Question

According to Newton, when the distance between two interacting objects doubles, the gravitational force is

Answer 1

Answer:

1/4 of its original value

Explanation:

Newton's law of universal gravitation states that when two bodies of masses M₁ and M₂ interact, the force of attraction (F) between these bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between these bodies. i.e

F ∝ $\frac{M_1 M_2}{r^2}$       ------------(i)

From the equation above, it can be deduced that;

F ∝ $\frac{1}{r^2}$

=> F = G $\frac{1}{r^2}$       -----------(ii)

Where;

G = constant of proportionality called the gravitational constant

Equation (ii) can be re-written as

Fr² =  G

=> F₁r₁² = F₂r₂²              -----------(iii)

Where;

F₁ and r₁ are the initial values of the force and distance respectively

F₂ and r₂ are the final values of the force and distance respectively

From the question, if the distance doubles i.e;

r₂ = 2r₁,

Then the final value of the gravitational force F₂ is calculated as follows;

Substitute the value of r₂ = 2r₁ into equation (iii) as follows;

F₁r₁² = F₂(2r₁)²

F₁r₁² = 4F₂r₁²          [Divide through by r₁²]

F₁ = 4F₂                 [Make F₂ subject of the formula]

F₂ = F₁ / 4              [Re-write this]

F₂ = $\frac{1}{4} F_1$

Therefore the gravitational force will be 1/4 of its original value when the distance between the bodies doubles.