Kepler's third law provides a relationship between a planet's

Question

BlackTom AI · Accepted Answer

Kepler's third law states P^2 ∝ a^3, so the square of a planet’s orbital period is proportional to the cube of the semimajor axis of its orbit.

1. Relevant knowledge points and concepts
- Semimajor axis (a): half the longest diameter of an elliptical orbit; it characterizes the orbit’s size.
- Orbital period (P): the time a body takes to complete one orbit.
- Kepler’s third law (classical form): P^2 = k a^3 for bodies orbiting the same central mass, where k is the same constant for that system. Newton’s generalization gives P^2 = (4π^2/ G(M + m)) a^3, showing the dependence on central mass M (and negligible planet mass m typically).

2. Step-by-step reasoning
- Kepler observed for planets orbiting the Sun that the ratio P^2/a^3 is constant. That directly links P and a, not orbital shape, physical size, or rotation on axis.
- Starting from Newtonian gravity, equate gravitational force and centripetal acceleration for near-circular motion and integrate for elliptical orbits to get P^2 = 4π^2 a^3 / G(M + m). For planets, m << M, so P^2 ≈ 4π^2 a^3 / GM, reaffirming P^2 ∝ a^3.

3. Intermediate steps (algebraic)
- From Newton: GMm/r^2 = m v^2 / r. For circular orbit v = 2πr / P.
- Substitute v: GM / r^2 = (4π^2 r) / P^2 ⇒ P^2 = 4π^2 r^3 / GM. Replace r by semimajor axis a for elliptical orbits: P^2 = 4π^2 a^3 / G(M + m).

4. Why the correct answer is correct
- The law explicitly connects orbital period and semimajor axis. Observationally and theoretically, changing a changes P according to P^2 ∝ a^3.

5. Why the other options are incorrect
- "Orbital eccentricity and length of semimajor axis.": Eccentricity describes orbit shape; Kepler’s third law does not relate eccentricity to a in determining period.
- "Physical radius and its time of rotation on its axis.": That describes a planet’s spin and size, unrelated to orbital motion about the Sun.
- "Mass and orbital period.": While Newton’s form includes the central mass M in the constant, the standard statement of Kepler’s third law relates period to semimajor axis; planetary mass (m) does not control the period in the simple Keplerian statement.

6. Conclusion
The correct choice is “orbital period and length of semimajor axis.” This matches the provided answer: orbital period and length of semimajor axis.

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