The questions are on the orbits and escape speed test. Take it under real conditions before reading the mark scheme.
Quick answer: Gravitational potential energy is U = −GMm/r, with zero at infinity. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics C test on orbits and escape speed, from field strength to Kepler’s third law, with the equivalent answers to accept.
The model answers show where each mark is earned. Method marks are awarded for correct equations even when the arithmetic is wrong. Derivations must show the integral and its limits, and explanations should state cause and effect.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 7
Question 1 Field strength (2 marks)
g ∝ 1/r², so g = 9.81 / 3² = 1.09 m/s².Marking: inverse-square idea (1); answer with unit (1).
Question 2 Low Earth orbit (3 marks)
a. GMm/r² = mv²/r.b. r = 6.97 × 10⁶ m, v = √(GM/r) = 7.56 km/s.
c. T = 2πr/v = 5.79 × 10³ s (96.6 min).
Marking: one mark per part.
Question 3 Effect of drag (2 marks)
Drag removes mechanical energy, so E = −GMm/2r becomes more negative and r decreases. Since v = √(GM/r), a smaller r gives a larger v. Kinetic energy K = −E rises.Marking: E decreases so r decreases (1); v = √(GM/r) conclusion (1).
Question 4 Escape from the Moon (3 marks)
Escape needs ½mv² − GMm/R = 0, so v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 7.35 × 10²² / 1.74 × 10⁶) = 2.37 km/s.Marking: energy condition (1); formula (1); answer (1).
Question 5 Energies of an orbit (3 marks)
a. K = GMm/2r = 1.33 × 10¹⁰ J.b. U = −GMm/r = −2.65 × 10¹⁰ J.
c. E = K + U = −1.33 × 10¹⁰ J.
Marking: one mark per part, with the correct sign.
Question 6 Comet speeds (3 marks)
Gravity is a central force, so angular momentum about the Sun is conserved: m v_p r_p = m v_a (4r_p). Hence v_a = v_p / 4 = 5.0 km/s. The speed is smaller far away.Marking: central force, L conserved (1); equation (1); answer (1).
Question 7 Deriving U (2 marks)
U(r) = −∫ from ∞ to r of F dr = ∫ from ∞ to r of GMm/r² dr = GMm[−1/r] from ∞ to r = −GMm/r.Marking: correct integral and limits (1); evaluation (1).
Question 8 Kepler’s third law (2 marks)
T² ∝ r³, so T ∝ r^(3/2) and the ratio is 4.0^(3/2) = 8.0.Marking: use of r^(3/2) (1); answer (1).
