Why the torque on Earth is zero if it's assumed to be spherical

Question

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.

Answer 1

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.

Answer 2

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).

Answer 3

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.