7.5 The distance between the objects is halved. 76 The distance between the objects are doubled. 7.7 One of the masses is doubled and the distance halved. 7.8 goth of the masses are doubled and the distance halved. 7.9 Both the masses and the distances are doubled 8.1 Calculate the gravitational aturaction force between a man with a mass of 100 kg and a child with a 20 kg if they are 100 mm apart.

Below are the step‐by‐step answers for each part. ───────────────────────────── 7.5 Distance Halved ───────────────────────────── The gravitational force is given by   F = G (m₁ m₂) / r² If the distance r is halved (r becomes r/2), then   r² becomes (r/2)² = r²/4. Thus, the new force (F′) becomes   F′ = G (m₁ m₂) / (r²/4) = 4 · G (m₁ m₂)/r² = 4F. So, the force is quadrupled. ───────────────────────────── 7.6 Distance Doubled ───────────────────────────── If the distance is doubled (r becomes 2r), then   r² becomes (2r)² = 4r². Thus,   F′ = G (m₁ m₂) / (4r²) = F/4. So, the force is reduced to one‐quarter of the original. ───────────────────────────── 7.7 One Mass Doubled and Distance Halved ───────────────────────────── Doubling one mass increases the numerator by a factor of 2. Halving the distance increases the force by a factor of 1/(1/2)² = 4. Thus, the overall change factor is   2 (from the mass) × 4 (from the distance) = 8. The force is increased eightfold. ───────────────────────────── 7.8 Both Masses Doubled and Distance Halved ───────────────────────────── Doubling both masses multiplies the numerator by 2×2 = 4. Halving the distance increases the force by a factor of 1/(1/2)² = 4. Thus, the overall factor is   4 × 4 = 16. The force is increased sixteenfold. ───────────────────────────── 7.9 Both Masses and the Distance Doubled ───────────────────────────── Doubling both masses increases the numerator by 2×2 = 4. Doubling the distance makes r become 2r so that r² becomes 4r². Thus, the new force becomes   F′ = (4 G m₁ m₂) / (4r²) = G m₁ m₂ / r² = F. There is no change in the force. ───────────────────────────── 8.1 Gravitational Force Between a Man and a Child ───────────────────────────── Given:   Mass of man, m₁ = 100 kg   Mass of child, m₂ = 20 kg   Distance, r = 100 mm = 0.1 m   Gravitational constant, G = 6.67×10⁻¹¹ N·m²/kg² Using the formula:   F = G (m₁ m₂) / r² First, compute the product of the masses:   m₁ m₂ = 100 kg × 20 kg = 2000 kg² Then, compute r²:   r² = (0.1 m)² = 0.01 m² Now substitute into the formula:   F = (6.67×10⁻¹¹ N·m²/kg² × 2000 kg²) / 0.01 m²     = (6.67×10⁻¹¹ × 2000) / 0.01     = (1.334×10⁻⁷) / 0.01     = 1.334×10⁻⁵ N Thus, the gravitational attraction force is approximately 1.33×10⁻⁵ newtons. ───────────────────────────── Summary Answers: 7.5: Force becomes 4F (quadrupled). 7.6: Force becomes F/4 (reduced to one‐quarter). 7.7: Force becomes 8F (increased eightfold). 7.8: Force becomes 16F (increased sixteenfold). 7.9: Force remains F (no change). 8.1: Gravitational force ≈ 1.33×10⁻⁵ N.