Back-of-the-envelope calculation: Earth’s orbital velocity

What is the orbital speed of the Earth? We know from Kepler’s second law that the answer won’t be unique since it will change as it traces out the orbit of an ellipse, but one of the first rules of approximating is to lower your standards so let’s do so.

Let’s assume that Earth’s orbit is circular. Why can we do that? Well, I remember reading that Mars’ orbit has a particularly high eccentricity (how much it deviates from a true circle) which is why Tycho was able to get good pre-telescope measurements for Kepler to use, but even still it wasn’t that big of a difference, so Earth’s should be even less. Since we’re assuming a circle, we can assume a constant speed because of spherical symmetry (why would the universe prefer one leg of the journey over another?). Luckily the time it takes to make one revolution is well known, that being a year. Now the year itself is a fairly complicated topic, but never mind that: We’re approximating, so no point in getting bogged down in the details.

How long is a year in seconds? When approximating we only care about three numbers: 1, a few, and 10, where \displaystyle (few)^2 = 10. That is, more or less \displaystyle few = 3. A year is complicated, so let’s divide and conquer: a year is composed of 365 days (leap years are too complicated, so forget them!), a day is composed of 24 hours, and a hour has 3600 seconds in it \displaystyle (60 \times 60). So

\displaystyle 1 \mbox{ year} = 365 \mbox{ days} \times 24 \frac{\mbox{hours}}{\mbox{day}} \times 3600 \frac{\mbox{seconds}}{\mbox{hour}} \\ = (3 \times 10^2) \times (3 \times 10^1) \times (3 \times 10^3) \mbox{ seconds} \\ = 30 \times 10^6 \mbox{ seconds since } 3^3 = 30 \\ = 3 \times 10^7 \mbox{ seconds}

which is within 5% of the true value. So far so good. Now, we need to turn to the geometry of the circle itself.

The circumference of a circle is given by the well-known equations \displaystyle C = 2 \pi r where r is the radius. This radius is what is known as the astronomical unit (AU). For astronomical calculations this is a quantity worth memorizing, and luckily it’s a fairly nice and round number: 150 million kilometers or \displaystyle 1.5 \times 10^{11} \mbox{ m.} But memorizing stuff is a pain, so how might we have come to that number “in the field”? Well, perhaps you can recall hearing that it takes light from the Sun about 8 minutes to reach Earth. “About 8 minutes” is 500 seconds, and light travels at \displaystyle c = 3 \times 10^8 \mbox{ m/s} (now that one you do want to memorize!) so multiplying the two together gives \displaystyle 500 \mbox{ s} \times 3 \times 10^8 \mbox{ m/s} = 1500 \times 10^8 \mbox{ m} = 1.5 \times 10^{11} \mbox{ m} as stated. So we have

\displaystyle C = 2 \pi r = 2 \times 3 \times 1.5 \times 10^{11} \mbox{ m} = 10 \times 10^{11} \mbox{ m} = 1 \times 10^{12} \mbox{ m.}

We’re nearly done! Speed is distance divided by time, so we take the orbital circumference C and divided it by period which is 1 year, counted in seconds to give

\displaystyle v = \frac{C}{T} = \frac{1 \times 10^{12} \mbox{ m}}{3 \times 10^7 \mbox{ s}} = \frac{1}{3} \times 10^{12 - 7} \mbox{ m/s} = 3 \times 10^4 \mbox{ m/s} = \boxed{30 \mbox{ km/s}}

which is quite accurate!

This post was fairly verbose, but you can see that the heart of the issue was in approximating Earth’s orbit as a circle with constant orbital speed thus turning a complicated celestial mechanics problem into a simple distance over time problem. With a little practice you can do all the necessary calculations fairly rapidly in your head, multiplying up the significands and using the above rule for the value few while adding up the exponents.

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