Newton's Law of Universal Gravitation
Q1. Calculate the gravitational force between two 5.0 kg masses separated by 0.30 m. G = 6.67 × 10−11 N m2 kg−2. 2 marks
Q2. The gravitational force between two spheres is 2.0 × 10−7 N. If the distance between their centres is doubled, what is the new force? 1 mark
Q3. Earth's mass is 5.97 × 1024 kg and its radius is 6.37 × 106 m.
(a) Calculate the gravitational field strength at Earth's surface. 1 mark
(b) Calculate the gravitational field strength at an altitude of 300 km. 2 marks
Q4. The gravitational force on a 70 kg astronaut on Earth is 686 N. Calculate the gravitational force on the same astronaut at:
(a) twice Earth's radius from the centre 1 mark
(b) three times Earth's radius from the centre 1 mark
Q5. The Moon has mass 7.35 × 1022 kg and radius 1.74 × 106 m.
(a) Calculate the gravitational field strength on the Moon's surface. 1 mark
(b) Calculate the weight of a 70 kg astronaut on the Moon. 1 mark
(c) Compare this to their weight on Earth as a percentage. 1 mark
Q6. Two masses of 4.0 kg and 6.0 kg are placed 0.50 m apart. A 2.0 kg mass is placed on the line joining them, 0.20 m from the 4.0 kg mass. Calculate the net gravitational force on the 2.0 kg mass. 3 marks
Q7. Consider two planets with masses M and 4M, separated by distance d.
(a) At what point between them is the net gravitational field zero? 2 marks
(b) Explain why this point is closer to the less massive planet. 2 marks
(c) State two model conditions needed when using F = GMm/r2 for these planets. 2 marks
Answer Key
Q1. F = Gm1m2/r2 = 6.67×10−11×25/0.09 = 1.85 × 10−8 N
Q2. F ∝ 1/r2, so new F = 2.0×10−7/4 = 5.0 × 10−8 N
Q3. (a) g = GM/r2 = 6.67×10−11×5.97×1024/(6.37×106)2 = 9.81 m/s2 (b) r = 6.67×106 m, g = 6.67×10−11×5.97×1024/(6.67×106)2 = 8.94 m/s2
Q4. (a) F = 686/4 = 172 N (b) F = 686/9 = 76.2 N
Q5. (a) g = 6.67×10−11×7.35×1022/(1.74×106)2 = 1.62 m/s2 (b) W = 70 × 1.62 = 113 N (c) 113/686 × 100 = 16.5%
Q6. F4 = 6.67×10−11×4×2/0.04 = 1.33×10−8 N toward the 4 kg mass. F6 = 6.67×10−11×6×2/0.09 = 8.89×10−9 N toward the 6 kg mass. The forces oppose, so the net force is 4.44×10−9 N toward the 4 kg mass.
Q7. (a) GM/x2 = G(4M)/(d−x)2, so d−x = 2x and x = d/3 from M. (b) The balance point is closer to the less massive planet M because the object must be nearer that weaker source for its field magnitude to equal the field from 4M. (c) Treat each planet as a point mass, or as spherically symmetric with the field point outside it, and use centre-to-centre separations. The equations give magnitudes; the opposing vector directions must be assigned separately.
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