The problems are on the orbits and escape speed worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Gravitational potential energy is U = −GMm/r, with zero at infinity. These worked answers correct the 20 problems of the AP Physics C worksheet on orbits and escape speed, from force on a satellite to falling from rest: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit in bold. Use GM = 3.98 × 10¹⁴ N m²/kg² for Earth and R = 6.37 × 10⁶ m. Always measure r from the centre of the planet, and keep signs consistent when working with negative energies.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 7
Problem 1 Force on a satellite
1. F = GMm/r² = (3.98 × 10¹⁴ × 500) / (7.0 × 10⁶)² = 1.99 × 10¹⁷ / 4.9 × 10¹³ = 4.1 × 10³ N.2. g = F/m = GM/r² = 8.1 m/s².
Problem 2 Potential energy
U = −GMm/r = −(3.98 × 10¹⁴ × 500) / (7.0 × 10⁶) = −2.8 × 10¹⁰ J. The reference point is U = 0 at infinite separation.Problem 3 Orbital speed and period
v = √(GM/r) = √(3.98 × 10¹⁴ / 7.0 × 10⁶) = 7.5 km/s. T = 2πr/v = 2π(7.0 × 10⁶) / 7542 = 5.83 × 10³ s = 97 min.Problem 4 Scaling Kepler’s third law
1. T ∝ r^(3/2), so T₂ = 4^(3/2) T = 8T.2. v ∝ r^(−1/2), so v₂/v₁ = 1/√4 = 1/2.
Problem 5 Weightlessness
1. g = GM/r² = 3.98 × 10¹⁴ / (6.77 × 10⁶)² = 8.7 m/s².2. The astronauts and station are in free fall together, so there is no contact force between them: apparent weightlessness, not zero gravity.
Problem 6 Escape speed
v_esc = √(2GM/R) = √(2 × 3.98 × 10¹⁴ / 6.37 × 10⁶) = 11.2 km/s. In ½mv² = GMm/R the mass m cancels.Problem 7 Energies of a circular orbit
K = GMm/2r = (3.98 × 10¹⁴ × 1200) / (1.6 × 10⁷) = 3.0 × 10¹⁰ J. U = −2K = −6.0 × 10¹⁰ J. E = K + U = −3.0 × 10¹⁰ J.Problem 8 Raising the orbit
1. ΔE = (GMm/2)(1/r₁ − 1/r₂) = (3.98 × 10¹⁴ × 600 / 2)(1.4286 × 10⁻⁷ − 1.0 × 10⁻⁷) = 5.1 × 10⁹ J.2. v₁ = √(GM/r₁) = 7.54 km/s and v₂ = 6.31 km/s. The speed decreases although the energy increases, because U rises by twice the amount that K falls.
Problem 9 Geostationary orbit
1. r³ = GM T²/4π² = 3.98 × 10¹⁴ × (8.62 × 10⁴)² / 39.48 = 7.49 × 10²², so r = 4.22 × 10⁷ m.2. Altitude = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.58 × 10⁷ m.
3. v = 2πr/T = 3.07 km/s.
Problem 10 Period of the Moon
T = 2π√(r³/GM) = 2π√(5.66 × 10²⁵ / 3.98 × 10¹⁴) = 2π × 3.77 × 10⁵ = 2.37 × 10⁶ s = 27.4 days.Problem 11 Escape from orbit
1. v_esc = √(2GM/r) = √(2 × 3.98 × 10¹⁴ / 6.77 × 10⁶) = 10.8 km/s.2. The orbital speed is 7.67 km/s, so Δv = 10.85 − 7.67 = 3.2 km/s.
Problem 12 Energy to escape
E_min = GMm/R = 3.98 × 10¹⁴ × 2.0 × 10³ / 6.37 × 10⁶ = 1.25 × 10¹¹ J.Problem 13 Perigee and apogee
1. Angular momentum is conserved: v_a = v_p r_p / r_a = 9000 × 7.0 × 10⁶ / 1.73 × 10⁷ = 3.6 km/s.2. a = (r_p + r_a)/2 = 1.2 × 10⁷ m. E/m = −GM/2a = −3.98 × 10¹⁴ / 2.43 × 10⁷ = −1.6 × 10⁷ J/kg.
Problem 14 Reading a potential energy graph
1. 62.5 MJ/kg (the surface value is U/m = −GM/R = −62.5 MJ/kg).2. E/m = −GM/(2 × 2R) = −62.5 / 4 = −15.6 MJ/kg.
3. ΔE/m = −15.6 − (−62.5) = 46.9 MJ/kg.
Problem 15 Spot the error
1. A circular orbit has K = −E, so losing energy increases K.2. Drag lowers the orbit and the satellite speeds up: v = √(GM/r) grows as r falls; the extra kinetic energy comes from gravitational potential energy.
Problem 16 Integrating to find the work
1. W = ∫ from R to 2R of GMm/r² dr = GMm(1/R − 1/2R) = GMm/2R.2. W = 3.98 × 10¹⁴ × 1500 / (2 × 6.37 × 10⁶) = 4.7 × 10¹⁰ J.
3. mgh = 1500 × 9.81 × 6.37 × 10⁶ = 9.4 × 10¹⁰ J, twice as large, because g falls with height and is not constant.
Problem 17 Comet orbit
1. a = (1.00 + 2.50) × 10¹¹ / 2 = 1.75 × 10¹¹ m.2. T = 2π√(a³/GM) = 2π√(5.36 × 10³³ / 1.33 × 10²⁰) = 3.99 × 10⁷ s = 1.26 years.
3. v² = GM(2/r − 1/a): at perihelion v = 4.36 × 10⁴ m/s; by angular momentum v_a = 4.36 × 10⁴ × 1.00 / 2.50 = 1.74 × 10⁴ m/s.
Problem 18 Mass of a planet from a moon
1. T = 1.77 × 86 400 = 1.53 × 10⁵ s.2. M = 4π²r³ / (G T²) = 39.48 × 7.515 × 10²⁵ / (6.67 × 10⁻¹¹ × 2.339 × 10¹⁰) = 1.9 × 10²⁷ kg.
Problem 19 Speed at infinity
1. ½mv₀² − GMm/R = ½m v∞².2. v∞² = v₀² − 2GM/R = 2.25 × 10⁸ − 1.25 × 10⁸ = 1.00 × 10⁸, so v∞ = 10 km/s.
Problem 20 Falling from rest
1. W = ∫ F dr = GMm(1/r₂ − 1/r₁) = 3.98 × 10¹⁴ × 250 × (1.4286 × 10⁻⁷ − 1.0 × 10⁻⁷) = 4.3 × 10⁹ J.2. ½mv² = W, so v = √(2 × 4.27 × 10⁹ / 250) = 5.8 km/s.
