I was thinking of this thread again (the tug-of-war definition) and thought of another interesting thing to calculate: What if I swap Pluto and Charon in the equation? Would it indicate that Pluto is also more influenced by Charon than the sun at times?

The answer is yes, meaning Pluto does not have a concave orbit either. The tug-of-war value focusing on Pluto as the primary body is 337.3 at perihelion, but focusing on Charon as the primary body it is 39.3. Since both are greater than 1, this means Pluto and Charon cause each other to have an occasionally convex orbit.

So perhaps this definition would still find Pluto/Charon to be a binary planet, but in a different way than the Earth/Moon are. Compare the numbers above with Earth/Moon: Earth-centric has a value of 0.45, Moon-centric has a value of 0.006. Both being less than 1, Earth and Moon always have concave orbits around the sun.

Add proper salting and an adequate number of iterations over the hashing function, and the attacker's computer will be necessarily slowed as well.

That definition would reduce the problem to the relationship of their masses.

There is no such thing as "Windows 9 developer preview". There's speculation that it will be called "Windows 9". There's speculation that it will have a developer preview. There's speculation about the system requirements. But that's all it is, speculation. No announcements and to my knowledge no leaks about any of those specific points have been made.

Until we see leaks or at least strong rumors by sources that have been correct in the past, nothing is "known". (Remember how many rumors we've had about Apple's iWatch and for how many years? Some even by credible sources? Yet we still don't have an iWatch.)

They could let the host venue determine whether to blackout. Some might, others wouldn't, and eventually the better outcome would become the most popular.

In retrospect I don't like how I phrased this:

Change it to the following:

Another reason I don't care for the barycentric approach is because it depends so highly on the radius of the larger body. What if the barycenter of the moon were right above the surface but well within the atmosphere? What about a gas giant where the definition of the radius is a bit fuzzier?

Agreed, that one is a bit far fetched. It's a many-body problem and all it takes is a bit of eccentricity or pull from other moons, planets, and the sun to destabilize.

But back to the first question, what if the barycenter moves in and out of the planet due to multiple moons? This would be akin to the solar system, where the barycenter moves in and out of the sun. I don't know if we could easily call it a ternary planet, quaternary planet, etc.

I think I prefer Isaac Asimov's tug-of-war definition of a binary planet. It would be considered a binary planet if the smaller body has a concave orbit around the sun; in other words, the two are both primarily orbiting the sun and just happen to be close to each other. This would, however, define the earth/moon system as a binary planet and Pluto/Charon would be a planet/moon.

Windows 8 has been optimized to run faster in most benchmarks than Windows 7. I am sure this trend will reverse itself eventually but for now I'll take your bet.

I'm just a guy who was interested enough to Google and throw together some calculations.

The gravitational forces between the earth and moon are major components of tides. However the barycenter doesn't seem to contribute directly. Moving the earth and moon farther apart (and thus moving the barycenter further away from the center of the earth) actually causes the tides to become weaker. This actually happens regularly as the moon gets closer and then farther from the earth in its orbit (the moon's orbit is not perfectly circular, but slightly elliptical). When the moon is at its closest, the tides are barely higher, and at its furthest the tides are barely lower.

Gravity accelerates two objects toward one another, based on their mass and their distance. It works the same whether the two objects are initially moving toward each other or away from each other (away from each other, we usually call "decelerating", but there's no difference in the math).

Escape technically occurs when the two objects are moving away from each other, but the deceleration due to gravity will never be enough to overcome their initial velocity at their initial distance from each other. Gravity diminishes as the objects move farther apart, which results in less deceleration over time. In the case of escape, the velocity will never reach zero.

The answer is "yes"... assuming the moon magically appeared at that new farther distance but traveling at the same velocity as it is currently. According to the Wikipedia link on escape velocity:

Also, the orbital velocity of an object decreases as its distance increases. So increasing the distance of the moon would decrease how fast it would need to be going to stay in orbit.

But remember that the orbital distance suddenly increased but the orbital velocity did not change...

Let's say the moon is orbiting at distance R with orbital velocity V. Thus, all we need to do is figure out at what new distance R2 the new orbital velocity V2 = V * (1 / square root of 2).

This page contains the formula we need. Solving for r, r is proportional to 1/(v^2). So R2 is proportional to 1 / (V2^2), and substituting the equation above we find that R2 is proportional to 2 / (V^2), which equals 2 * (1/V^2), which equals 2 * R. R2 = 2*R.

Thus the answer is "2 times the original orbital distance, 478,000 miles".

That's a bit unfair to compare Mac computers with Windows phones. Comparing computers, Microsoft has pretty much an unbeatable record when it comes to supporting old versions.

But iPhones are updated for somewhat longer than Windows Phones (my 3GS was supported from iOS 3 through iOS 6, so 4 years worth).

My 2006 Macbook hasn't been eligible for an OS upgrade since Lion (3 versions ago). To upgrade to the latest OS means buying a new Macbook (Air) at or around $1000 with tax.

Meanwhile I can still purchase the full latest version of Windows for that computer for $120-200 depending on edition. I'd bet that any Intel version of Windows Microsoft releases in the next 10, possibly 15 years will still run on it.

BTW, I actually did the full calculations and accounted for the radius of earth/moon in the distance. But according to the equation for calculating barycentric coordinates, the distance of the barycenter from the center of the primary is linearly proportional to the distance of the centers of mass of the two bodies... so a pretty close estimate would have been (1 / 0.75) * 239,000 miles.

320,000 miles

Defining it based on barycenter will lead to curious outcomes. What if the barycenter moves into and out from the planet (such as with multiple moons)?

And what if Pluto had a second moon, equal in mass and distance as Charon but always on the exact opposite side (L3)? The barycenter would be at the center of Pluto, but why does this change cause Pluto to become a "real" planet?