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

Question

State Kepler’s three laws of planetary motion. Using the third law, if Earth’s orbital period is 1 year and its distance from the Sun is $1$ AU, find the orbital period of a planet at distance $4$ AU from the Sun.

(CBSE 11 & NEET standard question)

Solution — Step by Step

First Law (Law of Orbits): Every planet moves in an elliptical orbit with the Sun at one focus.

Second Law (Law of Areas): The line joining the planet to the Sun sweeps equal areas in equal time intervals. This means the planet moves faster near the Sun (perihelion) and slower far from the Sun (aphelion).

Third Law (Law of Periods): The square of the orbital period is proportional to the cube of the semi-major axis: $T^2 \propto a^3$ .

For two planets orbiting the same star:

\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}

This is extremely powerful — it eliminates the need to know the mass of the Sun.

Planet 1 (Earth): $T_1 = 1$ year, $a_1 = 1$ AU

Planet 2: $a_2 = 4$ AU, $T_2 = ?$

\frac{1^2}{T_2^2} = \frac{1^3}{4^3} = \frac{1}{64}

T_2^2 = 64 \implies T_2 = \mathbf{8 \text{ years}}

Why This Works

Kepler’s laws are empirical — he discovered them from Tycho Brahe’s observational data, decades before Newton explained them using gravity. Newton showed that:

The first law follows from the inverse-square nature of gravity
The second law is equivalent to conservation of angular momentum
The third law can be derived as $T^2 = \frac{4\pi^2}{GM}a^3$

graph TD
    A["Kepler's Laws"] --> B["1st Law: Elliptical orbits"]
    A --> C["2nd Law: Equal areas in equal time"]
    A --> D["3rd Law: T² ∝ a³"]
    B --> B1["Newton: Follows from<br/>inverse-square gravity"]
    C --> C1["Newton: Angular momentum<br/>conservation"]
    D --> D1["Newton: T² = 4π²a³/GM"]
    D1 --> D2["For same star: T₁²/T₂² = a₁³/a₂³"]

The second law tells us something beautiful: angular momentum $L = mr^2\omega$ is constant. When $r$ decreases (planet closer to Sun), $\omega$ must increase — the planet speeds up. This is the same principle as an ice skater spinning faster when pulling arms in.

Alternative Method — Direct Formula for Circular Orbits

For a circular orbit (a special case of elliptical), we can derive the third law from $F = ma$ :

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

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

This confirms $T^2 \propto r^3$ and lets you calculate the actual period if you know $G$ and $M$ .

For JEE numericals involving satellites: the third law applies to any object orbiting a central body — planets around the Sun, moons around a planet, or artificial satellites around Earth. Just use the correct central mass $M$ .

Common Mistake

Students apply $T^2 \propto a^3$ to compare planets around different stars. The proportionality constant $4\pi^2/(GM)$ depends on the central mass. Two planets around different stars cannot be compared using the ratio method — you need the full formula with $M$ for each star. This trap appeared in JEE Main 2022.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Formula for Circular Orbits

Common Mistake

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