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

[Lotto]

TL;DR

The Earth is rotating around its axis, but it also orbits the Sun. So what is its total kinetic energy?

Because it rotates around its axis, then it has a rotational kinetic energy. But when it orbits the Sun, then it has an another rotational energy too. Would it be $E_\mathrm{rot} = \frac 12 (\frac 25 MR^2+ Md^2) {\omega}^2$, where $d$ is the distance between the Sun and the Earth. I calculated the Earth's moment of inertia using Steiner's theorem. Is that corect? Or should it be a translation energy instead? So instead of $E_\mathrm{rot}$ it would be just $\frac 12 M{\omega}^2d^2$?

 

[malawi_glenn]

Is the Earth a homogenous solid sphere?

 

[hmmm27]

Regarding the solar system, Earth/Moon has an orbital velocity. (There's also a few big rocks and lots of dust that orbit the Earth or L4/L5, but not much mass, relatively speaking).

The Earth's density changes from core to crust : this will affect your calculations for rotational energy.

 

[Lotto]

I am asking because I saw many times that Earth's kinetic energy is calculated as a rotational energy around its axis plus a translation energy caused by orbiting the Sun. That is not a translation motion. Only if we consider Earth to be a point. When we assume that it is a sphere, then each point has a different velocity. So how to determine its total kinetic energy? Can we use that Steiner's theorem or is it a nonsense?

 

[Nugatory]

When we assume that it is a sphere, then each point has a different velocity.

For the energy of the rotation around the axis, google for "moment of inertia".
This is a quantity that is calculated from the shape of density of any non-point object (and the calculation is particularly easy for a sphere, not much harder for a sphere with different densities at different depths). The rotational kinetic energy is $E_R=I\omega^2/2$ where $I$ is the moment of inertia and $\omega$ is the rate of rotation - the similarity to the translational kinetic energy $E_K=mv^2/2$ is not a coincidence.

For the translational energy of the earth orbiting the sun, do the calculation for yourself: What is the velocity of a point on the surface of the earth closest to the sun? Of a point on the earth farthest from the sun? Are they sufficiently different to justify doing an exact calculation instead of pretending that they are equal?
(Although if you do want to do the exact calculation, you can reasonably approximate the earth as spherically symmetric and you will have an integral is a good exercise in first-year calculus).

 

[Lotto]

For the energy of the rotation around the axis, google for "moment of inertia".
This is a quantity that is calculated from the shape of density of any non-point object (and the calculation is particularly easy for a sphere, not much harder for a sphere with different densities at different depths). The rotational kinetic energy is $E_R=I\omega^2/2$ where $I$ is the moment of inertia and $\omega$ is the rate of rotation - the similarity to the translational kinetic energy $E_K=mv^2/2$ is not a coincidence.

For the translational energy of the earth orbiting the sun, do the calculation for yourself: What is the velocity of a point on the surface of the earth closest to the sun? Of a point on the earth farthest from the sun? Are they sufficiently different to justify doing an exact calculation instead of pretending that they are equal?
(Although if you do want to do the exact calculation, you can reasonably approximate the earth as spherically symmetric and you will have an integral is a good exercise in first-year calculus).

I understand that we can consider Earth to be a point but as a wrote, if we do not do that, then Earth is rotating around the Sun, so my question is if we can describe its rotational energy with that equation in my original post.

I think we cannot do that but I do not know why.

 

[A.T.]

I am asking because I saw many times that Earth's kinetic energy is calculated as a rotational energy around its axis plus a translation energy caused by orbiting the Sun.

Yes, one can do this.

That is not a translation motion.

That's why you have to add the rotational energy (based on rotation around the Earth's center of mass) to the translational energy (based on the center of mass translation).

Can we use that Steiner's theorem

The decomposition of KE above is related to Steiner's theorem.

 

[Vanadium 50]

Is the Earth a homogenous solid sphere?

I don't think it matters.

Once a year the earth goes around the sun: that's 20π x 10⁷ miles in π x 10⁷ seconds. OK 20 miles/second. The earth at the equator is moving at 1000 miles per hour due to rotation, or 1/3 mile per second. Ignoring all the complications like 2/5 and the like, there is 60x the orbital velocity pr 3600 times the kinetic energy - the rotation makes an effect in the 3rd or 4th decimal place. (Even without the 2/5_

We know the earth's mass a little better than that, but not much.

It;s a tiny correction. Whether 2/5 should be 2/5 or some othe number will be tinier.

 

[Lotto]

I don't think it matters.

Once a year the earth goes around the sun: that's 20π x 10⁷ miles in π x 10⁷ seconds. OK 20 miles/second. The earth at the equator is moving at 1000 miles per hour due to rotation, or 1/3 mile per second. Ignoring all the complications like 2/5 and the like, there is 60x the orbital velocity pr 3600 times the kinetic energy - the rotation makes an effect in the 3rd or 4th decimal place. (Even without the 2/5_

We know the earth's mass a little better than that, but not much.

It;s a tiny correction. Whether 2/5 should be 2/5 or some othe number will be tinier.

I am confused by this: if we have a spherical pendulum of radius $r$ and mass $m$ hanged on a rope of lenght $l$, then what is its total kinetic energy?

I would say that $E_k =\frac 12 (\frac 25 mr^2+m(l+r)^2) {\omega}^2$. If we say that $d=l+r$, then we can write the energy as $E_k=\frac 12 \frac 25 mr^2 \omega^2 + \frac 12 md^2 {\omega}^2=\frac 12 \frac 25 mr^2 {\omega}^2+\frac 12 mv^2$. So that $v$ is a velocity of the centre of mass of the sphere, but not the velocity of a translational motion because the spehre is only rotating?

 

[vanhees71]

The decomposition of the motion in translational motion and intrinsic rotation, of course depends on the choice of the body-fixed reference point. Here you made the most convenient choice of the center of mass (of a homogeneous sphere) as the body-fixed reference point. The corresponding kinetic energy is $m v^2/2=m d^2 \omega^2/2$. The other part is due to the rotation of all the other mass elements of the sphere around the body-fixed reference point. It's easy to see that indeed the corresponding angular velocity is also $\omega$ and thus that's $\Theta_{\text{CM}} \omega^2/2$ with $\Theta_{\text{cm}}=2m r^2/5$. I thus think your expression is correct.

You can directly see this also by using the parallel-axis theorem, as indicated in your first ansatz for $E_k$.

 

[Vanadium 50]

Honestly, if the approximation of "all the weight is at the end of the pendulum" is not good enough, I would treat this entirely as a rotation: your pendulum has a certain moment of inertia and a certain restoring torque when rotated.

 

[vanhees71]

This he did with his first ansatz:
$$E=\frac{\Theta}{2} \omega^2$$
with
$$\Theta=\Theta_{\text{CM}}+m (l+r)^2.$$

 

[Lotto]

I think I know what confuses me: that axes of rotation.

When we have a sphere rolling without slipping on the ground, we can write its total kinetic energy as $E_k=\frac 12 J_{CM}{\omega}^2+\frac 12 mv^2$. That I understand. But I saw another derivation of it:
if I let the sphere rotate around a point where the sphere touches the ground, then its new moment of inertia is $J_{CM}+mr^2$ and then $E_k=\frac 12 (J_{CM}+mr^2){\omega}^2=\frac 12 J_{CM}{\omega}^2+\frac 12 mv^2$.

I don't understand why I can do this. Why I can let it rotate around that point? The whole sphere is rotating around its centre of mass and that is its only axis. But it seems that it is not. So are axes of rotation relative? And when I let it rotate around that point, how to imagine that? Because I am struggling with that.

 

[A.T.]

So are axes of rotation relative?

Yes, there are entrie lengthy threads about this here:
https://www.physicsforums.com/threa...ays-rotate-about-their-centre-of-mass.990571/

 

[vanhees71]

The somewhat mindboggling point of the confusion is that you can describe the rigid body by arbitrarily chosing a body-fixed frame of reference, i.e., one body-fixed reference point and a body-fixed (Cartesian) basis fixed at that reference point.

In your first argumentation you choose the center of the sphere as the body-fixed reference point. From the no-slipping condition you get that its velocity is related to the angular velocity by $v=r \omega$, and note that $\omega$ is always the angular velocity of the rotation of the sphere around the chosen body-fixed reference point. Now $J_{\text{CM}}=2/5 m r^2$ and you get
$$E_{\text{kin}}=\frac{m}{2} \left (\frac{2}{5} r^2+r^2 \right)\omega^2=\frac{m}{2} \frac{7}{5} r^2 \omega^2.$$
Now take the other point of view, where the momentary point at the ground is taken as the momentary body-fixed reference point. That point is momentary at rest, and you have only rotation of the sphere around it, but of course the momentum of inertia changes. According to the parallel-axis theorem it's given by
$$J=J_{\text{CM}}+m r^2 = m \left (\frac{2}{5} r^2 + r^2 \right)=\frac{7}{5} m r^2.$$
From this point of view there's only the rotational energy for the rotation around this point. It's easy to make sure that $\omega$ is the same as in the first approach (which again is due to the no-slipping condition). Of course, you get the same kinetic energy with both approaches, as it should be.

 

[Lotto]

The somewhat mindboggling point of the confusion is that you can describe the rigid body by arbitrarily chosing a body-fixed frame of reference, i.e., one body-fixed reference point and a body-fixed (Cartesian) basis fixed at that reference point.

In your first argumentation you choose the center of the sphere as the body-fixed reference point. From the no-slipping condition you get that its velocity is related to the angular velocity by $v=r \omega$, and note that $\omega$ is always the angular velocity of the rotation of the sphere around the chosen body-fixed reference point. Now $J_{\text{CM}}=2/5 m r^2$ and you get
$$E_{\text{kin}}=\frac{m}{2} \left (\frac{2}{5} r^2+r^2 \right)\omega^2=\frac{m}{2} \frac{7}{5} r^2 \omega^2.$$
Now take the other point of view, where the momentary point at the ground is taken as the momentary body-fixed reference point. That point is momentary at rest, and you have only rotation of the sphere around it, but of course the momentum of inertia changes. According to the parallel-axis theorem it's given by
$$J=J_{\text{CM}}+m r^2 = m \left (\frac{2}{5} r^2 + r^2 \right)=\frac{7}{5} m r^2.$$
From this point of view there's only the rotational energy for the rotation around this point. It's easy to make sure that $\omega$ is the same as in the first approach (which again is due to the no-slipping condition). Of course, you get the same kinetic energy with both approaches, as it should be.

OK, as I can see, when the axis has no velocity, then the translational part of a kinetic energy is zero.

But I still cannot imagine it. When the axis goes through the centre of mass, how should I imagine the motion to understand it intuitively? If I understand it correctly, the axis is a frame of reference, so can I imagine it as if I was that axis and the sphere was rotating around me? But then I would have a zero velocity since I am not moving, only the space around me, including the sphere.

I know that my thoughts are probably wrong, but this is how I see it. How do you imagine it?

 

[jbriggs444]

But I still cannot imagine it. When the axis goes through the centre of mass, how should I imagine the motion to understand it intuitively? If I understand it correctly, the axis is a frame of reference, so can I imagine it as if I was that axis and the sphere was rotating around me? But then I would have a zero velocity since I am not moving, only the space around me, including the sphere.

I am not seeing what you do not understand. So it is hard to know whether the following words will help.

You appear to be talking about an object rotating about an axis through its mass center. Yes, a reasonable way to think about that is to adopt a frame of reference where the axis is stationary. In an intuitive way, you are standing at the origin of a coordinate system and that origin is tied to the center of mass.

So yes, the sphere is rotating around you. You can calculate its moment of inertia about an axis passing through the center of mass. Given its rotation rate, you can calculate its angular momentum about the same axis.

So what if it has zero velocity in this frame of reference?

But that is not the scenario that @vanhees71 is describing.

@vanhees71 adopts a frame of reference where a solid sphere is rolling on a surface. The point of contact between sphere and surface is not slipping. The point on the sphere where this contact is made is, at least momentarily, stationary in the rest frame of the surface.

@vanhees71 adopts the rest frame of the surface. He places the reference axis (and the origin of his coordinate system if you like to think that way) at this momentary point of contact. This is the "instantaneous center of rotation".

Each and every point on the sphere is, at least momentarily, moving in the correct direction and at the correct speed to rigidly rotate the whole sphere about the instantaneous center of rotation. There is a rotation rate associated with this instantaneous rate of rotation. There is a moment of inertia associated with this instantaneous axis.

The instantaneous motions of every point on the sphere is completely described by the rotation alone. The axis is not translating. Accordingly, the kinetic energy of the sphere must consist entirely of the its rotational kinetic energy.

With the parallel axis theorem, we can see that the moment of inertia of the sphere about an axis tangent to its rim is equal to the moment of inertia of the sphere about its center plus $mr^2$ where $r$ is there sphere's radius and $m$ is its mass.

 

[A.T.]

If I understand it correctly, the axis is a frame of reference,

You seem to conflate two different choices:
- To describe the motion of a rigid body you can choose from infinitely many frames of reference
- For each of those reference frames you can choose from infinitely many ways to decompose the motion of the rigid body in that frame into rotation (around an axis) and translation (of the axis)

 

[Lotto]

OK, I think I get this (although it still appears weird to me), but what I do not get is when I have the pendulum and let it rotate around its centre of mass. How is that possible? It rotates around a pivot, so then its kinetic energy is $\frac 12 (J_{CM}+md^2) {\omega}^2$. And when it rotates somehow around its axis, it has the kinetic energy written as $\frac 12 mv^2 + \frac 12 J_{CM} {\omega}^2$, where that ${\omega}$ is an angular velocity around the pivot. But why around the pivot when it rotates around its axis? This is almost the same problem like the one in my original post.

 

[jbriggs444]

OK, I think I get this (although it still appears weird to me), but what I do not get is when I have the pendulum and let it rotate around its centre of mass. How is that possible? It rotates around a pivot, so then its kinetic energy is $\frac 12 (J_{CM}+md^2) {\omega}^2$. And when it rotates somehow around its axis, it has the kinetic energy written as $\frac 12 mv^2 + \frac 12 J_{CM} {\omega}^2$, where that ${\omega}$ is an angular velocity around the pivot. But why around the pivot when it rotates around its axis? This is almost the same problem like the one in my original post.

Kinetic energy is not invariant. It changes with a choice of reference frame. If you pick a reference frame where the pendulum's center of mass is (momentarily) motionless then the kinetic energy associated with the motion of the center of mass goes *poof* in a puff of coordinates.

 

[A.T.]

...where that ${\omega}$ is an angular velocity around the pivot. But why around the pivot when it rotates around its axis?

Its not the same ${\omega}$ in general, just in special cases like a rigid body pendulum or tidally locked orbit.

 

[vanhees71]

Maybe a complete treatment of the rigid-body dynamics helps. Here's my try to explain it:

https://itp.uni-frankfurt.de/~hees/pf-faq/spinning-top.pdf

 

[Lotto]

But when I describe the pendulum's kinetic energy as $E_k=\frac 12 (J_{CM}+md^2){\omega}^2$, then how is it possible that at the maximum point, the pendulum has a zero velocity? Because I suppose that the angular velocity is constant, so there is nothing to change in the equation. And I think that this equation should be valid in every point the pendulum goes through.

I could you please explain me why is the pendulum's motion translational when all points don't move on the same trajectories? If the pendulum was a mass point, then I understand and the motion is translational, but this just doesn't make sense to me.

When I have a rolling object, then the motion is translational and also rotational, but in case of pendulum, I see no reason why it should be translational. I can denscibe its translational kinetic energy by using the velocity of the the centre of mass, but I can only do it when all points making the body have the same magnitude of velocities. And this is not the case.

Those are the things that confuses me most.

 

[jbriggs444]

But when I describe the pendulum's kinetic energy as $E_k=\frac 12 (J_{CM}+md^2){\omega}^2$, then how is it possible that at the maximum point, the pendulum has a zero velocity? Because I suppose that the angular velocity is constant, so there is nothing to change in the equation. And I think that this equation should be valid in every point the pendulum goes through.

What makes you think that the angular velocity of a pendulum is constant? The "maximum point" for angular velocity would be at the bottom of the stroke, right?

Please slow down and define your variables before you write down the equations. I suspect that $J_\text{CM}$ is the moment of inertia of the pendulum about its center of mass, that $E_k$ is the total kinetic energy of the pendulum, $m$ is the mass of the pendulum, $d$ is the distance from the pendulum's fixed pivot to the CoM and $\omega$ is the pendulum's current angular velocity.

One expects a pendulum with a short stroke to approximately conform to simple harmonic motion in which $\omega$ will vary in an approximately sinusoidal pattern.

 

[Lotto]

But what about the formula for angular velocity of a rigid body pendulum? That is $\omega=\sqrt{\frac{g}{L}}$ and it is valid in every point the pendulum goes through. So why does it change?

 

[vanhees71]

But when I describe the pendulum's kinetic energy as $E_k=\frac 12 (J_{CM}+md^2){\omega}^2$, then how is it possible that at the maximum point, the pendulum has a zero velocity? Because I suppose that the angular velocity is constant, so there is nothing to change in the equation. And I think that this equation should be valid in every point the pendulum goes through.

I could you please explain me why is the pendulum's motion translational when all points don't move on the same trajectories? If the pendulum was a mass point, then I understand and the motion is translational, but this just doesn't make sense to me.

When I have a rolling object, then the motion is translational and also rotational, but in case of pendulum, I see no reason why it should be translational. I can denscibe its translational kinetic energy by using the velocity of the the centre of mass, but I can only do it when all points making the body have the same magnitude of velocities. And this is not the case.

Those are the things that confuses me most.

Why should $\omega$ be constant? The equation of motion of the pendulum is
$$J \ddot{\varphi}=-\sqrt{\frac{g}{l}} \sin \varphi,$$
where $l$ is the distance from the pivot point to the center of mass. The only stable solution with $\omega=\dot{\varphi}=\text{const}$ obviously is $\varphi=0=\text{const}$ and thus $\omega=0$.

 

[Lotto]

Why should $\omega$ be constant? The equation of motion of the pendulum is
$$J \ddot{\varphi}=-\sqrt{\frac{g}{l}} \sin \varphi,$$
where $l$ is the distance from the pivot point to the center of mass. The only stable solution with $\omega=\dot{\varphi}=\text{const}$ obviously is $\varphi=0=\text{const}$ and thus $\omega=0$.

OK, sorry, now I see my mistake. I thought it is the angular velocity of the moving pendulum, but it seems to be only an angular velocity of a moving phasor, such in case of an oscillating spring.

So now it makes sense that the kinetic energy changes. But I still do not get why I can write the kinetic energy as $\frac 12 mv^2+\frac 12 J_{CM} {\omega}^2$, where the velocities correspond to instantaneous mangitudes.

 

[jbriggs444]

But I still do not get why I can write the kinetic energy as $\frac 12 mv^2+\frac 12 J_{CM} {\omega}^2$, where the velocities correspond to instantaneous mangitudes.

It is an algebraic fact of life, probably some named theorem that the total kinetic energy of a rigid body can be decomposed into the bulk kinetic energy associated with the motion of the center of mass plus the kinetic energy associated with rotation about the center of mass.

Personally, I find that intuition is easy while algebra is hard. Let us try a proof without any algebra.

Suppose that rotational kinetic energy (about the center of mass) were not independent of an object's linear velocity. Perhaps a spinning object has more rotational kinetic energy when it is moving rapidly down the highway than when it is spinning at rest.

We could take a gyroscope at rest, spin it up to a high rotation rate, and accelerate it down the highway. This would require some amount of energy.

We could then spin the gyroscope down, harvesting the increased rotational kinetic energy and bring it to rest, harvesting the kinetic energy from linear motion.]

*voila* -- perpetual motion. But we know that energy is conserved. So we have a proof by contradiction that rotational kinetic energy cannot increase with linear velocity.

For completeness, we could speculate that kinetic energy decreases with linear velocity. But we can run the cycle the other way around and still get perpetual motion and a proof by contradiction that such cannot be so.

 

[vanhees71]

I've no clue what a phasor is. The equation describes the motion of a pendulum with $\varphi$ the angle between the direction of the constant gravitational force and the line connecting the pivot point with the center of mass of the pendulum.

The kinetic energy follows from the parallel-axis theorem:
$$T=\frac{J}{2} \dot{\varphi}^2 = \frac{1}{2} (m l^2 +J_{\text{CM}}) \dot{\varphi}^2=\frac{m}{2} v^2 + \frac{J_{\text{CM}}}{2} \dot{\varphi}^2,$$
where $v=|\vec{v}|=|\dot{\varphi}|l$.

 

[jbriggs444]

I've no clue what a phasor is

As I construct it, he is considering the angular frequency of a system undergoing a cyclical state change such as simple harmonic motion rather than the angular velocity of a pendulum within such a system.