How Meteor Speeds Measure the Earth’s Speed

[Note: the final section of this article, which includes some detailed predictions, is still a bit of a work in progress. The predictions shown there are crudely made, and will be improved at a later time.]

When trying to understand a natural phenomenon, I often try first to imagine a simple way of capturing its essence that involves a “toy model” — a cartoon of the phenomenon that’s obviously overly simplistic. Toy models are a central tool in physics and other sciences, and I teach their use to the students I’m training.

We’ll start by using a toy model here, before moving on to more sophisticated ways of comprehending what’s going on.

Meteors occur when little rocks cruising through space (meteoroids) strike Earth’s atmosphere (and if a piece of the rock reaches the ground, it is called a meteorite.) As a “toy model” cartoon of real meteors, let’s imagine that all meteoroids hitting the Earth, and the meteors created by them

• have exactly the same speed (relative to the Sun), and
• come from completely random directions (relative to the Sun).

This is most certainly not the case! But let’s see what we can learn from this oversimplified scenario.

(Note that I used a variant of this scenario, replacing the meteoroids striking the Earth by raindrops hitting a car, in my original blog post concerning meteors and the distance to the Sun.)

A Simple “Toy Model” of Meteors

Let’s call the meteors’ speeds, relative to the Sun, v_m, and the Earth’s speed around the Sun v_E. We can consider three situations.

• The meteors’ speed is much faster than Earth’s speed (v_m >> v_E )
• The meteors’ speed is much slower than Earth’s speed (v_m << v_E )
• The meteors’ speed is comparable to Earth’s speed (v_m ~ v_E )

These give us three qualitatively different possibilities for how these equal-speed meteors will look from the Earth’s perspective.

In the first case, the Earth is so slow that we can view it as (almost) stationary, and then the perspective from Earth is (almost) the same as from the Sun: the meteors come from all directions at (almost) the same speed, v_m . Observing the meteors tells us nothing obvious about v_E.

In the second case, the meteors are so slow that we can view them as (almost) stationary, as though they were a cloud of dust or bugs that a fast car was plowing through, or like raindrops as seen by a pilot of an airplane. In this case, (almost) all the meteors seem to come toward the Earth’s front side, and the speed at which they strike the Earth is (approximately) v_E.

In the third case, both meteor and Earth motions matter, and there is interesting directional dependence. From Earth’s perspective, those meteors that strike the Earth’s front side have speed v_m + v_E , while those that strike its back side have speed v_m – v_E. Those that come from other directions have speeds in between — faster to the extent they arrive at the front side. and slower to the extent they arrive at the back. In this case, we can extract v_E from how meteor shower speeds differ depending on the location in the sky from which they arrive.

As shown in Fig. 2, something like the third case is seen in the pattern of meteor showers hitting the Earth. Each red dot represents a shower. The location of the dot shows the meteors average speed on the vertical axis and the meteors’ direction of arrival on the horizontal axis. You can see that most showers, and all of the fastest showers, are arriving from the post-midnight side — the front side — of the Earth. The meteor showers that come from the back side (the pre-midnight side) are the slowest. So this suggests that our simplistic toy model might actually be revealing what’s going on — that the Earth’s speed relative to the Sun might be comparable to the meteors’ speeds relative to the Sun.

The data shown in Fig. 2, with precise speeds, was obtained by experts. But the trend it shows is so dramatic that you, too, would notice it, even if you just made rough estimates of meteors’ speeds during the best meteor showers of the year.

And yet, we know our toy model is completely wrong: meteors do not arrive from entirely random directions, nor do they have the same speed as seen from the Sun. Why might it nevertheless capture the basic phenomenon?

It turns out that although meteors’ speeds (as viewed from the Sun) are not all the same, they are capped by a certain maximum. This fact is implied by Newton’s understanding of gravity and motion, as learned in the first few months of a physics course. I’ll explain how that works in the rest of this article.

The logic I’m about to follow has some assumptions and loopholes, which I’ll point out as we go. Do not fret, however; by the end of this article, the assumptions will be verified and the loopholes closed.

The Relation Between Escape Velocity and Circular Motion

What we’ll do first is compare the speeds of two small objects orbiting the same large object through its gravity.

• One object is in a circular orbit of radius R at a certain speed v₀.
• The second, on an orbit that takes it very far away, is just now crossing the first object’s orbit
  • Fact: whether the second object is coming or going, it has the same speed as it crosses that orbit.
  • Fact: that speed is almost enough for the object to escape altogether, and thus is just below the central object’s escape speed v_esc .

Remarkably, it turns out the two speeds differ by a factor of the square root of 2!

• v_esc = √2 v₀ ≈ 1.4 v₀

This fact follows from Newton’s laws of gravity and motion and/or the conservation of energy.

Meanwhile, there’s a well-known formula for v₀ in terms of the gravitational pull of the central object. If the mass of the central object is M and the radius of the circular orbit is R, then

• v₀² = GM/R

where G is Newton’s gravitational constant. The red equation above then implies

• v_esc² = 2GM/R

You can take my word for these statements, if you like. Or if you want to know more,

click here to see a proof that these statements are true.

In circular motion, there is a relation (discussed here) between acceleration a, speed v₀ and radius R, namely:

• a/v₀ = v₀/R — or equivalently, v₀² = a R

The acceleration of the orbiting object of mass m due to the gravity of the central object of mass M is

• a = F / m = ( G M m /R² ) / m = G M / R²

which implies the formula quoted above for v₀ :

• v₀² = a R = GM/R

Now that we know this, we can observe something interesting. The motion-energy (i.e., kinetic energy) of the orbiting object is

• 1/2 m v₀² = 1/2 GM/R

but its potential energy [i.e., energy from gravity due to its location] is

• – G M / R

so its kinetic energy is -1/2 of its potential energy.

That means that if its kinetic energy were doubled, its kinetic energy would equal minus its potential energy, and its total energy — their sum — would be precisely zero. But zero energy is exactly what is needed to escape the gravity of the central object. [Suppose the smaller object is able to reach a very large distance R_max, where it nearly stops and then begins to fall back toward the central object. Well, its kinetic energy is very small (because it almost stopped) and its potential energy – G M / R_max is very small because R_max is very large… and thus its total energy is tiny… and the larger is R_max, the tinier it becomes. So zero total energy is exactly what is needed to escape.]

Thus, doubling the object’s kinetic energy allows it to escape. Since kinetic energy is proportional to speed-squared, escape requires doubling the square of the speed, and thus

v_esc² = 2 v₀²= 2GM/R

This proves the relation between escape speed at a particular radius and the speed of circular motion at that same radius.

Meteors are Not Orbiting the Earth

For a satellite in near-Earth orbit, like the space station, v₀ is 5 miles (8 km) per second. To escape the surface of the Earth (or to escape the near-Earth satellite region) then requires a speed that exceeds this by the square root of 2: about 7 miles (11 km) per second. Any object passing the Earth at a higher speed (relative to the Earth) can escape the Earth, and thus cannot be in an orbit around the Earth!

Since, by measuring meteors’ speeds, you can easily find that most move at tens of km per second (see Fig. 2), it is clear they move much faster than Earth’s escape speed, and so we can conclude that most meteors are not from objects in orbit around the Earth. They must be in orbit around the Sun, or coming from deep space far beyond the Sun, or some of both.

To learn the speed required to escape the Sun from the region of Earth’s orbit, which I’ll call v_esc,S , requires knowing the Earth-Sun distance R_ES ; but I want to assume this distance is unknown, which means v_esc,S is also unknown. Even then, we do still know, from the discussion above, that v_esc,S is √2 ≈ 1.4 times larger than Earth’s orbital speed v_E, whatever that might be.

• v_esc,S = √2 v_E ≈ 1.4 v_E

How to Find Earth’s Speed from the Fastest Meteors

Now, let’s make an assumption, which we’ll check before we’re done. Let’s assume that meteoroids that create meteor showers are all bound to the Sun. That is, they orbit the Sun, rather than coming in from deep space.

All bound objects orbit in ellipses, as Kepler and Newton found. The further out the elliptical orbit goes, the faster is the meteor’s speed when it reaches Earth. (In short, the further out it goes before it stops moving away from the Sun, the harder it has to fall back toward the Sun to reach our neighborhood.)

Under this assumption, every meteor passing the Earth has a speed that is less than — possibly just a tiny bit less than — the speed necessary to escape the Sun from the Earth’s orbit,v_esc,S. (Recall that, as in Fig. 3, an incoming object has the same speed as it crosses the Earth’s orbit as it will have when outgoing.) The fastest meteors go the furthest from the Sun before they return to Earth; the further they go, the closer their speed as is to v_esc,S as they pass Earth.

Furthermore, all the meteors in a shower must be on the same orbit, and therefore must have almost identical direction and speed. If they weren’t, then the stream of meteoroids that creates the shower would quickly disperse across space, and there would be no reason for a host of them to arrive at the Earth at around the same time, causing a distinct increase in the number of meteors compared to usual nights. The directional alignment is confirmed by the existence of the radiant for each shower, and the common speed can already be estimated by naked eye viewing. It is only because each shower (not each meteor in a shower) has a definite direction and speed that we can even make the plot in Fig. 2.

For any given meteor, we can measure the direction it is coming from and its speed relative to Earth. But we know (under our assumption that meteoroids are bound to the Sun) that its speed relative to the Sun is less than v_esc,S = √2 v_E . From this, we can already make a very simple statement. If the meteor comes from a point directly ahead of the Earth in its orbit and travels as fast as it is allowed to, at or just below the Sun’s escape velocity v_esc,S, then the relative speed of the meteor and Earth — the speed with which, from our atmosphere’s point of view, the meteor enters the atmosphere and tracks across it — will be as fast as it can possibly be. That fastest meteor speed v_fastest will be given by adding the meteor’s speed relative to the Sun to the Earth’s speed relative to the Sun, giving

• v_fastest = v_esc,S + v_E = (√2 + 1) v_E ≈ 2.4 v_E

So measuring the Earth’s speed around the Sun is very easy. We just measure all the meteors in all the meteor showers that impact the Earth nearly head-on in its orbit, and find the shower whose meteors have the fastest speed. Then we will know

• v_E = v_fastest / (√2 + 1) = 0.4 v_fastest

One night does not suffice to measure the fastest meteors, unfortunately. We need to find the shower with the fastest meteors, and so we have to spend one year measuring the speeds of meteors in each shower that the Earth encounters. Once we find the shower with the fastest meteors relative to the Earth, we just multiply its meteors’ speed by 0.4 to get the Earth’s speed around the Sun. (Caution: remember we are assuming the meteoroids are all bound to the Sun, so we have to confirm that later; and also, we have neglected effects of Earth’s gravity, but we will see later that they are small.)

It’s quite easy for non-experts, using the naked-eye methods described on this webpage, to estimate that meteors’ speeds are always in the tens of km (miles) per second, not in the hundreds or thousands of km per second, and not just a few km per second. So we learn easily that the Earth, too, moves at tens of km per second. This gets us a rough but correct estimate of the distance to the Sun (via a method explained at the end of this blog post.)

But if we work a little harder, we can do a lot better. The more accurately and precisely we measure the fastest meteors’ speeds, the more precisely and accurately we can measure the Earth’s speed, and the circumference and radius of Earth’s orbit. As you can see in Fig. 2, experts have found that the fastest meteors move at about 72 km per second, and so

• v_E = v_fastest / (√2 + 1) = 0.4 x 72 km/sec = 30 km/sec = 19 miles/sec

With this information, you can correctly extract the distance to the Sun to within a few percent, as shown in the same blog post. And even without expert knowledge and equipment, you yourself could probably get the answer to within 20% or better, using a relatively small number of observations across a year during the most prominent meteor showers.

Checking Predictions Carefully, and Closing Loopholes

Attentive readers may have noted some loopholes here.

First and foremost is our assumption that meteoroids are all bound to the Sun. If they’re not, then we might be in case 1 of our toy model instead of case 3 (see Fig. 1), with the Earth moving very slowly and with meteoroids moving at much higher speeds. However, this would require strange correlations. The meteoroids, in this case, would all have to be coming from outer space far, far beyond the Sun and its planets. Why, in this case, would the pattern in Fig. 2 occur? The relation between the meteor’s speeds (relative to the Sun and Earth) and their incoming direction in the sky would have to be intrinsic to the meteor showers themselves, rather than a consequence mostly of Earth’s motion. But as the Earth orbits the Sun, the direction of overhead-at-midnight is constantly changing. How would the meteor showers know about it? Such a correlation between meteor directions and speeds and the Earth’s direction of motion would imply a relation between something happening outside the solar system (which determines where the meteors are coming from) and something happening inside the solar system (Earth’s orbital orientation and location.) Things inside the solar system may easily be correlated with each other, because they all formed together; that’s why all planets, dwarf planets and asteroids orbit around the Sun in the same direction. But meteoroids coming from outside the solar system would have to have formed in some completely different way that has nothing to do with how the Sun spins. Relative to the Sun, one would expect there to be just as many meteors hitting the Earth at 50 to 70 km per second from the back side as from the front side. In fact, there are no fast meteors on the back side. So this possibility just doesn’t make much sense.

But we can do better than an implausibility argument, and along the way, close another loophole. Under the assumption that all meteors are bound to the Sun, we can predict, as a function of the Earth’s speed, a relation between the angle with which a meteor shower strikes the Earth (i.e., the location of the radiant as seen in Earth’s sky) and the maximum speed that those meteors should have. If our assumption is wrong, this prediction should simply be irrelevant, and meteors’ maximum speed should not respect it.

In Fig. 4 is the result of that prediction, shown as a blue curve upon the data of Fig. 2, under the assumption that all meteors are bound to the Sun and therefore that the Earth moves at 30 km per second around the Sun (indicated by the black horizontal line in Fig. 4). [This first-try prediction was not made very carefully, and there are known omissions and possibly some errors in it; please point out any mistakes that you find.] The peak of the curve occurs, as it should, at v_fastest = v_esc,S + v_E = (√2 + 1) v_E. Everything looks very nice on the right half of the plot; for all meteor showers coming from a direction that is overhead after midnight, their speed lies on or below the predicted curve. Those that sit just along the curve are ones that, according to our assumption, have meteors moving almost at escape velocity, on elliptical orbits that stretch far outside Earth’s orbit.

Moreover, we can close a loophole. I assumed, in deriving the Earth’s speed, that the fastest meteors that we observe are indeed the fastest possible meteors. We see from Fig. 4 that this is actually true: there is a meteor shower located at almost the highest point in the curve, meaning that its meteoroids strike the Earth head-on, with speed (relative to the Sun) close to v_esc,S . These are the famous Leonid meteors that occasionally spark astonishing meteor storms, where the sky fills with meteor trails for a short time; by pure chance, these same Leonids teach us the distance to the Sun.

Things work less well on the left-hand side of the plot; the data points lie above the prediction instead of on or below it. This is a minor detail, but if you want an explanation (which at this point is still partial — I’m not 100% sure I know all the reasons for it) click here

The meteor showers that come from the pre-midnight sky are predicted to be quite slow, below 10-30 km per second. At such speeds, the fact that the meteors are actually falling into the Earth’s gravity, and accelerating as they do so, becomes important. Roughly, the observed speed of the meteors should be increased, relative to their speed v_m in the absence of Earth’s gravity, by an amount that is correlated with the escape velocity of the Earth v_esc,E . Specifically, their speed as they enter the atmosphere, by conservation of energy, will be increased as follows:

• v² ~ v_m² + v_esc,E²

This effect is negligible for a fast meteor; if it has intrinsic speed (relative to Earth) of 70 km per second, then the Earth’s gravity increases its speed by only 1 km per second. But for a slow meteor moving at 15 km per second relative to Earth, it’s a much bigger effect. This is shown in Fig. 4a, where the green curve corrects the blue curve, bringing it closer to the data points on the left-hand side of the plot.

Another thing I haven’t done carefully, in making the prediction, is that I’ve assumed that the meteor showers’ radiants lie more or less on the celestial equator. This isn’t true, and should be accounted for. I believe that leaving this out makes a bigger difference for the slower meteor showers, though I haven’t worked out the math (yet).

It’s useful to compare these curves to what the predictions would look like if the Earth moved twice as fast or half as fast. You can see in Fig. 5 that these other predictions look awful; the data don’t follow them at all. This comparison shows clearly that the Earth’s speed has been nailed down tightly to 30 km/sec by the meteor showers.

If you’re interested in how I generated Figs. 2, 4 and 5 — how I obtained and managed the data on meteor showers, and how I predicted the curve — click here for an explanation.

The table of meteor showers given here has all the information necessary. Though the meteors’ observed speeds are given in the table, locating the radiant in Earth’s sky requires some effort. Since the sky is two-dimensional, this is really two-dimensional information, but you will notice I reduced this to one-dimensional information, which is crude and leads to some imperfections in the predictions.

Specifically, I used Right Ascension, one of the two coordinates on the sky, and adjusted it for the date of the year on which the meteor shower appears, to obtain the longitude-like angle of the meteor shower on the sky. I expressed this as a fraction of 24 hours, with 0 meaning that the radiant is at 0 longitude [the north-south line that passes through the point overhead] at midnight. This is what is plotted versus the meteors’ speeds in Figs. 2, 4 and 5. Declination, the latitude-like angle, was ignored in the plotted points and in the corresponding prediction.

As for the prediction, the method is the following:

• Consider a meteor moving at the maximum possible speed v_esc,S = √2 v_E relative to the Sun, at an angle 𝛳_S relative to the Earth’s orbital path, as seen from the Sun.
• Compute the observed speed v_obs and angle 𝛳_obs of the same meteor as seen on Earth; that gives a rough prediction for the maximum speed at the angle 𝛳_obs, (at least if 𝛳_obs is small enough relative to the direction of the Earth’s motion.)

The formulas that result from this method are

• 𝛳_obs = tan^-1 [√2 sin 𝛳_S / (1 + √2 cos 𝛳_S ) ]
• v_obs² = v_E² [ (1 + √2 cos 𝛳_S )² + (√2 sin 𝛳_S )² ] = v_E² [ 3 + 2√2 cos 𝛳_S ]

The relation between v_obs and 𝛳_obs is what is plotted in Figs. 4 and 5. Again, this is not entirely right; one should do this more carefully than I did. But clearly, my crude approach already captures rather well what is going on for the meteor showers coming from the front side (the post-midnight side) of the Earth.

Why, by the way, are meteors in meteor showers all on elliptical orbits? It is because they are left behind by comets that travel on elliptical orbits. These comets repeatedly pass the Sun, partially evaporating during each pass, each time leaving a trail of rocks behind them on similar orbits. Comets on hyperbolic orbits, which pass only once, do not renew the trail of debris behind them, and thus do not create meteor showers.

Finally, not all meteors are in meteor showers; there are so-called “sporadics“. One could also learn the speed of the Earth from them too; their speeds are constrained in the same way as the meteors in showers. But precisely because they are sporadic, it’s not as easy to gather the necessary data.