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Kepler's second law is usually learnt as a sentence: equal areas in equal times. Exam questions then ask for speeds at different points, and students write vr = constant everywhere. This post builds the law from angular momentum, puts real numbers into it using NASA's data for Earth and Mars, and shows where the shortcut breaks.
Where is it in NCERT and the syllabus?
The 2026-27 reprint of the NCERT Class 11 chapter Gravitation states the law of areas in Section 7.2 and derives it. In a short time Δt, the planet's radius vector sweeps an area ΔA = ½(r × vΔt) (Eq. 7.1), so ΔA/Δt = L/(2m) (Eq. 7.2), where L is the angular momentum. Example 7.1 then relates speeds at perihelion and aphelion. The chapter's Points to Ponder add that angular momentum conservation "holds for any central force", not just inverse-square gravity.
The Syllabus for JEE (Main) 2026 and the NEET (UG) 2026 syllabus both list "Kepler's law of planetary motion" in the gravitation unit. Check the syllabus for your exam year.
Why does a central force give equal areas?
Gravity on a planet points along the line to the Sun, parallel to r. Its torque about the Sun is r × F = 0, so L = m(r × v) stays constant. Since dA/dt = L/(2m), the areal velocity is constant too.
Two useful forms follow:
Over a whole orbit: dA/dt = (area of ellipse)/T = πab/T, with b = a√(1 − e²).
At any point: dA/dt = ½ r v sin θ, where θ is the angle between r and v. At perihelion and aphelion θ = 90°, so v = 2(dA/dt)/r.
Worked example 1: Earth's areal velocity
NASA's Earth fact sheet gives a semi-major axis of 149.598 × 10⁶ km, an eccentricity of 0.0167 and a sidereal period of 365.256 days.
a = 1.49598 × 10¹¹ m.
b = a√(1 − 0.0167²) = 1.49577 × 10¹¹ m. The orbit is very nearly a circle.
T = 365.256 × 86400 = 3.1558 × 10⁷ s.
dA/dt = πab/T = π × 1.49598 × 10¹¹ × 1.49577 × 10¹¹ / 3.1558 × 10⁷ ≈ 2.23 × 10¹⁵ m²/s.
So Earth's radius vector sweeps about 2.2 billion square kilometres every second. Per kilogram of Earth, the angular momentum is L/m = 2 dA/dt ≈ 4.46 × 10¹⁵ m²/s.
Worked example 2: speeds at perihelion and aphelion
The same fact sheet gives perihelion at 147.095 × 10⁶ km and aphelion at 152.100 × 10⁶ km.
v_p = 2(dA/dt)/r_p = 2 × 2.228 × 10¹⁵ / 1.47095 × 10¹¹ ≈ 3.029 × 10⁴ m/s = 30.29 km/s.
v_a = 2 × 2.228 × 10¹⁵ / 1.52100 × 10¹¹ ≈ 2.929 × 10⁴ m/s = 29.29 km/s.
NASA lists Earth's maximum and minimum orbital speeds as 30.29 km/s and 29.29 km/s. The second law, with nothing but a, e and T, reproduces both.
Worked example 3: how long on each half of the orbit?
NCERT's Example 7.1 asks whether the planet takes longer to traverse the far arc BAC or the near arc BPC, where B and C are the ends of the minor axis. We can put a number on it.
The minor axis BC passes through the centre of the ellipse. The Sun sits at a focus, a distance ae from the centre on the perihelion side.
The area swept while going from B through perihelion to C is the half-ellipse on that side minus the triangle formed by the Sun, B and C: πab/2 − ½ × 2b × ae = πab/2 − abe.
Divide by the whole area πab. The fraction of the period spent on the near half is ½ − e/π.
For Earth, ½ − 0.0167/π = 0.4947, so 180.7 days on the near half and 184.6 days on the far half. For Mars, NASA's Mars fact sheet gives e = 0.0935 and a period of 686.980 days: ½ − 0.0935/π = 0.470, about 323 days near and 364 days far.
Common mistakes and how to avoid them
Using v₁r₁ = v₂r₂ at a general point. Use r v sin θ, or work only at perihelion and aphelion.
Thinking the law needs inverse-square gravity. Any central force conserves L. The inverse square matters for the elliptical shape and for the third law.
Using ½rvΔt for the area at an angle. The triangle's area is ½ r v sin θ Δt.
Assuming equal halves of the orbit take equal times. Equal areas, not equal arcs or equal angles.
Our post on Johannes Kepler and how he found his laws tells the history, and the guide to angular momentum of a point particle covers r × p in detail. If you would like to watch a planet speed up near the Sun, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
A practice routine
Derive dA/dt = L/2m from ΔA = ½(r × vΔt) without notes.
Repeat examples 1 and 2 for Mars using NASA's fact sheet, and compare with its listed maximum and minimum speeds.
Derive the ½ − e/π result once, with a sketch.
Solve one problem where the velocity makes an angle with the radius, using r v sin θ.
Try it now: a comet's speed is 50 km/s at perihelion, 0.5 AU from the Sun. Find its speed at aphelion, 5 AU away.
