Calculate the gravitational force of attraction between two lead spheres of mass 50 kg and 80 kg whose centers are separated by a distance of 2.0 m. (Universal gravitational constant G = 6.67 × 10⁻¹¹ N·m²/kg²)
Answer and explanation
Correct answer: D. 6.67 × 10⁻⁸ N
- Step 1: Newton's gravitational formula: F = G m1 m2 / r². Step 2. Substitute: F = (6.67 × 10⁻¹¹ × 50 × 80) / 2.0² = (6.67 × 10⁻¹¹ × 4000) / 4 = 6.67 × 10⁻¹¹ × 1000 = 6.67 × 10⁻⁸ N. Answer: 6.67 × 10⁻⁸ N. Why not the others: 1.33 × 10⁻⁷ N fails to square the distance; 3.34 × 10⁻⁸ N halves the result.
Practise more Mechanics, Thermal Physics and Waves.
Free questions with an explanation for every one, and full timed Unified Tertiary Matriculation Examination mock tests.
More Mechanics, Thermal Physics and Waves questions
- A car accelerates uniformly from rest at a constant rate of 2.5 m/s² for 8.0 seconds along a straight highway. What total distance does the…
- A stone is projected horizontally with an initial speed of 15 m/s from the edge of a vertical cliff 45 m high. Calculate its horizontal rang…
- A block of mass 12 kg rests on a rough horizontal concrete floor. If the coefficient of static friction between the surfaces is 0.35, calcul…
- An artillery shell of mass 8.0 kg is fired horizontally at 250 m/s from a mobile gun of mass 1,000 kg on a frictionless track. Determine the…
- A uniform meter rule is balanced horizontally on a knife edge placed at the 40 cm mark when a mass of 60 g is suspended from the 10 cm mark.…
- An electric crane lifts a steel beam of mass 500 kg vertically upwards through a height of 24 m in 12 seconds at constant speed. What is the…