Back from sort of a mini-break this summer — it’s time to dive into some more physics topics! In honor of the spectacular success so far of the JWST mission, and not wanting to re-hash all the excellent writing about the initial images themselves, maybe it’s worth spending a little time talking about where the thing actually is, and why it’s there.
The previous generation’s Space Telescope, the Hubble, is only barely in space. It speeds around the Earth just outside the atmosphere about 340 miles up — only about a tenth of the radius of the Earth:
There are some advantages to being so close. Maybe most importantly, it’s close enough to be serviced by actual humans in case something goes wrong or for basic maintenance. This was really important in Hubble’s early days when a “corrective” lens was installed by shuttle astronauts to sharpen images. But it’s a fast-moving camera, so long time exposures are tricky, especially if the Earth, Moon, or Sun gets in the way.
Soon after the Hubble launch, planning was underway for its successor/partner, the JWST. A large part of this new mission was to be able to observe extremely distant galaxies, which means that the light would have been traveling an extremely long time, which means that we would be looking at them as they were a very long time ago. Now, the Universe is expanding (stretching!), and it turns out that this stretch of space also stretches the light that travels through the space in the same way, towards red/infrared wavelengths. All this implies that if you want to see the (formerly) visible light given off by a very distant galaxy, you need a telescope able to see into the infrared part of the spectrum. In fact, the “stretch” (called the redshift, z) for these galaxies can be about a factor of 12 or 13, so the JWST is specially designed to be just such an infrared telescope.
Now, an associated problem is that to sensitively detect infrared light the telescope needs to be in a very cold environment. A telescope that is 150 K will emit infrared light that peaks right in the JWST spectrum! So if you want your instruments to be free of contamination from local heat, they’ll need to be below about 40 K (all the way down to 7 K for the MIRI infrared instrument). For this telescope, that means putting it far away from Earth in a stable orbit where it can shield itself from the Sun’s energy, but yet stay close enough for easy communication. In other words, at a Lagrange Point!
Orbital Ideas
Sometimes there’s a misconception that the L-Points are regions where the gravitational forces from two or more objects balance or cancel out. Not quite! It’s a dynamical equilibrium, where the key idea is that objects placed at these points will remain at the same relative position to Earth. In the language of physics equations, what we require is that they have the same period of revolution around the Sun as the Earth does. Here’s how the period of an orbit works for the Earth (not to scale!):
An orbiting object (E) at some distance r away from a much more massive thing (S) will feel only the gravitational force from it, so Newton says that we ought to sum the forces on it (there’s only 1!) and set that equal to the orbiting object’s mass times how it accelerates — moving in a circle is called the centripetal acceleration:
So using Newton’s Universal Law of Gravitation and the expression for centripetal acceleration,
from this, we can calculate the orbital speed of an object in a circular orbit:
For the Hubble circling the Earth, that would mean it’s skimming around at about 7800 m/s, or 17,000 mph! But there’s something we’ll come back to in that last equation — look how, if the distance r gets larger, the orbital speed is smaller. Thinking about this in terms of the period of revolution, an object farther out travels a larger circle all the way around, so the time taken to go around once increases faster than just linearly. In fact, we can see how that works by just replacing the speed with a quantity involving the period (T). Since speed is just distance/time, and the distance all the way around the orbit is the circumference of a circle,
That last equation (it follows after a bit of algebra!) is the modern statement of Kepler’s 3rd Law, which tells us that the square of the period is proportional to the cube of the distance. It was something of a mystery to Kepler, who noticed this relationship for every known planet but didn’t understand the physical reason (gravity) underlying it. Newton was the one who, with his concept of gravity and force, derived the behavior of bodies in the heavens.
By the way, I can’t resist pointing out that we use the above ideas in astrophysics all the time to do mind-blowing things. We know, for example, the mass of the Earth and Sun, which seem like very odd things to know precisely. How can you “weigh” them on any kind of scale? But look again at that equation — what we can easily observe of an orbiting body is the distance from the central object and the time taken to go around once. For the Earth, the orbiting body is the Moon, so if you plug in the Moon’s distance and time taken to go around once (27-ish days) you’ve now weighed the Earth! For the Sun, we know the Earth’s distance and orbital period, so the Sun’s mass is immediately known. We do the same thing for distant binary stars too!
Finding the Lagrange Points L1 and L2
Ok, here’s the general problem in finding stability points — let’s find the one between Earth and Sun (this is called L1).
At some instant in time, our little mass m feels the gravitational tug from the Sun pulling to the left and a smaller force pulling to the right. Let’s set this up like we did earlier, summing those two forces and setting them equal to the centripetal acceleration:
… notice that the direction of the centripetal acceleration is always towards the center of the motion for m, which in this case is to the left. And for this mass, its orbital speed is related to its period as we did before:
So we can plug this in to the above, and we’ll have a factor of T squared in the denominator. Ok, and here’s the condition: this is a Lagrange point if the orbital period T for this mass equals the Earth’s orbital period! If so, then it will stay in the same relative position with respect to the Earth all the way around. We’ve already found that factor above, so plug that in too:
(here, I canceled out common terms like G and m and flipped all the signs to make it look nicer)
What we’re after is that L, the distance away from Earth that will be a stable point. This is no algebraic walk in the park, so lemme outline what I did to get the answer between the dividers. Feel free to skip!
A standard way to attack things like this is to see if the problem involves “small quantities”. For example, in the above equations notice that we’d expect the distance r from the Sun to Earth to be quite a bit larger than the distance L from the Earth to our point in question since the Sun’s mass is 300,000 times larger than the Earth’s mass. The thing that’s small, then, is the ratio L/r. A crazy useful mathematical trick is the Binomial Approximation, which says
This turns nasty quantities raised to some power into just adding and subtracting terms! Awesome. And n doesn’t even have to be an integer, so things like this pop up in physics all the time:
aaaaaanyway, I can massage our last equation into a form where I can take advantage of this:
then using the Binomial Approx.,
Now this algebra is much easier — the leading terms with “1” just cancel:
and then
So L is close to r, except it’s multiplied by a factor involving the ratio of the masses of the two other bodies involved (which kind of makes sense). Just working it out roughly in your head, the Sun’s mass is about 300,000 times the Earth’s mass, so 3 times that is about a million. The cube root of a million is a hundred, so L is about 1% of r, or about 1.5 million km away from the Earth towards the Sun (about 4 times the Moon’s distance!) The SOHO solar observatory spacecraft is here so it can hang out at that point and monitor the Sun.
Let’s pause for a sec to see if this makes conceptual sense. L1 is closer to the Sun, so normally we’d expect that it would orbit faster as well as having a shorter path all the way around, making its period smaller. But Earth is “behind” it, sort of pulling it back in just such a way as to keep it in the same relative position. Same for L2, except that Earth tugs it a bit “forward” to help it keep up. The position of L2, by the way, we get exactly the same way as for L1, but in the beginning equations we’d write the distance to it as (r+L) instead of (r-L) and flip the sign of the gravitational force due to the Earth — we get the same distance, just on the other side.
All 5 Lagrangian Points!
You might imagine there’s another one, sort of a “mirror Earth” position on the other side of the Sun, and we call this one L3. An excellent place to launch an alien invasion force, since being exactly opposite to the Sun, we’ll never see it!
L4 and L5 are, as they say, “beyond the scope of this article”, but they’re found (interestingly) at the other corner of equilateral triangles formed by the Sun and Earth.
On Deck:
The next article I’m working on, in honor of the US Senate passing a first step in collectively addressing climate change, is an executive summary of the scientific evidence — is the Earth warming? Why? How much? How do we know if it’s us causing it? It seems to me that too often we lose the forest for the trees, so to speak, and those with a vested interest in confusing the issue love to lead us down deceptive paths!
If you’re a student/teacher and want to see lots of worked examples that I like to include in my classes when I teach the “standard” University Physics 1 and 2 courses, feel free to browse the (growing) collection of 150+ videos at
And if something is especially cool and you’re inclined to leave a “tip” I’m not above coffee or pizza:
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