Year 12 Physics · Module 5

Escape Velocity

Lesson 16 Worksheet
Name
Date
Class
Section 1, Skill Practice

Q1. Earth's mass = 5.97 × 1024 kg, radius = 6.37 × 106 m.

(a) Calculate the escape velocity from Earth's surface. 1 mark

(b) Calculate the escape velocity from 1000 km altitude. 2 marks

Q2. The Moon has mass 7.35 × 1022 kg and radius 1.74 × 106 m. Calculate its escape velocity. 1 mark

Q3. Explain why escape velocity does not depend on the mass of the escaping object. 2 marks

Section 2, Problem Solving

Q4. Jupiter has mass 1.90 × 1027 kg and radius 7.15 × 107 m.

(a) Calculate the escape velocity from Jupiter's surface. 1 mark

(b) Compare this to Earth's escape velocity. 1 mark

(c) Explain why Jupiter retains hydrogen while Earth does not. 2 marks

Q5. A rocket is launched from Earth's surface at 8.0 km/s.

(a) Will it escape Earth? Show your reasoning. 1 mark

(b) Calculate the maximum distance it reaches from Earth's centre. 2 marks

Section 3, Extended Response

Q6. Consider a black hole with mass 5 solar masses (1 solar mass = 1.99 × 1030 kg).

(a) Calculate the Schwarzschild radius (radius where escape velocity equals c = 3.0 × 108 m/s). 2 marks

(b) Explain what happens to light at this radius. 2 marks

(c) Discuss why the concept of escape velocity from a black hole is misleading. 2 marks

Answer Key

Q1. (a) vesc = √(2GM/R) = √(2×6.67×10−11×5.97×1024/6.37×106) = 1.12 × 104 m/s = 11.2 km/s   (b) r = 7.37×106 m, v = √(2×6.67×10−11×5.97×1024/7.37×106) = 10.4 km/s

Q2. v = √(2×6.67×10−11×7.35×1022/1.74×106) = 2.38 × 103 m/s = 2.38 km/s

Q3. vesc = √(2GM/R): mass m cancels in derivation. Both kinetic energy (1/2 mv2) and gravitational potential energy (GMm/R) are proportional to m, so m cancels.

Q4. (a) v = √(2×6.67×10−11×1.90×1027/7.15×107) = 5.96 × 104 m/s = 59.6 km/s   (b) About 5.3 times Earth's   (c) Jupiter's high escape velocity means it can retain light gases like H and He; Earth's lower vesc allows these to escape over geological time.

Q5. (a) No; 8.0 km/s < 11.2 km/s   (b) 1/2 mv2 − GMm/R = −GMm/rmax. v2/2 − GM/R = −GM/rmax. 32×106/2 − 6.25×107 = −4.09×107 = −GM/rmax. rmax = 9.71×106 m = 9710 km from centre, altitude = 3340 km.

Q6. (a) Rs = 2GM/c2 = 2×6.67×10−11×9.95×1030/(9×1016) = 1.47 × 104 m = 14.7 km   (b) Light cannot escape; gravitational field is so strong that all paths lead inward; event horizon.   (c) For normal objects, escape velocity assumes you could reach that speed; nothing can exceed c, so no object can escape; black hole is better described by spacetime curvature in general relativity.

hscscience.com.au · Worksheet for Lesson 16 · Year 12 Physics Module 5: Advanced Mechanics
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