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Let's assume, the mass of Earth = $M$, the radius of Earth = $r$, the moment of inertia due to revolution of Earth = $I$, angular momentum due to the revolution = $L$, distance between the center of Earth and Sun = $R$, the angular velocity of Earth due to its revolution around the sun = $\omega$

According to the parallel axis theorem, $I = \frac{2}{5}Mr^2 + MR^2$

So, $$L = I\omega$$ $$= \frac{2}{5}Mr^2\omega + MR^2\omega$$ $$= \frac{2}{5}Mr^2R^2\omega\cdot\frac{1}{R^2} + MR^2\omega$$

According to Kepler's second law of planetary motion, the constant areal velocity of Earth = $\frac{1}{2}R^2\omega$

Hence, $$L = \frac{p}{R^2} + q$$ where, p and q both stays constant over time.

Since $R$ changes over time, $L$, the angular momentum is not conserved. Rather, it's a quadratic function of $\frac{1}{R}$

So why do we say that the torque acting on Earth due to Sun is $0$ when the angular momentum is not conserved?

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  • $\begingroup$ I have deleted my answer because I don't think I have explained myself very well. When I have more time I'll make the answer better. $\endgroup$ – BioPhysicist Jun 26 at 4:31
  • $\begingroup$ @BioPhysicist parallel axis theorem I think, there is no such requirement of using parallel axis theorem (like, every point must stay equidistant from the axis). If the points of a body stay equidistant from the axis, why even bother using this theorem!! Please explain this when u have time $\endgroup$ – Imtiaz Kabir Jun 26 at 4:33
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    $\begingroup$ What is really conserved is the full angular momentum. Kepler's Law just assumes that $(r/R)^2$ is negligible. $\endgroup$ – Felipe Jun 26 at 4:59
  • $\begingroup$ @Felipe What does full angular momentum even mean! If u r trying to say that $\frac{1}{R^2}$ is almost zero (since R is so huge), then keep in mind that we are scaling $\frac{1}{R^2}$ with $p = \frac{2}{5}Mr^2R^2\omega$, So $\frac{p}{R^2}$ is not negligible at all! This will evaluate very different value at perihelion and aphelion! $\endgroup$ – Imtiaz Kabir Jun 26 at 5:04
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    $\begingroup$ Let's not approximate anything. Then Kepler's Law does not hold. The quantity conserved is $R^2\omega\left[1+\frac 25\left(\frac rR\right)^2\right]$, not $R^2\omega$ (Kepler). $\endgroup$ – Felipe Jun 26 at 6:19
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Parallel axis theorem is for the rotation of a body about a single axis that does not pass through the center of mass.

But the earth's motion cannot be simply described in this way in the frame where the sun is at rest. It is more usually described as a combination of rotation and orbital movement, where each of these motions has different angular speeds. Because earth's rotational speed is not equal to $\omega$, the stated equation does not match. If it were true, then the earth would have the same angular momentum regardless of the rotational speed (since that parameter does not appear). Two identical orbits with different rotational speed must have different angular momenta.

We can instead sum the rotational and orbital angular momenta into the similar formula:

$$L = L_{\text{rot}} + L_{\text{orbit}} = \frac 25 Mr^2\omega_{\text{rot}} + MR^2\omega_{\text{orbit}}$$

When this formula is manipulated, allowing $\omega_{\text{rot}}$ to remain constant, the angular momentum also remains constant.

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Torque on earth due to sun is 0 as gravitational force always acts towards center.

In the formula of torque of torque which is $\tau = \ rF sin\theta$. Now $\theta$ becomes 0 as Force and r vector acts on same line (r is perpendicular distance from the force F)and hence overall torque becomes 0. As net torque is 0 about earth's center, therefore angular momentum is conserved at center only.

And i don't see why did you parallel axis theorem ?

The angular momentum of earth around sun is just $Mr^2\omega$(R =radius of earth) as rotation of earth on it's own axis is caused by moon and not by sun. Kepler law states $T^2$ is proportional to $r^3$. Hence to calculate time period of earth around sun you only need to consider I = M$r^2$ Also note r is usually taken to be $\frac{Apogee+Perigee}{2}$

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  • $\begingroup$ How is $r$ zero? And the moment of inertia of a point mass is $Mr^2$. The inertia of earth (a solid sphere) cant be the same. To find the moment of inertia of a solid sphere around an axis outside the body, we must integrate all the moment of inertia of all points of Earth, hence the parallel axis theorem. Just because I am using parallel axis theorem, does not mean I am talking about the rotation around Earth's own axis, Rather it is the proper way to calculate the moment of inertia of a solid sphere, no matter whether its rotating around its own axis or not $\endgroup$ – Imtiaz Kabir Jun 26 at 4:19
  • $\begingroup$ @ImtiazKabir I edited my answer. $\endgroup$ – Binod Jun 26 at 4:22
  • $\begingroup$ It's an elliptical orbit. In a circular orbit, the tangent of any point on the perimeter is perpendicular to the radius. According to Kepler's first law, the orbit is elliptical. So the tangent is not necessarily perpendicular to the radius. And thus r is not perpendicular to F $\endgroup$ – Imtiaz Kabir Jun 26 at 4:24
  • $\begingroup$ @ImtiazKabir Gravitational force will still acts towards center. We are only concerned about torque at the center hence r =0 $\endgroup$ – Binod Jun 26 at 4:27
  • $\begingroup$ I also agree that the force acts towards center. Yet $r$ is not $0$. (And please try to understand that $r$ being perpendicular and $r$ being $0$ are 2 different things) $\endgroup$ – Imtiaz Kabir Jun 26 at 4:36

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