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India's Aditya-L1 solar observatory does not orbit the Earth, and it does not fly to the Sun. It loops around a point in space about 1.5 million km away, where it can keep the Sun in view all the time. That point is named after Joseph-Louis Lagrange, an 18th-century mathematician who never saw a rocket. This post covers who he was, what a Lagrange point is, and how to estimate the distance to L1 using only Class 11 physics.
Who was Joseph-Louis Lagrange?
Lagrange was born in Turin on 25 January 1736 and died in Paris on 10 April 1813, according to his MacTutor biography. He learned much of his mathematics on his own. In August 1755, aged 19, he sent his results on the tautochrone problem to Leonhard Euler, who wrote back saying how impressed he was. Those ideas fed into what became the calculus of variations.
On 6 November 1766 Lagrange succeeded Euler as Director of Mathematics at the Berlin Academy, and he stayed in Berlin for 20 years. In that time he regularly won the prize of the Paris Académie des Sciences, including a share of the 1772 prize on the three-body problem with Euler. His Mécanique analytique, written in Berlin and published in 1788, turned mechanics into a branch of analysis. In its preface he warned readers that they would find no figures in the book, only algebraic operations.
What problem was Lagrange working on?
Newton's laws give an exact orbit for two bodies, such as one planet and the Sun. Add a third body and there is no general formula. Lagrange looked for special arrangements that could be solved exactly. NASA's explainer What is a Lagrange Point? says he considered this problem in his prize-winning Essai sur le problème des trois corps of 1772, and describes the result:
There are five special points where a small mass can move in a fixed pattern with two larger masses.
At these points the gravitational pull of the two large masses equals the centripetal force the small object needs to move with them.
L1, L2 and L3 lie on the line through the two large masses and are unstable. NASA notes that L1 and L2 are unstable on a timescale of about 23 days, so spacecraft there need regular course corrections.
L4 and L5 sit at the tips of two equilateral triangles with the large masses at the other corners. They are stable as long as the ratio of the two large masses is more than 24.96.
Why does Aditya-L1 sit at L1?
ISRO's Aditya-L1 mission page explains that the spacecraft is in a halo orbit around L1 of the Sun-Earth system, about 1.5 million km from Earth, which gives it a continuous view of the Sun without occultations or eclipses. PSLV-C57 launched it on 2 September 2023, and it was inserted into its halo orbit on 6 January 2024. Our post on what Aditya-L1 studies covers its instruments.
How far away is L1? A worked estimate
Lagrange points are not named in the JEE (Main) 2026 syllabus or the CBSE syllabus, but you can find L1 with tools you already have: Newton's law of gravitation, circular motion and the binomial approximation.
Let the Sun have mass M, the Earth mass m, and the Earth-Sun distance be a. The Earth goes round the Sun with angular speed ω, where ω² = GM/a³. Put a spacecraft a distance r from the Earth, on the side facing the Sun. To go round the Sun once a year, in step with the Earth, the net inward pull per kilogram must equal ω²(a − r):
GM/(a − r)² − Gm/r² = GM(a − r)/a³
The Sun pulls the craft inward; the Earth pulls it outward. Because r is much smaller than a, use (1 − r/a)⁻² ≈ 1 + 2r/a:
GM/a² × (1 + 2r/a) − Gm/r² ≈ GM/a² × (1 − r/a)
The GM/a² terms cancel, leaving 3GMr/a³ = Gm/r², so r³ = (m/3M)a³ and r = a × (m/3M)^(1/3).
Now the numbers. NASA's Sun fact sheet gives the Sun's mass as 332,900 Earth masses, and the Earth fact sheet gives a semi-major axis of 149.598 million km.
3M/m = 3 × 332,900 = 998,700.
The cube root of 998,700 is about 99.96.
r ≈ 149.598 million km ÷ 99.96 ≈ 1.497 million km.
That matches ISRO's "about 1.5 million km", which is about 1% of the way to the Sun.
Common mistakes in this calculation
Balancing forces on a body that is accelerating. Setting Gm/r² = GM/(a − r)² gives r = a/(1 + √332,900) ≈ 149.598 million km ÷ 578 ≈ 259,000 km. That point exists, but nothing stays there, because nothing provides the centripetal force.
Using the binomial approximation the wrong way. (1 − x)⁻² ≈ 1 + 2x, not 1 − 2x. Check by putting x = 0.01 into both sides on a calculator.
Thinking L1 is a parking spot. It is a balance point, not a stable one. Aditya-L1 is in a halo orbit around it and needs small corrections, as NASA describes for L1 and L2.
The same force-on-a-circle thinking appears in our guide to satellite and gravitation problems. If you find the moving-frame picture hard to see, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than to read.
What can a student take from Lagrange's way of working?
Shrink the problem until it can be solved. He could not solve every three-body problem, so he found the special cases that could be solved exactly. In an exam, try a limiting or symmetric case first.
Trust the algebra, then check it. His book had no figures, but our r ≪ a approximation is only safe because we checked the answer against ISRO's figure.
Show your work early. At 19 he sent his results to the leading mathematician of the day and got an answer. Asking a teacher to look at your method works the same way.
Try the estimate again for the Earth-Moon L1 point, using the Moon's mass and distance from NASA's Moon fact sheet, and see how far from the Moon it lands.
