Question

Two solid spheres of radius R made of the same

type of steel are placed in contact, as shown in
the figures above. The magnitude of the gravitational force that they exert on each other is Fi. When two other solid spheres of radius 3R made of this steel are placed in contact, what is the magnitude of the gravitational force that they exert on each other?

Answer 1

The magnitude of the gravitational force Fᵧ between two solid spheres of radius 3R made of the same steel and placed in contact is (3/4)Fᵢ, which is three-fourths the original force Fᵢ between two spheres of radius R.

Step-by-step explanation:

The question revolves around Newton's Law of Universal Gravitation and the calculation of the gravitational force between two solid spheres of different radii. Newton's Universal Law of Gravitation equation, F = G(M₁M₂)/R², where G is the gravitational constant, M₁ and M₂ the masses, and R the separation distance, provides the key to solving the problem.

For spheres with radius R, we have a force Fᵢ. When the radius is increased to 3R, the volumes and, therefore, masses of the spheres increase by a factor of 3³ or 27, since volume scales with the cube of the radius for spheres. Subsequently, Fᵧ will be (27M₁ * 27M₂)/R² times greater than Fᵢ, with R being the original separation distance of 2R (as they are in contact). However, given the contact, the new separation distance Rᵧ is 6R, leading to (27M₁ * 27M₂)/(6R)². Factoring out the original force Fᵢ, the new force Fᵧ can be determined by multiplying Fᵢ by (27/36), simplifying to (3/4)Fᵢ.