doesn't arise during those measurements. It only arises when calculating apparent solar time. $\endgroup$ Commented Apr 18, 2018 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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