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### The Planet X Saga: Orbital Math

As I say on the page about orbits, at a year away Planet X must be roughly as far as Saturn (and therefore closer as you read this page). The distance from Earth in May 2002 should be about 900 million kilometers. How do I know this?

I used what we know about how gravity works and how objects behave under its influence. For example, the usual sort of elliptical orbit has certain characteristics. As Kepler showed centuries ago, the period of the orbit depends on the semimajor axis of the orbit (the major axis of ellipse is the length across it longways, and the semimajor axis is half that distance). Specifically, the cube of the semimajor axis equals the period squared:

a3 = p2
where a is the semimajor axis, and p is the period. In this case, a is measured in units of the Earth-Sun distance, called an Astronomical Unit, or AU for short. 1 AU is about 150 million kilometers. The period is measured in years.

Planet X is alleged to have a period of 3600 years. By plugging it into that equation, we can solve for a. In this case, a = 235 AU, or 35 billion kilometers. That's a long way! Pluto, on average, is about 35 or so AU away, or only 5 billion kilometers. Planet X, if it orbits on an ellipse, can get as far as 235 AU away, 7 times the distance of Pluto.

But wait! Most people claim it's on a very elliptical orbit. That means it can get close to Earth. The diameter of the orbit must be 2 x a = 470 AU. If it gets as close to the Sun as the Earth does, 1 AU, then it must get about 469 AU away from the Sun at its farthest point. It then takes half a period, or 1800 years, to come back into the solar system from that point.

An ellipse stretched this thin is practically a straight line. At aphelion, or farthest point from the Sun, it's velocity is so low that it might as well be zero. It then spends 1800 years falling toward the Sun, and whips around it tremendously fast, only to head out into deep space again.

Now I will digress very briefly into escape velocity. When you throw a rock in the air, it goes up some distance, slowing all the way up. It stops, then falls back down, accelerating all the way. If you throw it harder, it goes faster and gets higher, but it will always fall back down.

Unless, that is, you throw it really hard. Remember, the gravity of the Earth weakens as you get farther from it. If you throw that rock high enough, the gravity it feels from the Earth will get weaker. If you throw it hard enough, the weakening gravity will not be able to get the rock to fall back to the Earth. The velocity at which this happens is called the escape velocity.

For the Earth, this velocity is about 11 kilometers/second. A rock thrown this hard will go up, slowing all the time. But the gravity of the Earth is never quite enough to make it stop, so it always slows, but never falls back. In a sense (mathematically this is exact) the rock will stop at an infinite distance from the Earth.

The critical thing here is that the reverse is true as well. If you go really really far away from the Earth and drop a rock, it will impact at escape velocity! The equation for this is exactly reversible; it doesn't care whether rock is going up or down.

Escape velocity depends on your distance form an object too. It's much easier to throw a rock hard enough to go on forever if you are already a long way from the Earth. Earth's gravity is weaker far away, so the escape velocity is lower. In fact, escape velocity is mathematically represented as

escape velocity = square root(2 x G x mass/distance)
where G is Newton's constant (just a number you plug into the equation), mass is the mass of the object and distance is your distance from it. You can see that if the distance gets bigger, the escape velocity drops, just like in my example above.

So why is this important to Planet X? Because at 470 AU, Planet X would be essentially infinitely far away from the Sun. If it fell toward the Sun from this distance, it will round the Sun at just a hair under the Sun's escape velocity at that distance. This means that we can always know the velocity of Planet X everywhere in its orbit: its just a teeny bit less than the Sun's escape velocity at that distance. At its most distant, that's about 1 kilometer/second. At 1 AU (the Earth's distance), it's about 42 kilometers/second.

In general, if you know how fast something is moving, you can calculate its distance. For example, say you are in a car moving 100 kilometers per hour. How far will you move in an hour? Easy: 100 kilometers! So if a car is an hour away at that speed, you can say it is 100 km away.

A planet, though, is not moving at a constant speed. It's accelerating as it heads toward the Sun. If you assume constant speed to get the distance, you'll get the wrong answer. The only way to do this is with calculus. Bear with me here.

The method is described by Dr. Joseph Gallant, Assistant Professor of Physics at Kent State University, on this page. He solves the equation for time, given the mass of the Sun and the distance a planet must travel. He gets this: In this case, a is the maximum distance of Planet X, or 469 AU. R is the distance at which you want to know the time; in this case, when Planet X passes the Earth, or 1 AU. r is a dummy variable, which means it has no value to plug in, it's the variable you are integrating. M is mass and G is again Newton's constant.

What we want to do is assume a time of 1 year (that is, when Planet X is one year away, in May 2002) and see what R is. But first we have to integrate the equation.

It's a difficult equation to integrate. However, computers do this kind of stuff pretty well. I used a program to integrate this for me. Just to check, I used an R of 470 AU (Planet X's max distance from the Sun), and got a time of 1781.4 years, which is pretty close to the "correct" value of 1800 years (half the period, since it is only going half an orbit), only off by about 1%. So it looks good.

Then I put in an R of 1 AU. It got a time of 1781.2789 years. That's how long Planet X takes to fall from the tippy top of its orbit all the way to Earth. So now all I had to do was try different values of R until I got a time of 1780.2789 years, which would put it one year from the Earth (1781.2789 - 1780.2789 = 1 year). I did this, and the best value I got was about 5.9 AU, or just under 900 million kilometers away.

So there you have it! When I originally did this calculation, I made a bad assumption and got a distance of 550 million kilometers. I assumed it was starting from there, and so it took a year to fall to Earth. But it already has built up some velocity from it's long fall inward from 470 AU away, so at 550 million kilometers form Earth its velocity is not zero! However, note that when I correct for this, it only gets farther away by a factor of two. That's because even after falling all that way, it still isn't moving terribly fast when it passes 550 million kilometers away.

One more thing. We can use this equation to figure out how far away Planet X will be at a given time from closest approach. Here are my results:

 Time from closest approach Distance (AU) Distance (millions of km) 6 months 2.9 435 3 months 1.7 255 2 months 1.2 180 1 month 0.64 96 3 weeks 0.45 68 2 weeks 0.3 45 1 week 0.17 26 3 days 0.07 11 2 days 0.05 7.5 1 day 0.024 3.6

As you can see, even a six months from closest approach, it would be pretty far away. But that's misleading. Even at that distance of over 400 million kilometers, it would outshine every other object in the sky save the Sun, Moon and Venus. As I have said before, even now it should be tremendously bright due to its distance, yet we see nothing at all. My conclusion: there is nothing to see.

Back to the main Planet X page