# Tag Info

5

Consider a particle with mass m, orbiting in a circle a body with mass M. The gravitational force must be the centripetal force causing circular motion, so $$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ Cancelling the $m$ and $r$ and square rooting gives: $$\frac{\sqrt{GM}}{\sqrt{r}} = v$$ So you see the velocity is inversely proportional to the square root of ...

1

The page you link to and the paper refer to PCM (the Point Circle Method), and the PCM treats the orbits of two objects as circular, concentric, and coplanar. If planets in a system had odd (and probably unstable) orbits, this might not be the case. The PCM method treats the planet's position at any given time as a uniform probabilistic distribution. That ...

2

[This is complementary answer to Justin Tackett answer. I felt somewhat compelled in light of the comments (even after edits), and to illustrate my example (referenced in the answer) with a diagram.] I say there are 2 naïve situations where the innermost planet is not the closet. In the attached illustration there are only 2 planets and I consider the ...

0

I think the best way to answer your question is to "ask Skyfield" to check it for you! See .separation_from() added back in 2016. Sticking with Skyfield's methods ensures that issues like timeframes and space frames are handled correctly. The developer @BrandonRhodes (who also maintains PyEphem) has work painstakingly for years to ensure the ...

2

Lets take a more general look at the situation in systems with one star: What is the average distance to a planet? Break this down into two components, the distance along a line from the planet you are measuring from (which I will call O), through the barycenter of the planetary system which I will call X, and the distance the planet is offset from this ...

7

As has been mentioned earlier, the fully general version of this question requires rather tricky analysis to deal with orbits that aren't circular and coplanar. It's even trickier if you want to include the gravitational interactions between the planets, although in most cases those small perturbations won't be large enough to affect which planet has the ...

2

In a binary star system, there are two types of orbits which planets can have. A planet in an S-type or non circumbinary orbit will orbit one of the two stars. Thus the distance between the two stars must be at least several times as great as the semi-major axis of the planet's orbit, if the planet is to have a stable orbit. A planet in an p-type or ...

5

Not necessarily The authors of the article linked above assume that the planets involved are in roughly circular, concentric, and coplanar orbits Suppose Mercury and Earth had circular co-planar orbits with Mercury at 0.4 AU and Earth at 1 AU from the Sun. With a quick program and Monte Carlo analysis, we get the same result as the above referenced ...

1

A modification to this question that might clarify it: Is the average distance between a planet in a standard orbit (one planet per circular-ish orbit) and its primary, always less than the average distance between the planet and any other planet orbiting the primary? The nearer a planet is to the primary, the nearer its average distance to any other planet ...

13

For a planet with another planet at their Lagrange L4 or L5 point, those would be each others' closest neighbours; as you can see from the simulations, Mercury has an average distance a few % more than the average distance between the Sun and the planet (e.g. 1.04 AU for Earth), and the Lagrange points form equilateral triangles with the largest planet and ...

6

Not for double planets, for which each body is always the nearest planet to the other. Pluto and Charon disputably fit the bill. This 2006 International Astronomical Union report states Q: Is Pluto a planet? A: Yes. In fact, Pluto’s large companion named Charon is also large enough and massive enough to satisfy the definition of “planet”. Because Pluto and ...

7

The reason Mercury is the closest planet to all of the planets on average is in part because of the fact that, because of Kepler's laws, period scales with the semimajor axis of an orbit, and so big orbits are going to be slow and spend lots of time far away from any given object because the orbits are so slow. At opposition, for any two objects, they will ...

Top 50 recent answers are included