As we know the Saturn's shape is not spherical. In fact, its polar radius is almost 10% smaller than equatorial radius. This flattening is caused by the rapid rotation of this planet.

Saturn's non spherical shape

Image credit: NASA, Voyager 2.

Question: how can we explain this flattening with forces?

Here is my unfinished attempt of explaining it. If we consider a fixed mass of material on the surface of the planet, the centripetal force is

$$F_c = a_cm,$$

where $a_c$ is the centripetal acceleration of this material. Using formula for centripetal acceleration $a_c = \frac{v^2}{r}$ and replacing tangential velocity with angular velocity $v=r\omega$ we get:

$$F_c = r\omega^2m \tag{1}.$$

Now we compare two identical masses located on the surface of the planet shown on the image below. One mass $A$ is on the equator and the other mass $B$ is near the pole. Mass $A$ will have bigger distance $r_a$ to the axis of rotation and, according to Equation 1, will require bigger centripetal force to keep it on the surface than mass $B$.

This is where I get stuck. While I understand the flattening intuitevely, I struggle to understand it in exact physics terms using the centripetal acceleration and forces. Can anyone help me finish my explanation, or suggest a better one?

This question is different from Why is the Earth so fat? because here I am seeking for a simple explanation of the flattening of Saturn in terms of centripetal forces. I am hoping to understand the principle — why is the flattening occurs — and not seeking for exact solution that gives me the shape of the planet. How can this question be answered that a high-school student can understand?

Masses located on the surface of rotating Saturn

Figure 1: Clumps of gas $A$ and $B$ on rotating Saturn.

Attempt #2 of explaining the flatness of Saturn using centripetal forces

Let us consider again two identical clumps of gas $A$ and $B$ on the surface of Saturn, shown on Figure 2. The distance from mass $A$ to the axis of planet's rotation is the radius of the planet $r$. If $\alpha_{lat}$ is the latitude angle of the clump $B$, then the distance from $B$ to the axis is $r\cos{\alpha_{lat}}$.

Distances from two masses to the axis of planet's rotation

Figure 2: Distances from two clumps to the axis of planet's rotation.

Let's draw a diagram of forces acting on clump $B$ (see Figure 3). One force will be the force of gravity $F_{grav}$ pointing to the center of the planet. Here we only care about the horizontal direction, therefore, the horizontal component of the force of gravity will be $F_{Hgrav} = F_{grav} \cdot \cos{\alpha_{lat}}$.

Force diagram for $B$

Figure 3: Forces acting on clump $B$.

Another force acting on $B$ is the force of tension $T$. This force of tension is the key idea and it is worth explaining what it means. Force $T$ arises if we view the planet as a rigid object where all different clumps of matter are stuck together. Those clumps of matter act like a 'rod' that connects $B$ to the axis of rotation. We will use the tension $T$ and the 'rod' as simplifications in order to understand the flattening of Saturn with simple concepts.

As Saturn rotates, clump $B$ naturally wants to move in straight line (Newton's 1st law). However, clump $B$ is tied to the center of the planet with this imaginary 'rod' that prevents $B$ from flying off the surface of the rotating planet.

If we imagine Saturn rotating faster, the tension $T$ of this rod will increase, since the centripetal acceleration of $B$ will be higher and more force will be needed to keep $B$ on the surface. If, on the other hand, Saturn stops rotating, then the tension $T$ will disappear and gravity will be the only force that acts on $B$.

We can use Newton's 2nd law to write the centripetal acceleration $a_c$ in terms of the forces of gravity, tension and the mass of a clump.

$$ a_c = \frac{F_{Hgrav} + T}{m}. $$

Using the formula for centripetal acceleration $a_c = \frac{v^2}{r}$ and replacing the tangential velocity $v$ with angular velocity $v=r\omega$ we get:

$$ r\omega^2 = \frac{F_{Hgrav} + T}{m}. \tag{2} $$

We rearrange Equation 2 and solve for the force of tension $T_a$ of the clump $A$ located on Saturn's equator:

$$ T_a = mr\omega^2 - F_{Hgrav}.\tag{3} $$

Similiarly, the force of tension $T_b$ for clump B will be:

$$T_b = m \cdot r\cos{\alpha_{lat}} \cdot \omega^2 - F_{grav} \cdot \cos{\alpha_{lat}}\\ = \cos{\alpha_{lat}} (m r\omega^2 - F_{grav}).$$ Therefore, $$ T_b = T_a \cdot \cos{\alpha_{lat}}. \tag{4} $$

From Equation 4 it follows that

$$ |T_b| < |T_a|. \tag{5} $$

In fact, this absolute value of tension gets smaller for clumps that are closer to the poles, since latitude angle $\alpha_{lat}$ gets larger. This means that less tension $T$ is needed to keep clump $B$ on the surface, as planet rotates, and more tension is needed for clump $A$ on the equator. If we use Newton's 3rd law, we can see that clump $B$ pulls less on the imaginary 'rod' that connects it to the axis of rotation. As a result, this 'rod' gets less stretched for clump $B$.

On the other hand, tension $T$ is higher for clump $A$, which stretches its 'rod' more and makes it longer. This increases the length of the radius in the direction of the equator of Saturn and makes the planet fatter along the equator.

Does this explanation make sense?

  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/8074/2451 , physics.stackexchange.com/q/69562/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 4, 2016 at 9:49
  • 2
    $\begingroup$ The shape of a rotating planet in hydrostatic equilibrium, also known as the reference geoid, is defined as the level surface of equal effective potential U by: $\frac{GM}{r}\left(1-\sum_{n=1}^{\infty}\frac{a^{2n}}{r^{2n}}J_{2n}P_{2n}cos(\theta)\right)+\frac{1}{2}\omega^{2}r^{2}sin(\theta)$ where a is the equatorial radius of the geoid and P, the time period of rotation. J is defined as saturns harmonic coefficient. You could find the force resulting from this potential for different angles and then see the data for any significant result. $\endgroup$ Dec 4, 2016 at 10:16
  • $\begingroup$ Do you understand the difference between centrifugal & centripetal force? $\endgroup$
    – Qmechanic
    Dec 4, 2016 at 23:52
  • $\begingroup$ @Qmechanic, centripetal is the force that is pointing towards the center of rotation, like the force of gravity in our context. Centrifugal force is a fictitious force observed in an accelerating frame of reference? Is that correct? $\endgroup$
    – Evgenii
    Dec 5, 2016 at 6:38
  • 1
    $\begingroup$ Right, they are in principle equivalent formulations. Now try to ask your question in each formulation. Can you solve your question in one of the formulations? $\endgroup$
    – Qmechanic
    Dec 5, 2016 at 7:33