Kepler's three laws — statement, derivation hints, and application

Question

What are Kepler’s three laws of planetary motion? How do we derive the third law from Newton’s gravitation, and how do we apply these laws to solve problems?

(CBSE 11, JEE Main, NEET — Kepler’s third law numerical is a recurring question, and the second law connects to angular momentum conservation)

Solution — Step by Step

Every planet moves in an elliptical orbit with the Sun at one of the two foci.

For most problems, we approximate the orbit as circular (eccentricity is small for most planets). The key takeaway: the Sun is not at the centre of the orbit — it is at a focus. This means the planet-Sun distance varies throughout the orbit.

Perihelion: closest point to the Sun. Aphelion: farthest point from the Sun.

The line joining the planet and the Sun sweeps out equal areas in equal time intervals.

\frac{dA}{dt} = \text{constant}

This means the planet moves faster at perihelion (closer to Sun) and slower at aphelion (farther from Sun).

Connection to physics: This law is a direct consequence of conservation of angular momentum. Since the gravitational force is central (always directed toward the Sun), it produces zero torque about the Sun, so $L = mvr\sin\theta$ is constant. At perihelion, $r$ is smaller, so $v$ must be larger.

T^2 \propto a^3

For circular orbits: $T^2 = \frac{4\pi^2}{GM}r^3$

where $T$ = orbital period, $a$ = semi-major axis (or $r$ for circular orbits), $M$ = mass of the central body.

Derivation for circular orbit: For a planet in circular orbit, gravitational force provides centripetal force:

\frac{GMm}{r^2} = \frac{mv^2}{r} = \frac{m \cdot 4\pi^2 r}{T^2}

T^2 = \frac{4\pi^2 r^3}{GM}

This shows $T^2 \propto r^3$ — the farther the planet, the longer its orbital period. Also note: $T$ depends on $M$ (mass of the Sun) but not on $m$ (mass of the planet).

Problem: Two satellites orbit Earth. Satellite A has orbital radius $r$ and satellite B has orbital radius $4r$ . Find the ratio of their time periods.

Using Kepler’s third law:

\frac{T_B^2}{T_A^2} = \frac{(4r)^3}{r^3} = 64

\frac{T_B}{T_A} = \sqrt{64} = \mathbf{8}

Satellite B takes 8 times longer to complete one orbit.

flowchart TD
    A["Kepler's Laws"] --> B["1st Law: Elliptical orbits, Sun at focus"]
    A --> C["2nd Law: Equal areas in equal times"]
    A --> D["3rd Law: T² ∝ r³"]
    C --> E["Consequence of angular momentum conservation"]
    D --> F["Derived from: gravitational force = centripetal force"]
    D --> G["Application: compare periods of different orbits"]
    G --> H["T₂/T₁ = (r₂/r₁)^(3/2)"]

Why This Works

Kepler’s laws are empirical observations that Newton later explained through his law of gravitation. The first law follows from solving the orbit equation for an inverse-square central force (which gives conic sections). The second law is angular momentum conservation. The third law comes from balancing gravitational and centripetal forces.

The power of the third law is that it connects orbital size to orbital period through a simple power law — no need to know the velocity, acceleration, or force explicitly.

Common Mistake

When using $T^2 \propto r^3$ to compare two orbits, students sometimes forget that this proportionality holds only for orbits around the same central body. You cannot compare a satellite orbiting Earth with a planet orbiting the Sun using the same proportionality constant. The constant $4\pi^2/(GM)$ is different because $M$ is different (Earth vs Sun).

The ratio form is fastest for comparisons: $T_2/T_1 = (r_2/r_1)^{3/2}$ . You do not need $G$ , $M$ , or $\pi$ . JEE Main almost always asks Kepler’s third law in ratio form — practise converting radius ratios to period ratios and vice versa.

Question

Solution — Step by Step

Why This Works

Common Mistake

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