Gravitation — High-Yield Notes & Problem-Solving Hacks

Newton's Law of Gravitation 1. Core Concept: Newton's Law of Gravitation The gravitational force between two point masses m1 and m2 separated by a distance r is: F = (G × m1 × m2) / r² Where: • G = 6.67 × 10⁻¹¹ N·m²/kg² • m1 and m2 = Masses of the two objects • r = Distance between their centers Vector Form F12 = -(G × m1 × m2 / r²) r̂12 The gravitational force is always attractive and acts along the line joining the centers of the two masses. Superposition Principle The net gravitational force on a body is the vector sum of all individual gravitational forces. Fnet = F1 + F2 + F3 + ... ──────────────────────── 2. Acceleration Due to Gravity (g) and Its Variations At the Earth's surface: g = GM / R² Where: • M = Mass of the Earth • R = Radius of the Earth Approximate value: g ≈ 9.8 m/s² Variation with Height (h) Exact Formula: gh = g × (R / (R + h))² Approximation (Only when h is much smaller than R, generally h < 500 km): gh ≈ g × (1 − 2h/R) ──────────────────────── Variation with Depth (d) At a depth d below the Earth's surface: gd = g × (1 − d/R) Important Note: At the center of the Earth (d = R), g = 0 ──────────────────────── 3. High-Yield Solved Example Example 1: Finding the Point of Zero Net Gravitational Force Question: Two point masses M and 4M are fixed at a distance L apart. A third point mass m is placed on the line joining them such that the net gravitational force on it becomes zero. Find its distance from mass M. Solution: Let the distance of mass m from mass M be x. Then the distance from mass 4M is: L − x For zero net force: (G × M × m) / x² = (G × 4M × m) / (L − x)² Cancel the common terms: 1 / x² = 4 / (L − x)² Taking the square root on both sides: 1 / x = 2 / (L − x) Cross multiply: L − x = 2x Therefore: 3x = L So, x = L / 3 Answer: The point is located at a distance L/3 from mass M. ──────────────────────── Topper's Pro Tip / Common Pitfall Common Trap Do not use the approximation: gh ≈ g × (1 − 2h/R) when the height is comparable to Earth's radius (for example, h = R/2 or h = R). JEE Advanced frequently tests this concept. Instead, always use the exact formula: gh = g / (1 + h/R)² whenever: h > 0.05R Quick Shortcut Compare Gravitation directly with Electrostatics. Replace: • G with 1/(4πϵ₀) • Mass (m) with Charge (q) Most force, field, and potential equations become identical. The only difference is that gravity is always attractive, while electrostatic force can be either attractive or repulsive. ──────────────────────── Closing Note Keep practicing! Try applying the electrostatic analogy while studying shell theorems in gravitation. It can save valuable time during exams. If any step feels unclear, ask your questions in the Ocoviz Open Learning Community and keep learning together.