Physics · Gravitation · NEET
R is NOT the radius of the planet. R is the semi-major axis of the orbit (the average distance of the planet from the Sun). For a circular orbit the semi-major axis is just the orbit radius. Never use the planet's own radius here.
The constant is k = 4π²/(GM), where G is the universal gravitational constant and M is the mass of the central body (the Sun for planets, the Earth for satellites). Because k depends only on M (not on the small orbiting body), every planet around the same Sun shares the SAME k. That is why T²/R³ is equal for all planets of one system.
Set gravity equal to the centripetal force. GMm/R² = mv²/R gives v² = GM/R. Since v = 2πR/T (distance in one revolution over the period), v² = 4π²R²/T². Equate: 4π²R²/T² = GM/R. Rearranging gives T² = (4π²/GM)R³, so T² ∝ R³.
No. The mass m of the orbiting planet cancels out during the derivation (it appears on both sides of GMm/R² = mv²/R). T² depends only on M (mass of the Sun) and R. A light planet and a heavy planet at the same distance have the same period.
It is T² ∝ R³ — square the period, cube the distance. Students often flip it. From this, T ∝ R^(3/2). In NEET the answer 'T proportional to R^(3/2)' is the direct-choice version of the same law.
In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to:
A satellite orbits just above the earth's surface with period T. If d is the mean density of the earth and G the gravitational constant, the quantity 3π/(Gd) represents:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The square of a planet's period is proportional to the cube of its mean orbital distance: T² ∝ R³. It is also called the Law of Periods.
T² = (4π²/GM)R³, where G is the gravitational constant and M is the mass of the central body (Sun or Earth).
Yes. NCERT derives T² = (4π²/GM_E)(R_E + h)³ for Earth satellites. The same law applies, with M being the Earth's mass and R the orbit radius from Earth's centre.
Because T²/R³ = 4π²/(GM) depends only on the Sun's mass M and constants. Since every planet orbits the same Sun, the ratio is identical for all of them.
Mostly as a ratio: (T1/T2)² = (R1/R2)³. Given two planets' distances you find the ratio of their periods, or the reverse. See the Mars–Mercury type problems for worked steps.
