Question

Two particles are separated by a certain distance. The force of gravitational interaction between them is F0. Now the separation between the particles is tripled. Find the new force of gravitational interaction F1. Express your answer in terms of F0.

Answer 1

Answer:

F₁ = $\frac{1}{9}$F₀

Explanation:

Newton's law of universal gravitation states that the force of attraction or repulsion, F, between two particles of masses M₁ and M₂ is directly proportional to the product of these particles and inversely proportional to the square of the distance, r, between the two particles. i.e

F ∝ M₁M₂ / r²

F = GM₁M₂ / r²            --------------------(i)

Where;

G is the constant of proportionality.

From equation (i), since the force is inversely proportional to the square of the distance, holding other variables constant, the equation can be reduced to;

F = k / r²

This implies that;

Fr² = k          -------------------(ii)

Now, according to the question;

F = F₀

Substitute this into equation (ii) as follows;

F₀ r² = k       ----------------(iii)

Also, when the distance of separation, r, is trippled i.e r becomes 3r;

F = F₁

Substitute these values into equation (ii) as follows;

F₁(3r)² = k

9F₁r² = k                ---------------(iv)

Substitute the value of k in equation (iii) into equation (iv) as follows;

9F₁r² = F₀ r²             --------------(v)

Cancel r² on both sides of equation (v)

9F₁ = F₀

Now make F₁ subject of the formula

F₁ = $\frac{1}{9}$F₀

Therefore, the new force F₁ = $\frac{1}{9}$F₀

Answer 2

Answer:

F'=1/9*F0

Explanation:

F0 is the gravitational force between the particles. When the distance is triplicated we have that

$F'=G\frac{m_1m_2}{(3r)^{2}}$

where r is the distance before the particles are separated, m1 and m2 are the masses their masses and G is the Canvendish's constant.

By some algebra we have

$F'=\frac{1}{9}G\frac{m_1m_2}{r^2}=\frac{1}{9}F_0$

hope this helps!!