Gravitation

Answer

Newton's Universal Law of Gravitation states: 'Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportiona…

Answer

Value of G = 6.67 × 10⁻¹¹ Nm²kg⁻² Unit: Nm²kg⁻² or m³kg⁻¹s⁻² The universal gravitational constant G was first determined experimentally by Henry Cavendish in 1798 using a torsion balance. The experim…

Answer

Gravitational field strength (E) at a point is defined as the gravitational force per unit mass experienced by a test mass placed at that point. Derivation: If a test mass m experiences force F = GmM…

Answer

Gravitational potential energy is the work done in bringing a mass from infinity to that point. U = -∫[∞ to r] F·dr = -∫[∞ to r] (GmM/r²)dr U = GmM∫[∞ to r] (1/r²)dr = GmM[-1/r][∞ to r] U = GmM[0 - (…

Answer

Consider a body of mass m on Earth's surface experiencing gravitational force. From Newton's law: F = GmM/R² (where M = Earth's mass, R = Earth's radius) From Newton's 2nd law: F = mg Equating: mg =…

Answer

At height h above Earth's surface: g_h = GM/(R+h)² where R is Earth's radius For small heights (h << R): g_h = GM/R² × 1/(1+h/R)² g_h = g × (1+h/R)⁻² Using binomial approximation: g_h ≈ g(1-2h/R) Th…

Answer

Given: Weight on surface W = 500 N, h = 3200 km, R = 6400 km Using: g_h = g(R²)/(R+h)² Weight at height h: W_h = W × (R²)/(R+h)² W_h = 500 × (6400)²/(6400+3200)² W_h = 500 × (6400)²/(9600)² W_h = 50…

Answer

At depth d below Earth's surface, only the sphere of radius (R-d) contributes to gravitational force. Mass of inner sphere: M' = (4/3)π(R-d)³ρ where ρ is Earth's density g_d = GM'/(R-d)² = G(4/3)π(R…