Respuesta :
Answer:
F'=1/9*F0
Explanation:
F0 is the gravitational force between the particles. When the distance is triplicated we have that
[tex]F'=G\frac{m_1m_2}{(3r)^{2}}[/tex]
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
[tex]F'=\frac{1}{9}G\frac{m_1m_2}{r^2}=\frac{1}{9}F_0[/tex]
hope this helps!!
Answer:
F₁ = [tex]\frac{1}{9}[/tex]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 = [tex]\frac{1}{9}[/tex]F₀
Therefore, the new force F₁ = [tex]\frac{1}{9}[/tex]F₀