Kepler's Laws — Application Problems

What Kepler’s Laws Actually Say

Three deceptively simple statements about how planets move shaped two centuries of physics. Kepler stared at Tycho Brahe’s data on Mars for years before realising the orbits were not circles — they were ellipses. From that single insight, he extracted three laws that describe every planet, satellite, and binary star in the universe.

For JEE and NEET, Kepler’s laws sit inside the Gravitation chapter. The questions almost always test orbital periods, areal velocities, or comparison between two orbits. The good news — the maths is direct, and most problems collapse to one or two formulas. The bad news — students rush, mix up “semi-major axis” with “radius,” and lose easy marks.

We will set up the three laws cleanly, derive the working formulas, and walk through five graded problems including a JEE Main 2024 favourite.

Key Terms & Definitions

Ellipse — a closed curve where the sum of distances from any point to two fixed points (foci) is constant. Planet orbits are ellipses with the Sun at one focus, not at the centre.

Semi-major axis ( $a$ ) — half the longest diameter of the ellipse. This is the “average” orbital distance and the only length that appears in Kepler’s third law.

Eccentricity ( $e$ ) — a number between 0 and 1 that measures how stretched the ellipse is. $e = 0$ is a circle; $e \to 1$ is a parabola.

Perihelion — closest approach to the Sun ( $r_{\min} = a(1 - e)$ ). Aphelion — farthest point from the Sun ( $r_{\max} = a(1 + e)$ ).

Areal velocity — the rate at which the radius vector sweeps out area, $\frac{dA}{dt}$ .

The Three Laws

First Law — Orbits are Ellipses

Every planet moves in an elliptical orbit with the Sun at one focus. The other focus is empty — there is no second mass there.

For circular orbits ( $e = 0$ ), the two foci coincide at the centre. JEE problems often start with circular orbits before moving to elliptical ones, so make sure you can switch between the two cleanly.

Second Law — Equal Areas in Equal Times

The line joining the planet to the Sun sweeps out equal areas in equal time intervals. Mathematically:

$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$

This is just conservation of angular momentum in disguise — gravity is a central force, so $L$ is conserved, and areal velocity equals $L/2m$ .

A direct consequence: when the planet is closer to the Sun, it moves faster. At perihelion the planet whips around quickly; at aphelion it crawls.

Third Law — Period Squared, Distance Cubed

The square of the orbital period $T$ is proportional to the cube of the semi-major axis $a$ :

$T^2 = \frac{4\pi^2}{GM}\, a^3$

For the solar system, the proportionality constant is the same for every planet because $M$ is the Sun’s mass. So $T^2 / a^3$ has the same value whether you pick Mercury or Neptune.

Deriving the Third Law (Quick Sketch)

For a circular orbit, gravity provides the centripetal force:

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

Solving for $T^2$ :

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

For elliptical orbits, replace $r$ with $a$ . The full proof requires integration but the result has the same form.

Solved Examples Graded Easy → Hard

Example 1 (Easy, CBSE-level)

A satellite orbits Earth at radius $r = 7000\,\text{km}$ . Find its period. Take $GM_{\text{Earth}} = 4 \times 10^{14}\,\text{m}^3/\text{s}^2$ .

Solution: Apply $T^2 = 4\pi^2 r^3 / (GM)$ .

$T^2 = \frac{4\pi^2 \times (7 \times 10^6)^3}{4 \times 10^{14}} = \frac{4\pi^2 \times 3.43 \times 10^{20}}{4 \times 10^{14}}$

$T^2 = \pi^2 \times 3.43 \times 10^6 \implies T \approx 5817\,\text{s} \approx 97\,\text{min}$

Matches typical low-Earth-orbit periods (ISS is at 92 minutes).

Example 2 (Easy, NEET-level)

If the orbital radius of a planet is doubled, by what factor does its period change?

Solution: $T \propto r^{3/2}$ . Doubling $r$ multiplies $T$ by $2^{3/2} = 2\sqrt{2} \approx 2.83$ .

Example 3 (Medium, JEE Main)

A planet has eccentricity $e = 0.5$ and semi-major axis $a$ . Find the ratio of speeds at perihelion and aphelion.

Solution: Conservation of angular momentum at the two extremes:

$m v_p r_p = m v_a r_a$

$\frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{a(1+e)}{a(1-e)} = \frac{1.5}{0.5} = 3$

So the planet moves three times faster at perihelion than at aphelion.

Example 4 (Medium, JEE Main 2023)

The orbital periods of two satellites of Earth are $T_1$ and $T_2 = 8T_1$ . The ratio of their orbital radii?

Solution: $T^2 \propto r^3$ , so $r \propto T^{2/3}$ .

$\frac{r_2}{r_1} = \left(\frac{T_2}{T_1}\right)^{2/3} = 8^{2/3} = 4$

Example 5 (Hard, JEE Advanced)

A comet has a highly elliptical orbit with perihelion $r_p = 0.5\,\text{AU}$ and period $T = 76\,\text{years}$ . Find its aphelion distance. (1 AU is Earth’s orbital radius, $T_{\text{Earth}} = 1$ year.)

Solution: First find the comet’s semi-major axis. Apply Kepler’s third law in scaled form:

$\frac{T_c^2}{T_E^2} = \left(\frac{a_c}{a_E}\right)^3$

$76^2 = a_c^3 \implies a_c = 76^{2/3}\,\text{AU} \approx 17.94\,\text{AU}$

Aphelion: $r_a = 2a_c - r_p = 35.88 - 0.5 = 35.38\,\text{AU}$ .

This is roughly Halley’s comet, which actually has $a \approx 17.8$ AU.

Exam-Specific Tips

JEE weightage — Gravitation typically yields 1 question worth 4 marks. Kepler’s laws appear in nearly every paper, often blended with energy conservation or angular momentum.

NEET weightage — Gravitation contributes 2 to 3 marks, with one Kepler-law question almost guaranteed.

When a problem mentions “average distance,” “mean radius,” or “semi-major axis,” they all mean $a$ in Kepler’s third law. Do not waste time hunting for distinctions.

For ratio problems, take the ratio of two Kepler equations. The constants $4\pi^2/GM$ cancel, leaving $(T_1/T_2)^2 = (a_1/a_2)^3$ . This is the most-used form in MCQs.

Common Mistakes to Avoid

Mistake 1: Using $r$ for elliptical orbits instead of $a$ . The third law uses semi-major axis, not the instantaneous distance.

Mistake 2: Forgetting that Kepler’s third law applies to bodies orbiting the same central mass. You cannot compare a satellite of Earth to a satellite of Mars using a single equation — $GM$ differs.

Mistake 3: Confusing perihelion and aphelion. Perihelion (peri = near) is the closest point. Aphelion (apo = far) is the farthest.

Mistake 4: Forgetting to convert units. AU and years are convenient because $T^2 = a^3$ becomes a clean numerical identity for solar-system bodies.

Mistake 5: Using $v = 2\pi r/T$ for elliptical orbits. That formula assumes constant speed (circular motion). For ellipses, speed varies, and you must use angular momentum conservation instead.

Practice Questions

Q1. Mars’s semi-major axis is 1.524 AU. Find its orbital period.

$T = a^{3/2} = 1.524^{3/2} \approx 1.88$ years. Matches the known value.

Q2. A satellite at altitude $h$ above Earth has period $T_0$ . What is the period at altitude $3h$ if $h$ equals Earth’s radius?

Original radius $= 2R$ , new radius $= 4R$ . Period ratio $= (4/2)^{3/2} = 2\sqrt{2}$ . So new period is $2\sqrt{2}\,T_0$ .

Q3. A comet’s areal velocity is $5 \times 10^{15}\,\text{m}^2/\text{s}$ at perihelion. What is it at aphelion?

Areal velocity is constant by Kepler’s second law. So it is also $5 \times 10^{15}\,\text{m}^2/\text{s}$ at aphelion.

Q4. Two planets orbit the same star. If $T_1 : T_2 = 1 : 8$ , find $a_1 : a_2$ .

$a \propto T^{2/3}$ , so $a_1 : a_2 = 1 : 8^{2/3} = 1 : 4$ .

Q5. An asteroid has $r_p = 2$ AU, $r_a = 4$ AU. Find its period.

$a = (r_p + r_a)/2 = 3$ AU. $T = 3^{3/2} \approx 5.196$ years.

FAQs

Why is the orbit an ellipse and not a circle? A circle is a special ellipse with $e = 0$ . Real orbits have small but nonzero eccentricity because of the way they form from collapsing dust clouds.

Does Kepler’s law work for binary stars? Yes — but $M$ in $T^2 = 4\pi^2 a^3/(GM)$ becomes the total mass $M_1 + M_2$ , and $a$ is the semi-major axis of the relative orbit.

Why doesn’t the Moon spiral into Earth? Same reason planets don’t fall into the Sun — they have just enough sideways speed to keep “missing” the central body. Orbits are continuous free fall.

What about the Sun’s motion? Both the planet and the Sun orbit their common centre of mass. Since the Sun is so massive, the centre of mass lies almost at the Sun’s centre, and the simple “Sun fixed” picture works.

Can Kepler’s laws fail? Yes — near very massive bodies (general relativity corrections kick in, as with Mercury’s perihelion precession), and when there are strong perturbations from other bodies.

How do satellites use Kepler’s laws? Geostationary satellites have $T = 24$ hours. Solving $T^2 = 4\pi^2 r^3/(GM)$ gives $r \approx 42,164$ km — a fixed altitude where every commercial geo-satellite parks.