Two objects are attracted to each other by a gravitational force F. If the distance between the objects is doubled, what is the new gravitational force between the objects in terms of F?

Question

BlackTom AI · Accepted Answer

Let's lay out the situation step by step.

- The gravitational force between two masses follows the inverse-square law: F ∝ 1/r^2. Therefore, if the distance r is doubled, the force becomes F' = F × (1/2)^2 = F/4.

- This means the new force is one quarter of the original force, regardless of the actual numeric value of F, as long as F refers to the force at the initial distance.

- Now examine the given answer choices:
  - Option a: 1/9 N. This would imply a reduction by a factor of 9, which is not consistent with the 1/r^2 relationship when r is doubled. 
  - Option b: 4 F. This would imply the force increases by a factor of 4, opposite to what the inverse-square law predicts for an increased distance. 
  - Option c: 1/4 N. This corresponds to F' = F/4. If the original force F were 1 N, then F' would be 1/4 N; and the relationship F' = F/4 holds generally for doubling r. 
  - Option d: 16 F. This would imply a massive increase, which again contradicts the 1/r^2 decline when r doubles. 
  - Option e: 9 F. This is another incorrect scaling and does not follow the inverse-square law for r → 2r.

- Conceptually, the key idea is that doubling the separation reduces the gravitational pull by a factor of four, since the force falls with the square of the distance. The numeric value given in the correct option matches the general relation F' = F/4 when the original force is F.

Two objects are attracted to each other by a gravitational force F. If the distance between the objects is doubled, what is the new gravitational force between the objects in terms of F?

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