Answer:
The force of gravitational will decreases by a factor of 4
Explanation:
The given parameters are;
The symbol that represents the given mass = m
The mass of the circular ring = M
The initial distance of the mass "m" located at the point P from the center O = h
The final distance of the mass "m" when moved further away from point O = 2·h
Given that h = r, we have;
[tex]The \ initial \ gravitational \ force, F_1 =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
Therefore, when OP = 2·h, we have;
OP = 2·r and the new gravitational force, F₂, is given as follows;
[tex]F_2 =G \cdot \dfrac{M \cdot m}{(2\cdot r)^{2}} = G \cdot \dfrac{M \cdot m}{4\cdot r^{2}} = \dfrac{1}{4} \times G \cdot \dfrac{M \cdot m}{r^2} = \dfrac{1}{4} \times F_1[/tex]
[tex]\therefore F_2 = \dfrac{1}{4} \times F_1[/tex]
Therefore, when the mass, m, is removed such that OP = 2h, the value of the gravitational force decreases by a factor of 4.
