# Can planetary rings distort the shape of a planet?

I was watching this Kurzgesagt video 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?

• cool video! :-)
– uhoh
Mar 13 at 2:52

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

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.

• 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. Mar 12 at 11:20
• @planetmaker I didn't think of this. will edit it. Mar 12 at 11:34

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.