<p>The gravitational field near the center of a thin narrow ring of mass, <span class="math-container">$M_r$</span>, and radius, <span class="math-container">$R_r$</span>, is <span class="math-container">$H_z = -GM_rz/{(z^2+R_r^2)^{3/2}}$</span> for the z-directed field on the z-axis. Near the center, <span class="math-container">$(z,r) \approx (0,0)$</span>, the field is <span class="math-container">$H_z \approx -GM_rz/R_r^3 $</span> and the gradient is  <span class="math-container">$\partial H_z/\partial z = -GM_r/R_r^3$</span>
The field locally must satisfy the Laplace equation <span class="math-container">$\partial H_z/\partial z = -\partial (rH_r)/r\partial r$</span>. This leads to <span class="math-container">$H_r = (1/r) \int r (\partial H_z / \partial z) dr = +GM_rr/2R_r^3$</span> for the radial field near the center.</p>
<p>From above, the gravitational potential <em>near the center</em> of the ring is <span class="math-container">$\Phi(z,r) \approx GM_r(2z^2 - r^2)/4R_r^3$</span></p>
<p>For a small non-rotating spherical planet mass, <span class="math-container">$M_p$</span>, radius, <span class="math-container">$R_p$</span>, at the center of the ring, we are interested in how the equipotental surface is perturbed. The field gradient at the planet's surface is <span class="math-container">$-GM_p/R_p^2$</span>. The perturbation, <span class="math-container">$\epsilon$</span>, to the equipotential surface by the potential from the ring is disturbing-potential divided by field-gradient <span class="math-container">$\epsilon = [GM_r(2z^2 - r^2)/4R_r^3]/[-GM_p/R_p^2]$</span>.</p>
<p>At the pole, <span class="math-container">$(R_p,0)$</span>, we have <span class="math-container">$\epsilon_{pole} = -(1/2)(M_r/M_p)(R_p^4/R_r^3)$</span>.</p>
<p>At the equator, <span class="math-container">$(0, R_p)$</span>, we have <span class="math-container">$\epsilon_{equator} = +(1/4)(M_r/M_p)(R_p^4/R_r^3)$</span>.</p>
<p>As expected the spherical planet becomes slightly oblate. This result requires a small planet and a big ring, <span class="math-container">$R_p &lt;&lt;R_r$</span>, and the ring should be thin and narrow.</p>

CC BY-SA 4.0
2022-03-14T06:41:53.167Z
2022-03-14T06:41:53.167Z
Roger Wood
<p><strong>Question:</strong> Could the gravitational force of Earthly rings distort the Earth to make it more oval shaped at the equator?</p>
<p><strong>Short Answer:</strong> Yes</p>
<blockquote>
<p>This makes sense to me as when the moon is whole, all the mass is in relatively one place so the force of gravity will be stronger at its center of gravity, and when it's turned into rings there is no center of gravity for the moon anymore as its constituent particles are now spread around the Earth. But they must still exert some gravity on Earth</p>
</blockquote>
<p>As you have said, The distortion will not be as much compared to the gravitational effects of the moon as whole, for example, the tides would still happen if the moon is replaced with rings, only thing is it will be reduced greatly, if the rings is not aligned to the equator, but it will still distort the shape of Earth. like @planetmaker says in this answer's comment;</p>
<blockquote>
<p>This makes sense to me as when the moon is whole, all the mass is in relatively one place so the force of gravity will be stronger at its center of gravity, and when it's turned into rings there is no center of gravity for the moon anymore as its constituent particles are now spread around the Earth. But they must still exert some gravity on Earth</p>
</blockquote>
<p>To say so, even a baseball can cause distortion or tides, but it is always depends on how greatly is affects.</p>

CC BY-SA 4.0
2022-03-12T11:20:28.683Z
A homogenous ring around the equator will NOT cause tides. It only has a chance to induce slight tides, if it was inclined wrt to the rotational axis of Earth. Otherwise it's gravitational field is temporally constant and axial symmetric. A uniform sphere does not excert tides or gravitational pull in its interior, a ring does do that in the radial direction (but only towards its plane). Being homogenous, it can only change the shape as it deforms the time-invariant gravitational field, but not excert tides as it is temporally uniform.
planetmaker
CC BY-SA 4.0
2022-03-12T11:34:03.310Z
@planetmaker I didn't think of this. will edit it.
Kavin Ishwaran
CC BY-SA 4.0
2022-03-12T09:46:15.860Z
2022-03-12T11:36:43.703Z
2022-03-12T11:36:43.703Z
Kavin Ishwaran
<p>I was watching <a href="https://www.youtube.com/watch?v=lheapd7bgLA" rel="nofollow noreferrer">this Kurzgesagt video</a> on what would happen if the moon crashed into the Earth, and it mentioned that at the very end the tidal forces would tear the moon apart and create planetary rings around Earth. What piqued my curiosity was that Earth would experience extreme tidal forces from the moon until it broke apart into rings.</p>
<p>This makes sense to me as when the moon is whole, all the mass is in relatively one place so the force of gravity will be stronger at its center of gravity, and when it's turned into rings there is no center of gravity for the moon anymore as its constituent particles are now spread around the Earth. But they must still exert some gravity on Earth, and I was wondering if it's possible for planetary rings to distort the shape of the planet if they are massive enough. I.e. say the rings were directly over the equator, the part of the Earth at the equator would be closer to the rings and therefore experience a greater force. Could this force distort the Earth to make it more oval shaped at the equator?</p>

CC BY-SA 4.0
2022-03-13T02:52:58.207Z
*cool video!* :-)
uhoh
<gravity><planetary-ring>
Can planetary rings distort the shape of a planet?
ROODAY
# Can planetary rings distort the shape of a planet?

(Asked by: username_0 on 2022-03-12)

I was watching [this Kurzgesagt video](https://www.youtube.com/watch?v=lheapd7bgLA) on what would happen if the moon crashed into the Earth, and it mentioned that at the very end the tidal forces would tear the moon apart and create planetary rings around Earth. What piqued my curiosity was that Earth would experience extreme tidal forces from the moon until it broke apart into rings.

This makes sense to me as when the moon is whole, all the mass is in relatively one place so the force of gravity will be stronger at its center of gravity, and when it's turned into rings there is no center of gravity for the moon anymore as its constituent particles are now spread around the Earth. But they must still exert some gravity on Earth, and I was wondering if it's possible for planetary rings to distort the shape of the planet if they are massive enough. I.e. say the rings were directly over the equator, the part of the Earth at the equator would be closer to the rings and therefore experience a greater force. Could this force distort the Earth to make it more oval shaped at the equator?

**Tags:** gravity, planetary-ring
> **Comments:**
> **username_1**: *cool video!* :-)  

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## 2 answers

### Answer by username_3 on 2022-03-12. Score: 5 | Accepted answer

**Question:** Could the gravitational force of Earthly rings distort the Earth to make it more oval shaped at the equator?

**Short Answer:** Yes

> This makes sense to me as when the moon is whole, all the mass is in relatively one place so the force of gravity will be stronger at its center of gravity, and when it's turned into rings there is no center of gravity for the moon anymore as its constituent particles are now spread around the Earth. But they must still exert some gravity on Earth

As you have said, The distortion will not be as much compared to the gravitational effects of the moon as whole, for example, the tides would still happen if the moon is replaced with rings, only thing is it will be reduced greatly, if the rings is not aligned to the equator, but it will still distort the shape of Earth. like @planetmaker says in this answer's comment;

> This makes sense to me as when the moon is whole, all the mass is in relatively one place so the force of gravity will be stronger at its center of gravity, and when it's turned into rings there is no center of gravity for the moon anymore as its constituent particles are now spread around the Earth. But they must still exert some gravity on Earth

To say so, even a baseball can cause distortion or tides, but it is always depends on how greatly is affects.

> **Comments:**
> **username_4**: A homogenous ring around the equator will NOT cause tides. It only has a chance to induce slight tides, if it was inclined wrt to the rotational axis of Earth. Otherwise it's gravitational field is temporally constant and axial symmetric. A uniform sphere does not excert tides or gravitational pull in its interior, a ring does do that in the radial direction (but only towards its plane). Being homogenous, it can only change the shape as it deforms the time-invariant gravitational field, but not excert tides as it is temporally uniform.  
> **username_3**: @planetmaker I didn't think of this. will edit it.  

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### Answer by username_2 on 2022-03-14. Score: 5

The gravitational field near the center of a thin narrow ring of mass, $M\_r$, and radius, $R\_r$, is $H\_z = -GM\_rz/{(z^2+R\_r^2)^{3/2}}$ for the z-directed field on the z-axis. Near the center, $(z,r) \approx (0,0)$, the field is $H\_z \approx -GM\_rz/R\_r^3 $ and the gradient is $\partial H\_z/\partial z = -GM\_r/R\_r^3$
The field locally must satisfy the Laplace equation $\partial H\_z/\partial z = -\partial (rH\_r)/r\partial r$. This leads to $H\_r = (1/r) \int r (\partial H\_z / \partial z) dr = +GM\_rr/2R\_r^3$ for the radial field near the center.

From above, the gravitational potential *near the center* of the ring is $\Phi(z,r) \approx GM\_r(2z^2 - r^2)/4R\_r^3$

For a small non-rotating spherical planet mass, $M\_p$, radius, $R\_p$, at the center of the ring, we are interested in how the equipotental surface is perturbed. The field gradient at the planet's surface is $-GM\_p/R\_p^2$. The perturbation, $\epsilon$, to the equipotential surface by the potential from the ring is disturbing-potential divided by field-gradient $\epsilon = [GM\_r(2z^2 - r^2)/4R\_r^3]/[-GM\_p/R\_p^2]$.

At the pole, $(R\_p,0)$, we have $\epsilon\_{pole} = -(1/2)(M\_r/M\_p)(R\_p^4/R\_r^3)$.

At the equator, $(0, R\_p)$, we have $\epsilon\_{equator} = +(1/4)(M\_r/M\_p)(R\_p^4/R\_r^3)$.

As expected the spherical planet becomes slightly oblate. This result requires a small planet and a big ring, $R\_p <, and the ring should be thin and narrow.


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