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Solving Kepler's Third Law Problems (Planet Time Periods)

Physics · Gravitation · NEET

Kepler's Third Law says the square of a planet's time period is proportional to the cube of its orbit radius: T² ∝ R³, or T₁²/T₂² = R₁³/R₂³. So to find one planet's year when you know another's, take the ratio: T₁/T₂ = (R₁/R₂)^(3/2). Memory hook: "Two goes up top, Three goes down below" — T-squared over R-cubed is always the same number for every planet around the Sun.
Kepler's Third Law: T² ∝ R³ (same Sun)SunMercuryMarsR4RT²R³T² vs R³is a straight lineslope = 4π²/GM
Left: Mars orbits at 4× Mercury's radius, so its period is 8× longer (4^(3/2) = 8). Right: plotting T² against R³ gives a straight line through the origin, with slope 4π²/GM the same for every planet around the Sun.

Your doubts, answered

How do I set up a Kepler's Third Law ratio problem?

Write the law as T₁²/T₂² = R₁³/R₂³. Put the planet you KNOW on one side and the planet you WANT on the other. Then take the square root of both sides so you get T₁/T₂ = (R₁/R₂)^(3/2). Plug in the numbers and solve for the unknown period. You never need the mass of the Sun or G for a ratio problem — they cancel out.

In T² ∝ R³, which one gets squared and which gets cubed?

Time period T is SQUARED (power 2). Orbit radius R is CUBED (power 3). A simple way to remember: T comes before R in the alphabet, and 2 comes before 3, so T→2 and R→3. So the constant that is the same for all planets is T²/R³.

Why can I ignore G and the Sun's mass in these problems?

The full form is T² = (4π²/GM) R³. The part (4π²/GM) is a constant that is the same for every planet orbiting the same Sun. When you divide one planet's equation by another's, this whole constant cancels, leaving only T₁²/T₂² = R₁³/R₂³. That is why ratio numericals are fast — no constants needed.

What does the ^(3/2) power actually mean here?

The 3/2 power (called 'three-halves') means: cube the ratio, then take the square root. For example (1/4)^(3/2) = √(1/4³) = √(1/64) = 1/8. Practise squares and cubes of small numbers so you can do this in your head during NEET.

Can I use Kepler's Third Law for satellites around Earth?

Yes. The same law works for any objects orbiting the same central body. For two satellites of Earth, T₁²/T₂² = R₁³/R₂³, where R is measured from Earth's centre (surface height h plus Earth's radius R_E). This is exactly how the NEET 2019 satellite-period question is solved.

⚠️ The NEET trap
✗ Students plug the height above the surface as R and forget to add Earth's radius, or they square the radius instead of cubing it.
✓ R in Kepler's law is the FULL orbit radius from the centre (height h + planet/Earth radius, if given from the surface). And R is cubed, T is squared: T² ∝ R³.
🧠 NTA loves giving distance 'from the surface' — always add the radius to get distance 'from the centre' before using R³.

Real NEET questions

2025

The radius of the Martian orbit around the Sun is about 4 times the radius of Mercury's orbit. The Martian year is 687 earth days. The length of one year on Mercury is about:

A · 172 earth days
B · 124 earth days
C · 88 earth days ✓
D · 225 earth days
Solution: Use Kepler's Third Law: T² ∝ R³, so T_mer/T_mar = (R_mer/R_mar)^(3/2). Given R_mar = 4 R_mer, so R_mer/R_mar = 1/4. Then T_mer = T_mar × (1/4)^(3/2) = T_mar × 1/8. So T_mer = 687/8 = 85.9 ≈ 88 earth days. (Mercury's real year is 88 days, confirming the answer.)
2026

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to:

A · R^(1/2)
B · R^(3/2) ✓
C · R^2
D · R^3
Solution: Kepler's Third Law: T² ∝ R³. Taking the square root of both sides gives T ∝ R^(3/2). So the time period is proportional to R^(3/2), which is option B.
2019

A satellite has period 24 h at height 6R_E from the earth's surface. The period of another satellite at height 2.5 R_E from the surface is:

A · 6√2 h ✓
B · 12√2 h
C · 24/√2.5 h
D · 12/√2.5 h
Solution: Orbit radius = height + R_E. First satellite: r₁ = 6R_E + R_E = 7R_E. Second: r₂ = 2.5R_E + R_E = 3.5R_E. By Kepler's law T² ∝ r³, so T₂ = T₁ × (r₂/r₁)^(3/2) = 24 × (3.5/7)^(3/2) = 24 × (1/2)^(3/2) = 24 × 1/(2√2) = 12/√2 = 6√2 h. Answer: 6√2 h (option A).

Solved Gravitation NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 27 Gravitation NEET PYQs ›
Next concept: Newton's Universal Law of Gravitation →Keep learning — 2 minFeeling ready? Solve the Gravitation NEET PYQs ›Or practice on your phone — get the free MedicNEET app ›

Frequently asked

What is the formula for Kepler's Third Law numericals?

For comparing two planets or satellites around the same body: T₁²/T₂² = R₁³/R₂³, which rearranges to T₁ = T₂ × (R₁/R₂)^(3/2). The full single-body form is T² = (4π²/GM)R³, but for NEET ratio problems the constant cancels.

How do you find the time period of a planet if you know its orbit radius?

Compare it to a known planet. Set T₁/T₂ = (R₁/R₂)^(3/2), put the known planet's T and R on one side, and solve for the unknown T. No value of G or the Sun's mass is needed.

Is Kepler's Third Law valid for both circular and elliptical orbits?

Yes. For elliptical orbits, R is replaced by the semi-major axis 'a', so T² ∝ a³. For circular orbits the radius is the semi-major axis, so it becomes T² ∝ R³. Both forms appear in NEET.

Why is Kepler's Third Law important for NEET Physics?

NEET regularly asks direct ratio numericals from Gravitation (2019, 2025, 2026). These are fast, guaranteed marks if you remember T² ∝ R³ and can compute a (3/2) power — often just a cube then a square root of small numbers.

Related concepts:
Kepler's Third Law (T² ∝ R³): Formula and DerivationTime Period of a Satellite: Formula and ProblemsKepler's First Law (Law of Orbits) ExplainedEarth Satellites: Orbital Velocity DerivationGravitation Formula Sheet and Quick Revision for NEET
Practise Physics for NEET:
Physics MCQs (free, chapter-wise)Gravitation PYQs