What would be the structural impact of an oscillating orbit on a planet?

Question

Notice : If you wish to know how to make an oscillating orbit with real-world physics, you can looky-look at this question which is following mine. My question focuses on the consequences of such orbit.

A wavy stellar system

I have a stellar system with quite unique properties. How this system has come to be is unknown, but its central body -which for sweetness we'll call a white-hole- has unique physical properties. It gives out a force similar to gravity but acting as the reverse of it with a "higher derivative strength" : By this mathematical word abomination, I mean it decreases faster over distance than gravity does. In other words, if you're close, you get pushed away, and if you're far, you get pulled in. As a consequence, there is a sphere/circle where you are in a state of weightlessness, where the gravity's force counteract the repulsing one. Later on, we'll call this line the neutral-line.

Around this white-hole stands a telluric planet, quite similar to Mars in terms of composition and orbital characteristics, excepted for one thing. Due to some yet-to-understand space history, its orbit is crossing 8 times the neutral-line, as it is oscillating from and to the pull-zone with contrary forces. It forms a pretty star-like shape, as you can see in the toy model I made¹ below :

The red planet's orbit forming a star-like shape around the white star-like body.

My issue

This model, however, doesn't help me solve the fact that planets have got lots of added complexity, notably regarding their structure. They're not "solid as rock" like pétanque balls (or bowling ones if you're more accoustumed to them).

So what can I expect the structure of such planet to be, relatively to Mars? Here, I ask :

• Its general shape? An oval, a pancake (yummy!), exactly like Mars, something else?
• Would there be visible structure changes over-time at its surface when seen from another planet? Like signs of very strong pressures which leads to cracks, volcanic and seismic activity?
• And as part of a reality-check, would it be structurally obliterated into dust by such oscillations?

Known data from the model

Here are some data I gathered from my model that I guess could be useful in understanding the thing. Although... Recall that it's a toy model with much, much smaller-scale data (we're talking in km, not AUs)! Therefore, I can only reasonably give relative differences, and there might be large scale differences I'm not aware of!

• The speed of the object changes overtime. In my model, it ranges from 1 unit to 2 units of speed, quite a lot if I dare say. As you can see on the image, the speed increases as the object gets nearer the white-hole, and slower as it gets further.
• The force varies over time, too. For 1 N of overall pull force I get at the furthest distance, I get 1.6 N of push back force at the closest one.
• As for the distance, the planet's distance to the white-hole ranges from ~80 to ~140% of the neutral line distance. We'll consider the neutral line distance white-hole to be the same of what Mars would have, so your orbit ranges from 80% to 140% of Mars's orbit.
• For time, well... My model's going over the board and making a loop takes less than 15 minutes, so :D... We'll take in that to make a full period, it's the same as Mars. The thing here is that because it's oscillating along the neutral line, I doubt the third Kepler's law to calculate orbital periods applies here. In fact, by adjusting the perpendicular initial velocity, you get different loop times, as you can see below :

A side-by-side race between two planets with different initial velocity. The yellow planet has half the initial perpendicular speed of the red one, and takes a lot more time to move around

Other data

Outside the model's results, know that the white-hole, apart from its physic changing properties, act like the Sun gravity-wise and energy emission-wise. If by chance you're missing something, consider it to be Mars or Sun-based approprilate... Appropriately.

Therefore, what would be the structural impact a planet would face by having an oscillating orbit?

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¹ : Here's the model formula, whose result is positive when the object is attracted towards the white hole, and repulsed when negative. I mainly used as a sketch-up, but if you need it :) :

$$ F = -A \frac{m_h m_o}{d_{ho}^{2.5}} + G \frac{m_h m_o}{d_{ho}^2} $$

With F being the force applied, m_h the hole's mass, and m_o the mass of the repulsattracted object, d_ho the distance between the hole and the object, G the gravitational constant and A another "convenient" constant to balance things out. I don't really care what's inside the white-hole and didn't want to make general relativity calculations, so d_ho=0's undetermined result is irrelevant. Same for the white-hole's own movement, which I waived away since it's not really significant.

Answer 1

It's not the sun you should fear, it's everything else around it

In isolation, your planet would be virtually identical to Mars, assuming it was allowed to form and cool in the first place. In fact, since you're near the bottom of the gravitational potential well, the tidal forces would likely be much weaker than they are on Mars, so it could very well be slightly rounder and less "mixed".

You may be forgetting that circular orbits are still possible (and energetically favourable) - you can set centripetal force $$F = mR \left(\frac{2\pi}{T}\right)^2 = G m \left[\frac{M}{R^2} - \frac{A}{R^{2.5}}\right]$$

where $m$ is the planet mass, $M$ is the white hole's regular mass, $A$ is the white hole's "antimass", $T$ is orbital period and $R$ is the equilibrium radius.

The main physical difference between this system and reality is that you have upset the standard relationship between angular momentum and orbital radius - there is now a unique, "magic" radius at which gravitational potential energy is minimised. Based on thermodynamics, I'd expect all of the system's garbage to end up there eventually, all in different oscillating orbits. The other remarkable thing is that any orbital period is possible at roughly the same radius - we actually don't have enough information to say what T is based on the number of wiggles. In fact, the other terrifying possibility is a stationary, stable orbit where the planet simply sits at $R_0 = (A/M)^2$, where the gravitational force is zero.

This is a very different situation to our Sun, where every orbit is basically the same (just warmer/colder). Imagine your system was filled with a primordial soup of gas and rock - eventually, friction would compress this soup into a thin ring around the sun centered slightly beyond the neutral line. Within this ring, large planets would quickly form, potentially on some crazy wiggly trajectory. In such a scenario, collisions would be commonplace - your Mars would be riddled with craters, probably molten and possibly inside a gas giant. It depends how far you want the reality check to go, but the formation of such a system would be very odd.

Answer 2

The planet is just a planet.

It does not share the strange gravity of the star.

The star's forces on the planet are rather weak, the same as gravity. So they affect the planet's path as a whole, but have almost negligible effect on the planet itself. The full extent of local effects will be.... tides. Higher tides when the planet is closer to the star, lesser when it is further away. Both gravity and repulsion make the tides, all that matters is the balance, direction and gradient of the resultant forces. I do believe gravity tide will point towards the star, as our tides do, but repulsion tides will be 90 degrees radial to these. The result may be some interesting sloshing.

Just calculate the period of oscillation between closest and furthest for the planet. These are its seasons. Summer when the planet is closer to the star, winter when it is further.
The cycle around the star is interesting to astronomers, and not much more. It will not affect weather or climate or day-to-day life on the planet at all, just the scenic view of the stellar neighborhood.

Answer 3

Summary:

If the planet has a "normal" day length and is strong enough to keep from flying apart under its own rotation, then it'll be a boring old oblate sphere.

First-order effects on planetary shape: none

Assuming that the modified gravity law applies the same way to all constituent atoms of the planet, then your new force law obeys the equivalence principle. In other words, since the planet is "freely falling" under this gravitational force, the physics in its neighborhood are entirely equivalent to those if the planet was isolated in deep space with no gravity at all. And if the planet was in deep space, then it would accrete into a roughly spherical shape under its own gravity, possibly with a small amount of oblateness due to its rotation.

Second-order effects on planetary shape: varying tidal forces

Of course, there are small effects that the Sun has on the Earth's shape. The side of the Earth towards the Sun will feel a bit more gravitational force than the side of the Earth away from the Sun, since it is closer to the Sun and gravitational force decreases with distance. The difference between these force leads to the tidal force, which is roughly $$ F_\text{tidal} \approx R \frac{d F_\text{grav}}{d (d_{ho})} $$ where $R$ is the physical size of the planet. In your case, this will work out to be roughly $$ F_\text{tidal} \approx R \left[ 2.5 A \frac{m_h m_o}{d_{ho}^{3.5}} - 2 G \frac{m_h m_o}{d_{ho}^3} \right] $$

In principle, you could calculate how large these forces are. However, I would expect them to be relatively ineffectual so long as your planet is rotating. So long as your planet is rotating, then the stresses on the planet's crust due to its rotation will almost certainly be larger than the stresses due to the tidal forces. For a point of comparison, the tidal forces on Mars due to the Sun result in a relative acceleration of about $10^{-14} \text{ m/s}^2$; but the centripetal acceleration due to its rotation is about $10^{-2} \text{ m/s}^2$ along the equator. The calculations for your planet will, I expect, lead to similar results.

So long as the structure of your planet is strong enough to keep from flying apart due to its own rotation, and the planet stays far enough away from the hole, it should be able to handle the tidal effects. Even if they're time-varying due to the planet's varying distance from the hole, they are so incredibly small that they shouldn't threaten its structural integrity.