Practice question

Given: A 1 kg mass is moved from Earth’s surface to a height of 3.2 × 10⁶ m . What is the change in gravitational potential energy? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: Δ V = -G M_E m (1/r_2 - 1/r_1). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r_1 = R_E = 6.4 × 10⁶ m, r_2 = R_E + h = 9.6 × 10⁶ m . Δ V = -6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1 (1/9.6 × 10⁶- 1/6.4 × 10⁶) . Δ V = -4.002 × 10¹⁴(1.0417 × 10⁻⁷- 1.5625 × 10⁻⁷) . Δ V = -4.002 × 10¹⁴ × (-5.208 × 10⁻⁸) approx 2.08 × 10⁷ J . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.