What is the total kinetic energy of the Earth relative to the Sun?

[jbriggs444]

And when we get back to my original question - the total kinetic energy of Earth - what would it be?

I would say that it is similar to the pendulum, so $E_k=\frac 12 J_{CM} {\omega_1}^2+\frac 12 (J_{CM}+ma^2){\omega_2}^2$, where that angular velocities correspond to velocities about Earth's own axis of rotation and a rotation about the Sun, $a$ is a distance of Earth's centre of mass from the Sun.

Is this equation correct?

Is the equation correct? Let us try to justify it. Note that it is your job to justify the equations that you write down. It is not our job to reverse engineer them.You have a contribution from the rotational kinetic energy of the earth about its own axis. This from its [sidereal?] rotation rate of $\omega_1$.

So far, so good.

Then you have $\frac{1}{2}J_\text{CM}\omega_2^2$. No. That is not right at all. You've already accounted for the rotational kinetic energy of the Earth about its own center of mass. Counting it again would be double dipping.

[Counting it once with rotation rate corresponding to a solar day and then adjusting with the orbital rate about the sun to obtain a sidereal day would also be wrong. It makes my head hurt to think about it. The rotation rate is squared. Corrections would not add linearly]

Finally, you have a term $\frac{1}{2}ma^2\omega_2^2$. On the face of it, this is silly. Kinetic energy depends on velocity, not acceleration. But maybe some terms cancel out helpfully. So let us see where this leads...

Let us try dimensional analysis first. We are after a result in units of energy: $kg \ m^2 / sec^2$. You've proposed a term with units of mass times acceleration squared. That's $kg \ m^2 / sec^4$. No, that cannot possibly be correct.

[You could get a correct formulation such as $\frac{1}{2}maR$ by exploiting the fact that $a = \frac{v^2}{R}$. That is where I halfway thought you were going].

You want the second term to be the translational kinetic energy of an Earth-like object orbitting the sun at the orbital velocity of the Earth while maintaining a fixed orientation with respect to the distant stars.

We could get that using $E = \frac{1}{2}mv^2$ where $m$ is the mass of the Earth and $v$ is its orbital speed. But since you seem to prefer using the angular velocity ($\omega_2$) of the Earth in its orbit about the sun, we can write $v$ as $\omega_2 R$. Where $R$ is the radius of the Earth's orbit. That would give us:$$\text{KE}_\text{translation} = \frac{1}{2}mv^2 = \frac{1}{2}m\omega_2^2R^2$$

For consistency, we could write the rotational kinetic energy of a hypothetical Earth of uniform density as $$\text{KE}_\text{rotation} = \frac{1}{2}I\omega_1^2 = \frac{1}{2}\frac{2}{5}mr^2\omega_1^2 = \frac{1}{5}mr^2\omega_1^2$$

 

[Lotto]

Is the equation correct? Let us try to justify it. Note that it is your job to justify the equations that you write down. It is not our job to reverse engineer them.You have a contribution from the rotational kinetic energy of the earth about its own axis. This from its [sidereal?] rotation rate of $\omega_1$.

So far, so good.

Then you have $\frac{1}{2}J_\text{CM}\omega_2^2$. No. That is not right at all. You've already accounted for the rotational kinetic energy of the Earth about its own center of mass. Counting it again would be double dipping.

[Counting it once with rotation rate corresponding to a solar day and then adjusting with the orbital rate about the sun to obtain a sidereal day would also be wrong. It makes my head hurt to think about it. The rotation rate is squared. Corrections would not add linearly]

Finally, you have a term $\frac{1}{2}ma^2\omega_2^2$. On the face of it, this is silly. Kinetic energy depends on velocity, not acceleration. But maybe some terms cancel out helpfully. So let us see where this leads...

Let us try dimensional analysis first. We are after a result in units of energy: $kg \ m^2 / sec^2$. You've proposed a term with units of mass times acceleration squared. That's $kg \ m^2 / sec^4$. No, that cannot possibly be correct.

[You could get a correct formulation such as $\frac{1}{2}maR$ by exploiting the fact that $a = \frac{v^2}{R}$. That is where I halfway thought you were going].

You want the second term to be the translational kinetic energy of an Earth-like object orbitting the sun at the orbital velocity of the Earth while maintaining a fixed orientation with respect to the distant stars.

We could get that using $E = \frac{1}{2}mv^2$ where $m$ is the mass of the Earth and $v$ is its orbital speed. But since you seem to prefer using the angular velocity ($\omega_2$) of the Earth in its orbit about the sun, we can write $v$ as $\omega_2 R$. Where $R$ is the radius of the Earth's orbit. That would give us:$$\text{KE}_\text{translation} = \frac{1}{2}mv^2 = \frac{1}{2}m\omega_2^2R^2$$

For consistency, we could write the rotational kinetic energy of a hypothetical Earth of uniform density as $$\text{KE}_\text{rotation} = \frac{1}{2}I\omega_1^2 = \frac{1}{2}\frac{2}{5}mr^2\omega_1^2 = \frac{1}{5}mr^2\omega_1^2$$

OK, I had an idea that I need to account that rotational energy only once. And btw., that $a$ is a distance of Earth's centre of mass from the Sun, it is written in my post.

I would understand everything, except the fact that we still talk about a translational motion.

Its definition is that all points have same velocities, but when a take a point on Earth with a distance from the Sun $d_1$, its orbiting velocity is $\omega_2 d_1$. And when I now take a different point I a distance $d_2$, that is not equal to the first one, then $\omega_2 d_2$ is not equal to the first velocity either. So how can I treat it as a translational motion? This still confuses me.

 

[jbriggs444]

I would understand everything, except the fact that we still talk about a translational motion.

Its definition is that all points have same velocities, but when a take a point on Earth with a distance from the Sun $d_1$, its orbiting velocity is $\omega_2 d_1$. And when I now take a different point I a distance $d_2$, that is not equal to the first one, then $\omega_2 d_2$ is not equal to the first velocity either. So how can I treat it as a translational motion? This still confuses me.

*sigh*. We are not claiming that anything is actually moving in pure translational motion.

We are making a claim that the total kinetic energy of all the moving bits is equal to the mathematical sum of the kinetic energy of all the bits moving in one pattern plus the kinetic energy of all the bits moving in a different pattern.

That is one way that mathematics works. If there are many ways to calculate what is provably the same result, it is efficient to pick the way that is easiest, even if the steps might not match up with anything physical.

 

[Lotto]

*sigh*. We are not claiming that anything is actually moving in pure translational motion.

We are making a claim that the total kinetic energy of all the moving bits is equal to the mathematical sum of the kinetic energy of all the bits moving in one pattern plus the kinetic energy of all the bits moving in a different pattern.

That is one way that mathematics works. If there are many ways to calculate what is provably the same result, it is efficient to pick the way that is easiest, even if the steps might not match up with anything physical.

So if I understand it correctly, then the motion of the Earth is not actually translational, I just concentrate its whole mass into its centre of mass and assume it is a mass point moving with a corresponding velocity (when talking about that translational part).

But I thought I can do it only when the motion is actually translational, but it seems that it is not a condition.

 

[jbriggs444]

So if I understand it correctly, then the motion of the Earth is not actually translational, I just concentrate its whole mass into its centre of mass and assume it is a mass point moving with a corresponding velocity (when talking about that translational part).

The translational kinetic energy of a non-rotating rigid object is identical to the translational kinetic energy of a point-like object of equal mass with the same velocity, yes.

But I thought I can do it only when the motion is actually translational, but it seems that it is not a condition.

I am running out of words to explain this. The total kinetic energy of an object is equal to the sum of the rotational kinetic energy of that object about its center of mass plus the translational kinetic energy that the object would have if it were translating but not rotating.

See you in another thread perhaps.

 

[Lotto]

The translational kinetic energy of a non-rotating rigid object is identical to the translational kinetic energy of a point-like object of equal mass with the same velocity, yes.I am running out of words to explain this. The total kinetic energy of an object is equal to the sum of the rotational kinetic energy of that object about its center of mass plus the translational kinetic energy that the object would have if it were translating but not rotating.

See you in another thread perhaps.

OK, I think I understand it now. Thank you for your patience.