# Wandering Autumn

Exploring change and the life that comes with it

# Around and Around

January 12, 2017

Something that has come up in my writing life is figuring out a culture’s calendar. To determine that, I need to know information about the celestial bodies nearby to the culture’s planet.

Arguably, my post about apparent magnitude might have been getting just a bit ahead of myself; I have a bad tendency to do that.

But I have run into a bit more fundamental of an issue: how long does it take for a smaller object to orbit a larger object? That is, a moon around a planet; or a planet around a star?

This is known as the orbital period.

Before Newton and Leibniz invented calculus, the astronomer Johannes Kepler figured it out. At least, kind of. Hopefully, we should be able to do this with just some deep thinking and some algebra and trigonometry.

For simplicity, we’re going to start with a small object (such as a planet) orbiting around a large object (such as a star) in a perfectly circular orbit. This will allow us to think through the problem.

So, let’s start with how these orbits work, which is gravity. You have a large mass, $m_1$ that has a smaller mass, $m_2$ orbiting around it. We know, empirically, that the force of attraction due to gravity between two masses at distance $d$ is:

Where $G$ is the gravitational constant, which until we actually need to calculate a number we can keep an algebraic symbol. Of note, $d$ is the distance between their centers of mass, not the surfaces of the masses.

So at any given point in the circular orbit, we know the force of gravity between our two masses.

However, the smaller mass must be moving so that it doesn’t fall into the larger mass, and its velocity $v$ must be large enough that its orbit doesn’t decay and therefore collide; and small enough that the orbit doesn’t escape.

So we need to know the centripetal force between the smaller mass and the larger mass. This is kind of a mysterious force that I struggled with back when I took physics in college, but the best way I can think of it is: pretend you have a brick on a rope that you’re swinging around your head for no apparent reason. The brick obviously wants to keep moving forward, but it’s pulled into a circle by the rope; that force is the centripetal force.

The formula for centripetal force is:

Where $m$ is the mass of the object being swung around, $v$ is its velocity, and $r$ is its distance from the center of the circle it’s traveling on (that is, the radius of the circle). This can be derived from the standard equation $F = ma$ and some geometry I leave as an exercise for the reader.

Because the centripetal force must be what’s keeping the second mass orbiting, and the gravitational force is what’s keeping the smaller mass orbiting, we can set them equal:

If we have a circle of radius $r$, we know its circumference is $2\pi r$. If an object is traveling a circling path at velocity $v$, then the amount of time it takes is:

That is, the time it takes to go somewhere is the distance over the velocity. Therefore, rearranging the terms a little:

Really, this should come as no surprise, because this is how distance, time, and velocity are all related. But we can substitute this back in our original equation (giving us the $T$ we want to get an equation for, as it turns out):

Which is the equation, as it turns out. Hooray!

When moving into ellipses or having the masses affect each other, you have to bring in some fancier math than I care to do, though the answer for an ellipse is the exact same, except that $d$ is the semi-major axis of the orbit’s ellipse. With masses that really affect each other, the denominator simply becomes their sum through wizardry I haven’t derived myself.

You’ll notice that in the equation I derived, the mass of the smaller object disappeared; as noted, the full equation is:

But when $m_2$ is orders of magnitude smaller than $m_1$, its effect practically disappears. For reference, the sun is 3 orders of magnitude more massive than Jupiter.

If we ignore the masses of the planets, an interesting thing happens: we can treat $m_1$ as a constant, which means we can also pull it and the gravitational constant out of the square root. In other words:

Or, as it’s commonly written:

Where $k$ is a constant that is always the same for a given star. This huge, and it means that we can think in terms of ratios. So if $T_1$ is the orbital period for the first planet, and $T_2$ is the orbital planet of the second planet, and $d_1$ and $d_2$ are the distance from the sun to the first and second planet respectively, we can do some algebra (an exercise left to the reader) and rearrange a bit:

And if the second planet is Earth, and you define its distance from the sun as $1$, and we know its $T_2$, if we can figure out $T_1$ for any given planet, we can figure out how far away from the sun it is as a multiple of Earth distance.

It’s figuring out how many days another planet takes to go around the sun that’s the tricky part, then. We have plenty of sky observations, but the problem is that Earth moves, so just because a planet returns to the same place it was doesn’t mean that it’s actually been exactly one year for that planet.

The difference between these is the synodic period and sidereal period. The synodic period is how long it takes the planet to get back to its original place relative to Earth; the sidereal is how long it takes to make one revolution of the sun.

If you’re thinking I’m bringing this up because there’s a way to relate the two, you’re right!

Let’s let $E$ be the sidereal period of the Earth in days; we know this, it’s about $365$ days. But we also know that each day, the Earth travels $360/E$ degrees per day. Let $S$ be the synodic period of another planet in days, because we can measure that. And finally, $P$ is the sidereal period of that planet in days, meaning it orbits $360/P$ degrees per day.

For simplicity, let’s say the other planet is superior, meaning it’s farther out from the sun, so we know that $P > E$.

When we measure the synodic period, we do it based on when the planet is in opposition, which means that Earth is smack dab in between that planet and the Sun. Or in other words, the planet is on the complete opposite side of the Earth from the sun. This is pretty easy to measure over the course of a lot of nights with observations at midnight each nights.

We know, since we’re talking about a superior planet, that the synodic period is going to be larger than one Earth year; so the difference is going to be $S - E$. This is how many days beyond one Earth year it takes for the planet to be in opposition again.

During that time, Earth has orbited an additional $\theta$ degrees. So we know:

But because the other planet has orbited the same angle (by virtue of it being in opposition), we also know that:

We can then do some algebra:

And for inferior planets—that is, planets closer to the sun—we just swap $P$ and $E$ because Earth would have the outer orbit and the other planet would have the inner orbit, and the math still works out:

Which means that with some astronomical observations—which we have—we can figure out how long any planet’s year is.

Like I said, this is huge.

This is why the astronomical unit is a big deal in astronomy. An astronomical unit (AU) is the distance from the Earth to the Sun. We don’t actually need to know how far that is in kilometers.1 We can use our knowledge of how long a planet’s year is to figure out how many AU away from the sun it is.

This is why planet distances from the sun are given in AU: we’re a lot more confident of that than we are of its distance in kilometers, because its distance in kilometers depends on our measurement of Earth’s distance from the sun in kilometers. On the other hand, its distance in AU simply depends on how long its year is relative to ours—and we’re pretty sure about the distance to our sun relative to the distance to our sun.

The lesson here is that unit choice matters, and affects your accuracy.

Also, it explains why sometimes people pick what seems like odd units, like AU. Or parsec—but I’ll get to that sometime in the future.

Anyways, this has been a bit of an exploration of orbital periods, which is a pretty cool thing, I think. Just another piece to the puzzle of building a fictional universe.

1. And didn’t for quite some time; we had to wait for a transit of Venus to even get close.

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