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Exam level6 past questions

Kepler's laws of planetary motion

Kepler's laws describe the motion of planets and satellites: orbits are ellipses (first law), equal areas are swept in equal times (second law), and the square of the period is proportional to the cube of the semi-major axis (third law).

Why this shows up in the exam

NEET tests your ability to apply these laws to planetary and satellite orbits, including time periods and velocities.

How NEET tests this

Numerical · 4 QsStatement analysisDirect recall

Learn the idea

Kepler’s third law links the time period of an orbiting body to the size of its orbit: the square of the period is proportional to the cube of the semi‑major axis. This single relation instantly tells whether a nearer or farther satellite has a larger period.

🧠 Memory hook: Kepler’s third law is a square‑cube dance: the period’s square partners with the orbit’s cube, so the period grows as the 3⁄2 power of the size.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • First law: planetary orbits are ellipses with the Sun at one focus","Second law: a line joining the planet and the Sun sweeps equal areas in equal times","Third law: T² ∝ a³ for any orbit (for a circular orbit a = r)
  • For a circular orbit the orbital speed v = √(GM/r) and the centripetal force is provided by gravity
  • The constant of proportionality in T² = (4π²/GM) a³ is the same for all bodies orbiting the same central mass

How to approach it

  1. 1Read the question and note whether the orbit is circular or elliptical – replace the semi‑major axis by the radius for a circle
  2. 2Write the ratio form of the third law: (T₁/T₂)² = (a₁/a₂)³
  3. 3Take square‑root to get the period ratio: T₁/T₂ = (a₁/a₂)^(3/2)
  4. 4Apply the ratio to the numbers given and choose the answer
  5. 5If only a qualitative trend is asked, remember: larger a ⇒ larger T (because of the 3/2 power)

Worked example — watch it click

Assertion (A): The time period of revolution of a satellite close to surface of earth is smaller than that revolving away from surface of earth. Reason (R): The square of time period of revolution of a satellite is directly proportional to cube of its orbital radius.

  • ✅If both assertion and reason are true, and reason is the correct explanation of assertion.
  • B)If both assertion and reason are true, but reason is not the correct explanation of assertion.
  • C)If assertion is true, but reason is false.
  • D)If both assertion and reason are false.

The concept behind this problem

The worked example checks whether you can translate the statement “closer satellite has smaller period” into the quantitative relation T² ∝ r³ and recognise that the reason given is exactly the law that produces the trend.

Step by step

  1. 1Kepler's third law: T² ∝ r³, so T ∝ r^(3/2).
  2. 2For smaller orbital radius (close to Earth), T is smaller.
  3. 3The reason correctly states T² ∝ r³ and explains why satellites closer to Earth have smaller periods.
  4. 4Both A and R are true, and R correctly explains A.

Watch out

Students often write T ∝ r³ instead of the correct T ∝ r^(3/2), leading to the wrong conclusion about how period changes with radius.

Common slip-ups that cost marks

  • •Confusing the linear proportionality T ∝ a³ with the correct T ∝ a^(3/2)
  • •Using the orbital radius instead of the semi‑major axis for an elliptical orbit
  • •Treating the constant of proportionality as 1 for planets and satellites without the central mass factor – it cancels only when comparing bodies around the same central mass

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 6

Assertion (A): The time period of revolution of a satellite close to surface of earth is smaller than that revolving away from surface of earth. Reason (R): The square of time period of revolution of a satellite is directly proportional to cube of its orbital radius.

Push further

More challenging

14 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 14

A satellite orbits Earth at an altitude equal to Earth's radius (R_E). Another satellite is in a geostationary orbit at an altitude of 5.6 R_E. If the orbital period of the geostationary satellite is 24 hours, what is the orbital period of the first satellite?

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