Question

Assume a planet with orbiting radius R and period of revolution T around the sun experiences a gravitational force which follows inverse cube law instead of inverse square law. In this case, the period of revolution is proportional to:

• A. \(\frac{1}{R}\)
• B. \(R^2\)
• C. \(R^{3/2}\)
• D. \(R^4\)

Hint:

For non-Newtonian gravitational laws, use \(F = m v^2/R\) to relate velocity and radius, then find period via \(T = 2 \pi R/v\).

Answer 1

The Correct Option is B

Step 1: Centripetal force condition.
F = mv²/R, and given F ∝ 1/R³, so mv²/R = k/R³.

Step 2: Expressing velocity.
v² = k/(m R²) → v = √(k/m) · 1/R.

Step 3: Relating period to velocity.
T = 2πR/v = 2πR / [√(k/m) · 1/R] = 2π √(m/k) R².

Step 4: Conclusion.
Hence, T ∝ R².