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Kleppner and Kolenkow Introduction to mechanics

In the approximation that the Earth is spherical, it experiences no torque from nearby bodies. Consequently, its angular momentum is constant—both its spin and its angular momentum always point in the same direction in space

How is the torque zero?

If I take any point in space to be my origin then it is not necessary that the torque on Earth about that is zero.

Below is a sketch, the torques are calculated from A, the two bodies are sun and earth perfect spheres.

enter image description here

Kashmiri
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    possible duplicate: https://physics.stackexchange.com/questions/561902/why-do-we-say-that-the-torque-on-earth-due-to-sun-is-zero – WwW Jun 26 '21 at 15:39

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If we assume the Earth to be a perfect sphere, then the force, with which the Sun (or the Moon, or any other planet for that matter) acts on the Earth is

$$\vec{F} \, =\, -\,\frac{GMm}{|\vec{r}|^3} \, \vec{r}$$

Where $\vec{r}$ is the vector pointing from the Sun to the Earth. The vector pointing from the Earth to the Sun is then $- \,\vec{r}$ and since the Sun acts with the force $\vec{F}$ from the position $- \, \vec{r}$ relative to the Earth, then the torque is \begin{align} \vec{T} \, =& \, (-\,\vec{r}\,) \times \vec{F} = (-\,\vec{r}\,) \times \left( -\,\frac{GMm}{|\vec{r}|^3} \, \vec{r} \right)\\ =& \, \vec{r} \times \left(\,\frac{GMm}{|\vec{r}|^3} \, \vec{r} \right)\\ =& \, \left(\,\frac{GMm}{|\vec{r}|^3} \, \right) \vec{r} \times \vec{r}\\ =& \, 0 \end{align}

Because by the properties of the cross-product $\vec{r} \times \vec{r} = 0$.

However, in reality, in terms of rigid-body dynamics, Earth is much better modelled as an ellipsoid of revolution and then the gravitational force is not radial, i.e. is not a multiple of the position vector $\vec{r}$ and in that case non-zero torques arise.

I wrote this post that derives in a fairly detailed mathematical way some models of Earth's rotational axis' dynamics under the torques exerted by the Sun and the Moon. Take a look at it if you are interested in the precession and nutation dynamics of Earth's axis of rotation.

Futurologist
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  • Thanks, I do agree the torques are zero if the point of Calculation is in line with the sun earth system. I was however asking about the torque calculated about any general point as I've written in my post. – Kashmiri Jul 04 '21 at 16:58
  • @Kashmiri You understand that if a torque is zero in one inertial coordinate system, it is zero in any other one. The vector between earth and sun is aligned with the force vector no matter where the center of the inertial coordinate system is. – Futurologist Jul 04 '21 at 17:04
  • the torques are equal in any system only when the sum of the forces is zero. Here there is a net force towards the sun. – Kashmiri Jul 05 '21 at 03:28
  • @Kashmiri That last statement is simply not true. If the force and the vector, to the point at which the force is applied, are aligned (i.e. they point in the same direction or in opposite directions), then the torque is zero, because the cross product between to aligned vectors is always zero, no matter with respect to which coordinate system you write them to. – Futurologist Jul 05 '21 at 04:39
  • But this says the opposite https://physics.stackexchange.com/questions/93698/why-if-the-torque-equals-zero-measured-from-one-point-in-space-it-equals-zero-me – Kashmiri Jul 05 '21 at 05:32
  • You are confusing torque and angular momentum measured with respect to a point and torque and angular momentum written in the coordinates of a given coordinate frame. – Futurologist Jul 05 '21 at 14:57
  • The authors discuss the torque with respect to the center of the earth, written in the coordinates of the inertial coordinate system of the solar system. nobody is claiming the torque with respect to any point in space is zero. The torque is zero with respect to any point in space, only when the forces, as vectors add up to zero in at least one inertial coordinate system. That's what the statement of the link you posted is. But that's not what the authors are claiming. And neither am I. – Futurologist Jul 05 '21 at 15:12
  • So the torques have been calculated from the centre of earth. The authors should have written that explicitly imo. – Kashmiri Jul 06 '21 at 04:47
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An external field (uniform or radial) will cause no net torque on a sphere relative to the center of the sphere ( and will not cause an angular acceleration about that center). If you consider the force on the sphere to be acting about some other (perhaps external) point, then there will be a torque and a change in the angular momentum (associated with the change in linear momentum) of the sphere relative to that other point. The moon can exert a torque about the center of the (non-spherical) earth in two ways: If the moon does not lie in the equatorial plane, it will pull harder on the near side equatorial bulge than on the far side bulge. This contributes to the slow precession of the earth's axis. Also the moon pulls the oceans up on the near side, and the earth away from the oceans on the far side. Then the rotation of the earth drags this tidal bulge to the East. Since the moon pulls stronger on the nearside tide, it causes a torque which is slowly decreasing the angular velocity of the earth (and a reaction which is increasing the velocity of the moon).

R.W. Bird
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  • R.W. Bird, NASA laser ranging measurements indicate that the moon is slowly receding away from the earth. A "higher" orbit of the moon would require a slower velocity if it is to remain in orbit. – David White Jul 04 '21 at 21:26
  • Thank you Mr R.W. Bird. The author of book said "In the approximation that the Earth is spherical, it experiences no torque from nearby bodies" so is he talking with his origin at the centre of earth? Because otherwise as you said the torques won't be zero. – Kashmiri Jul 05 '21 at 03:31
  • David, I'm not at all clear on the mechanism involved when the reaction force gives the moon a small continuous acceleration in the forward direction. I do understand that if you want to move the space shuttle to a larger circular orbit, you accelerate briefly. That puts it into an elliptical orbit. At the outer end of the ellipse, you accelerate again to get into a circle (with a smaller velocity). The loss of kinetic energy is smaller than the gain in potential energy. The forward acceleration of the moon does move it to a larger orbit. – R.W. Bird Jul 05 '21 at 13:52
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Suppose we model a sphere as two conjoined hemispheres, divided along the axis from the sphere's center to a nearby body.

Let the center of mass of the top hemisphere be located at radius R, while the center of mass of the bottom half of the hemisphere is located at radius -R.

The gravitational force on a sphere is such that the net force on the top half of the sphere is $\vec F_1$ and the net force on the bottom half of the sphere is $\vec F_2$, where $F_1 = F_2 = F$ and both vectors point towards the nearby body.

Then the forces exert torque $T_1 = FRcos(\theta)$ on the upper hemisphere and $T_2 = -FRcos(\theta)$ on the lower hemisphere where $\theta$ is the angle between the forces and the axis.

$\sum T = 0$

We can repeat this process for any subdivision of the sphere, each time finding that for each torque on an upper subdivision, there is an equal and opposite torque on the corresponding lower subdivision, summing to a net zero torque.

g s
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