Three laws: elliptical orbits, equal areas, T²/a³ = constant.

Kepler's Laws of Planetary Motion — T² ∝ a³

Question

State Kepler’s three laws of planetary motion. A planet takes 365 days to complete one revolution around the Sun. Another planet takes 730 days. If Earth’s average orbital radius is $a$ , find the orbital radius of the second planet.

Solution — Step by Step

First Law (Law of Orbits): Every planet moves in an elliptical orbit with the Sun at one focus — not the centre, 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 when closer to the Sun.

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 \quad \Rightarrow \quad \frac{T^2}{a^3} = \text{constant}

Since the constant is the same for all planets orbiting the same star, we write:

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

Here, Planet 1 is Earth: $T_1 = 365$ days, $a_1 = a$ . Planet 2: $T_2 = 730$ days, $a_2 = ?$

Rearrange to isolate $a_2^3$ :

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

a_2^3 = a^3 \cdot \left(\frac{730}{365}\right)^2 = a^3 \cdot (2)^2 = 4a^3

\boxed{a_2 = 4^{1/3} \cdot a = 2^{2/3} \cdot a \approx 1.587\, a}

Why This Works

Kepler’s Third Law comes directly from the balance between gravitational force and centripetal acceleration. For a circular orbit (a simplified but good approximation), $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , and since $v = \frac{2\pi r}{T}$ , working through the algebra gives $T^2 = \frac{4\pi^2}{GM} r^3$ . Everything in the bracket is constant for the same central body (the Sun), so $T^2 \propto r^3$ .

The Second Law is actually a statement about conservation of angular momentum — nothing to do with gravity specifically. Any central force (one directed always toward a fixed point) conserves angular momentum, and that’s exactly what makes the equal-area rule work.

The First Law — ellipses, not circles — requires solving Newton’s law of gravitation fully. NCERT Class 11 asks you to state and apply these, not derive them. For JEE, the derivation of the Third Law for circular orbits is fair game.

Alternative Method

For the numerical part, you can work with the ratio directly without isolating $a_2^3$ first.

\frac{a_2}{a_1} = \left(\frac{T_2}{T_1}\right)^{2/3} = \left(\frac{730}{365}\right)^{2/3} = (2)^{2/3}

Since $(2)^{2/3} = \sqrt[3]{4} \approx 1.587$ , we get $a_2 \approx 1.587\, a$ .

Memorise this form: $\dfrac{a_2}{a_1} = \left(\dfrac{T_2}{T_1}\right)^{2/3}$ . In NEET and JEE Main, the ratio approach is faster than setting up the full equation — saves 30 seconds per question.

Common Mistake

The most common error: writing $a_2 = 2a$ because $T_2 = 2T_1$ . Students confuse a direct proportion ( $a \propto T$ ) with the actual law ( $a^3 \propto T^2$ ). The relationship is not linear. If the period doubles, the radius increases by a factor of $2^{2/3} \approx 1.587$ , not 2. This exact trap appeared in NCERT exercise questions and has shown up in CBSE board papers.

Also watch out: the " $a$ " in Kepler’s Third Law is the semi-major axis of the ellipse, not the radius. For circular orbits they’re the same, but if the question mentions an elliptical orbit, use the semi-major axis. NCERT Class 11, Chapter 8 is explicit about this.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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