14

Leconte et al. (2015) suggested that the presence of an atmosphere could prevent or at least slow tidal locking. The star should exert two separate torques: one on the atmosphere and one on the solid body of the planet: $$T_a=-\frac{3}{2}K_ab_a(2\omega-2n),\quad T_g=-\frac{3}{2}K_gb_g(2\omega-2n)$$ where $$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\...


14

Pluto will never be a planet. There are a number of technical papers that give more precise meaning to the concept of "clearing the neighborhood". It's not just now, it's can the object in question clear the neighborhood of its path while the Sun is still a star. In the case of Pluto, Ceres, Eris, and a host of other not-quite-planet objects, that will not ...


7

The question says a few interesting things: The system orbit isn't on the ecliptic The system hasn't cleared its neighbourhood These are not going to change in the next few million years - or ever. Orcus is an interesting counter-example. It is in a similar orbit to Pluto - similar aphelion, perihelion and eccentricity, similar orbital period (to within ...


6

Tidal locking occurs because the planet deforms the satellite into an oval, with long axis pointing towards the planet. If the satellite is rotating the long axis will move away from being pointing towards the planet, and the gravity of the planet will tend to pull it back, slowing the rotation until one face is permanently facing the planet. Tidal locking ...


6

Hill sphere is named after John William Hill (1812–1879) and its simple logic follows from the presence of three bodies (let's assume Sun is the largest mass with Earth as the secondary mass and a satellite of negligible mass orbiting the Earth as the third mass), where the radius of the Hill sphere will be the largest radius at which a satellite could orbit ...


6

Yes: It has a companion planet or an excessively large moon, with the two bodies orbiting their common center of mass (much like the Earth and the Moon). They could be tidally-locked to each other, but they cannot be tidally-locked to their star.


5

The more likely case is actually a spin-orbit resonance that is not 1:1 but a half odd multiple, like the 3:2 case of our own Mercury. Having eccentricity in the orbit encourages this situation. I’ve been meaning to write this up on the Worldbuilding.SE but I have not re-found enough references. But see this video.


5

There may be some three-body periodic solutions that orbit around a common point, but in general they get a little crazy-looking and may not always contain an immediately obvious center of mass from casual observation. But of course it will always exist. If you watch closely, you'll see that whenever one body reaches the intersection point, all three are on ...


5

The sun still rises in the East and sets in the West. So you quickly identify the cardinal directions. By observing the point around which the stars rotate each night you can find the altitude of the pole, which gives you your latitude. You can't find your absolute longitude, but with good time keeping you can find your longitude relative to your starting ...


5

Your understanding of the illumination is correct. k is the ratio of the illuminated length BC to the diameter AC. New Moon has an illumination of 0, Full Moon an illumination of 1, and the quarter phase an illumination of 0.5. The position angle is measured counterclockwise from north (celestial north, along a line of right ascension) to the bright limb C. ...


4

I'm not sure it qualifies as "simple", but, using http://idlastro.gsfc.nasa.gov/ftp/pro/astro/hadec2altaz.pro (and some additional calculations/simplifications): $ \begin{array}{|c|c|c|c|} \hline \text{Event} & \text{Time} & \phi & Z \\ \hline \text{Any} & \text{t} & \tan ^{-1}(\cos (\lambda ) \sin (\delta )-\cos ...


4

If this is a possible periodic solution for a three-body problem? What you've posted here is the classic "figure 8" orbit for a 3 body system. Originally the 3-body system was an unsolvable mathematics problem, until people started using computers to do the math for us. Recently, a set of 13 "stable" 3-body orbits have been found which were published in ...


4

Which is the least massive object? Quoting Wikipedia, In the hierarchical, restricted three-body problem, it is assumed that the satellite has negligible mass compared with the other two bodies (the "primary" and the "perturber"), . . . This is the case studied in Kozai (1962), specifically, the case of asteroids being perturbed by Jupiter. While not ...


4

There's no reason why 2 bodies of equal mass couldn't have elliptical orbits around each other. There's an example of that here . The simple way to think about this is, if two bodies of similar mass approach each other, one of two things can happen, they either have sufficient velocity to pass each other with some hyperbolic curving of both objects ...


4

The Hill sphere is defined slightly differently to the Roche lobe, but the radius is approximated by the distance to the Lagrange points L1 and L2. For circular motion with angular velocity $\omega$ around the origin, we have: $$\ddot{\mathbf{r}} = - \omega^2 \mathbf{r}$$ The acceleration due to gravity from a point mass on another mass at position $\...


4

Wikipedia gives the formula $$t_{\text{lock}} \approx \frac{\omega a^6 I Q}{3 G m_p^2 k_2 R^5}$$ where $\omega$ is the initial spin rate expressed in radians per second, $a$ is the semi-major axis of the motion of the satellite around the central body (given by the average of the periapsis and apoapsis distances), $I\approx 0.4 m_s R^2$ is the moment of ...


4

Certainly, the figure only uses different masses because that's the more general possibility (after all, the two masses won't be exactly the same). In fact, if you look at the distribution over mass ratio q in binary systems in Figure 1b in https://arxiv.org/pdf/1304.3123.pdf, it seems that the cumulative number of binaries with mass ratio less than q is ...


4

Why would precession affect the motion of the other planets? First things first: That's an unreferenced portion of a wikipedia article. That said, a perfectly spherical body acts exactly like a point mass in Newtonian mechanics. A non-spherical body does not. The Earth's equatorial bulge has a significant effect on satellites in low Earth orbit. Sun ...


4

The solar perturbations on most of the satellites of Uranus are on a very small scale indeed, which may explain the absence of the instabilities noted in the question. A perturbational effect depends on the scale of the solar perturbing accelerations relative to the ordinary inverse-square attraction of the primary body. The scale factor (often designated ...


4

That depends on the accuracy you want to work with. To zeroth order, as outlined in Murray & Dermott, "Solar System dynamics", Chapt 3., you can do the following: The zero-velocity contours that are plotted in your image will not be coinciding with particle orbits to infinite precision, but they're a good zeroth order approximation for objects with low ...


4

It's a good question! I went to the JPL Horizons web page and looked up 3753 Cruithne and saw that the current orbital solution is based on observations from 1973 to 2018, about 45 years. That's a fraction of an orbit, but very careful measurements of it's position on the celestial sphere (RA, Dec) over time will allow astronomers to calculate its position ...


4

The short answer is that they don't. If an even larger SMBH passed through the centre of the Milky Way and "swept up" the existing SMBH, all the stars and planets that are close to it (within a few tens of light-years, maybe more) would experience huge disruption and might end up orbiting the new black hole and being carried away, or flying free, or whatever....


3

The easiest approach is to do a numerical integration of the equation of motion of your system. For that you need to remember that the gravitational force of $a$ on $b$ is $$\vec{F}_{a\rightarrow b} = -G\frac{M_a M_b}{r_{a\rightarrow b}^{3}} \vec{{r_{a\rightarrow b}}}$$ where $\vec{{r_{a\rightarrow b}}} = \vec{x_b} - \vec{x_a}$ and that $$ M\vec{a} = \sum \...


3

2010 TK7 has a diameter of about 300 meters (1,000 ft). Its movements do not bring it any closer to Earth than 20 million kilometers (12.4 million miles), which is more than 50 times the distance to the Moon. (from the wiki entry for it) That should make it pretty clear? Can you see a fly from 200 km away?


3

The most simple Lidov-Kozai model is that of a massless object ($m_1$ in your diagram) rotating a massive object ($m_0$), which is itself in an orbit with another massive object ($m_2$). This is a hierarchical 3-body system ($m_2$ is assumed to always be far enough from $m_0$ and $m_1$). It is easier to look at as two 2-body orbits: Inner orbit - $m_0$ and ...


3

Since I'm a hobbyist, I usually wait to see if someone a bit smarter wants to answer first, but I can give a couple thoughts on this. Hot Jupiters are thought to have migrated inwards, implying that another giant planet has been ejected in order to conserve the orbital momentum of those planetary systems. In the article you posted (I'll pull the ...


3

The calculations are not trivial, but they are encapsulated in software libraries such as pyephem, which has examples of finding rise, set and transit If you want to understand how these are calculated, you can readnthe source, which is based on the xephem application. One detail which complicates the calculation is the refraction caused by the atmosphere, ...


3

To expand a little on James's answer: The pole angle method doesn't care what time of year it is, the celestial pole isn't going anywhere. ;) However, if you're in the tropics, the altitude of the celestial pole is rather low, which can make accurate observation difficult. Of course, if you're doing this on another planet, (or even in Earth's southern ...


3

The L1, L2, L3 points are not stable. (full stop) Small deviations grow exponentially, even in the perfectly circular restricted three-body problem. In reality, we have a non-circular many-body problem, when (1) these points are not strictly well-defined and (2) deviations from the simple case have to be accounted for to obtain the exact trajectories (and to ...


3

The Year 0 does not exist in BCE/CE and BC/AD notation. It jumps from 1 BCE to 1 CE directly. So if we look at the sequence of years in BCE/CE notation, we have: 3 BCE 2 BCE 1 BCE 1 CE 2 CE 3 CE If you want to calculate the number of years between, say, January 1, 2 BCE and January 1, 2 CE, you can add the year numbers to get 4, but then you need to ...


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