Q4. In the torsion balance of Example 1, what happens to the angle of twist if the large spheres are moved further away, so that the centres are 0.40 m apart (twice the distance), everything else unchanged?
In the simplified nearby-pair model, with the forces still perpendicular to the bar, the force varies as 1/d, so it becomes ¼ as large, and so does the torque. Since the restoring torque is κθ, the angle of twist also becomes ¼ as large. (The spheres cannot be brought much closer than in Example 1: lead spheres of 100 kg and 1 kg have radii of about 0.13 m and 0.03 m.)
Q5. Large spheres of 158 kg and small spheres of 0.73 kg have their centres 0.23 m apart. Find the gravitational force between a pair. Take G = 6.67 × 10 N m kg.
F = 6.67 × 10 × 158 × 0.73/0.23. The top line is about 7.69 × 10 and 0.23 = 0.0529, so F is about 1.45 × 10 N, about the weight of a 0.015 mg speck of dust (taking g = 9.8 m s).
Q6. If the value of G were suddenly to double, what would happen to your weight, standing on the Earth?
If the Earth’s mass and radius remain fixed and rotation is ignored, your gravitational weight is GMEm/RE2, proportional to G, so it would double, although your mass would not change.