Year 12 Physics · Module 5

Phase 3 Consolidation

Lesson 18 Worksheet
Name
Date
Class
Section 1, Skill Practice

Q1. A projectile is launched from ground level at 35 m/s at 40° above horizontal. Calculate:

(a) the maximum height 1 mark

(b) the range 1 mark

(c) the speed at maximum height 1 mark

Q2. A satellite orbits Earth at radius 10,000 km. Calculate:

(a) the orbital speed 1 mark

(b) the total energy per unit mass 1 mark

(c) the escape velocity from this orbit 1 mark

Q3. A 0.50 kg mass moves in a vertical circle of radius 1.5 m. At the top its speed is 5.0 m/s. Calculate the tension in the string. 2 marks

Section 2, Problem Solving

Q4. A planet has mass 8.0 × 1024 kg and radius 7.0 × 106 m.

(a) Calculate the gravitational field strength at the surface. 1 mark

(b) Calculate the escape velocity. 1 mark

(c) Calculate the orbital period of a satellite 2000 km above the surface. 2 marks

Q5. Using Kepler's third law, show that T2/r3 = 4π2/(GM) for a circular orbit. 3 marks

Section 3, Extended Response

Q6. A spacecraft is in a circular orbit around Earth. Its engines fire briefly to increase its speed.

(a) Explain why the spacecraft enters an elliptical orbit rather than a larger circular orbit. 2 marks

(b) Explain where in the new orbit the spacecraft is moving fastest. 2 marks

(c) Describe the manoeuvre needed to circularise the orbit at the new apogee. 2 marks

Answer Key

Q1. (a) h = (35 sin 40°)2/(2×9.8) = 25.7 m   (b) R = 352 sin(80°)/9.8 = 123 m   (c) v = 35 cos 40° = 26.8 m/s (horizontal only)

Q2. (a) v = √(6.67×10−11×5.97×1024/107) = 6.31×103 m/s   (b) E/m = −GM/(2r) = −1.99×107 J/kg   (c) vesc = √2 × v = 8.92×103 m/s

Q3. T + mg = mv2/r, T = 0.50(25/1.5 − 9.8) = 8.33 − 4.9 = 3.43 N

Q4. (a) g = 6.67×10−11×8×1024/(7×106)2 = 10.9 m/s2   (b) v = √(2×6.67×10−11×8×1024/7×106) = 1.23×104 m/s   (c) r = 9×106 m. T = 2π√(r3/(GM)) = 2π√(729×1018/(5.34×1014)) = 7330 s = 122 min

Q5. GMm/r2 = mv2/r, v2 = GM/r. T = 2πr/v, T2 = 4π2r2/v2 = 4π2r2/(GM/r) = 4π2r3/(GM). Therefore T2/r3 = 4π2/(GM).

Q6. (a) Increased speed raises total energy; only one circular orbit exists for each energy. Higher energy means larger orbit, but speed is wrong for circular at that radius. Ellipse has varying speed to match energy.   (b) Fastest at perigee (closest approach to Earth) by conservation of angular momentum / Kepler's second law.   (c) Fire engines retrograde (against motion) at apogee to reduce speed to circular orbit value for that radius.

hscscience.com.au · Worksheet for Lesson 18 · Year 12 Physics Module 5: Advanced Mechanics
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