Kepler's Laws & Orbital Mechanics
Q1. State Kepler's three laws of planetary motion. 3 marks
Q2. Mars has an orbital period of 687 days and semi-major axis 2.28 × 1011 m. Earth's orbital period is 365 days.
(a) Verify Kepler's third law by calculating T2/r3 for both planets. 2 marks
(b) Use Earth's data to find the mass of the Sun. 2 marks
Q3. An asteroid has a semi-major axis of 4.2 AU. Calculate its orbital period in years. 1 mark
Q4. A comet has an elliptical orbit with perihelion 0.59 AU and aphelion 35 AU.
(a) Calculate the semi-major axis. 1 mark
(b) Calculate the orbital period. 1 mark
(c) Calculate the eccentricity. 1 mark
Q5. A satellite in low Earth orbit (r = 6700 km) uses a Hohmann transfer to reach geostationary orbit (r = 42,200 km).
(a) Calculate the semi-major axis of the transfer ellipse. 1 mark
(b) Calculate the transfer orbit period. 2 marks
(c) Calculate the speed at perigee of the transfer orbit. 2 marks
Q6. Kepler's second law states that a line joining a planet to the Sun sweeps out equal areas in equal times.
(a) Explain the physical reason for this law in terms of conservation of angular momentum. 3 marks
(b) Explain why planets move faster at perihelion than at aphelion. 2 marks
(c) Discuss how Newton's law of gravitation explains all three of Kepler's laws. 3 marks
Answer Key
Q1. 1st: Planets orbit in ellipses with Sun at one focus. 2nd: Line from Sun to planet sweeps equal areas in equal times. 3rd: T2 ∝ r3 for all planets.
Q2. (a) Earth: (3.156×107)2/(1.496×1011)3 = 2.97×10−19 s2/m3. Mars: (5.936×107)2/(2.28×1011)3 = 2.97×10−19 s2/m3. Same. (b) M = 4π2r3/(GT2) = 4π2/(G×2.97×10−19) = 1.99 × 1030 kg
Q3. T2 = (4.2)3 = 74.1, T = 8.6 years
Q4. (a) a = (0.59 + 35)/2 = 17.8 AU (b) T = √(17.83) = 75.1 years (c) e = (35 − 0.59)/(2×17.8) = 0.967
Q5. (a) a = (6700 + 42200)/2 = 24450 km (b) T = 2π√(a3/(GM)) = 2π√((2.445×107)3/(6.67×10−11×5.97×1024)) = 3.81×104 s = 10.6 h (c) vp = √(GM(2/rp − 1/a)) = √(6.67×10−11×5.97×1024(2/6.7×106 − 1/2.445×107)) = 9.96×103 m/s
Q6. (a) Angular momentum L = mrv⊥ is conserved (gravity is central force, no torque). Area swept per time = 1/2 rv⊥ = L/(2m) = constant. (b) At perihelion r is smallest; since L = mrv is constant, v must be largest. (c) Inverse-square force produces conic section orbits (ellipse for bound orbits); central force conserves angular momentum (2nd law); T2 ∝ r3 follows from combining F = GMm/r2 with centripetal force and v = 2πr/T.
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