Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now

Explanation:

Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.

Mathematically, F = GMaMb/R² ... 1

G is the gravitational constant

Ma and Mb are the masses of the balls

R is the distance between the balls

If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;

F₂ = GMa(3Mb)/(3R)²

F₂ = 3GMaMb/9R² ... 2

Dividing equation 1 by 2 we will have;

F₂/F = (3GMaMb/9R²)/GMaMb/R²

F₂/F =  3GMaMb/9R² * GMaMb/R²

F₂/F = 3/9

F₂/F = 1/3

F₂ = 1/3 F

This shows that the magnitude of the new attractive force is one-third that of the initial attractive force

think its the third option

answer(20) psi;