Answers to Chapter 5 problems

5.1.1 (a) 6.4 × 10²⁰ N (b) 2.3 × 10²⁰ N

5.2.1 (a) 5.9 × 10⁻³ m s⁻² (b) 3.3 × 10⁻⁵ m s⁻². However tides are influenced by the gradients of the gravitational field strengths of the Sun and Moon rather than their magnitudes. These gradients are, respectively, − 7.9 × 10⁻¹⁴ s⁻² and − 1.7 × 10⁻¹³ s⁻² so that the influence of the Moon is greater.

5.2.1 Solution

5.5.1 6.39 × 10⁻¹⁰ N

5.5.1 Solution

5.5.2 \( { 2GmM\over R^2} \left[ {1- {d\over \sqrt{R^2+d^2}} } \right] \)

5.5.2 Solution

5.5.3 85 minutes

5.5.3 Solution

5.6.1 14.4 m s⁻¹

5.6.1 Solution

5.6.2 76°

5.6.2 Solution

5.7.1 (b) 6,516 s ; 1,678 m s⁻¹ (c) The spring balance reads 0 N throughout the oscillation.

5.7.1 Solution

5.7.2 (a) √(2gR(1 − cosq) (b) − 2g(1 − cosq) (c) gsinq ; 48.18°

5.7.2 Solution

5.7.3 5.94 × 10⁶ J

5.7.3 Solution

5.7.4 (a) − 4.2 × 10⁸ J (b) − 7.8 × 10⁸ J (c) − 6.7 × 10⁹ J

5.7.4 Solution

5.9.1 85 kJ

5.9.1 Solution

5.9.2 27.0 N

5.9.2 Solution

5.9.3 (a) 2.5 s (b) 88 mJ (c) 0.11 m

5.9.3 Solution

5.10.1 3,569 m

5.10.1 Solution

5.10.2 (a) 0.3 nm (b) 9.7 × 10⁻⁷ N

5.10.2 Solution

5.10.3

5.10.3 Solution

5.10.4 2.1 m

5.10.4 Solution

5.10.5 \( m_2 \sqrt{{2G \over m_1+m_2} \left( {1 \over r} - {1 \over r_0} \right)} \) and \( m_1 \sqrt{{2G \over m_1+m_2} \left( {1 \over r} - {1 \over r_0} \right)} \)

5.10.5 Solution

5.11.1 (a) 1 × 10⁻³⁴ J s (b) 2.7 × 10⁴⁰ J s (c) 0.2 J s

5.11.1 Solution

5.12.1 (a) 2,149 m s⁻¹ (b) 112 km (c) 560 m s⁻¹

5.12.1 Solution

5.12.2 (a) 455 km (b) 7,645 m s⁻¹ (c) 5,609 s

5.12.2 Solution

5.12.3 42,300 km

5.12.3 Solution

5.12.4 0.67 m s⁻¹; 0.02 J

5.12.4 Solution

5.12.5 (a) 7.3 km s⁻¹ (b) 8.1 × 10¹³ J s (c) − 4.0 × 10¹⁰ J (d) 2.6 × 10⁹ J