Question

the gravitational force between two objects is 2000 n. the mass of each object is reduced to one - third of its original mass. how must the distance between the objects change to keep the gravitational force between them 2000 n? ∘ the distance must be one - ninth the original distance. ∘ the distance must be one - third the original distance. ∘ the distance must be three times greater. ∘ the distance must be nine times greater.

Explanation:

The formula for gravitational force is $F = G\frac{m_1m_2}{r^2}$, where $F$ is the gravitational force, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them. Initially, $F_1=G\frac{m_{1i}m_{2i}}{r_{i}^2} = 2000\ N$. After the mass of each object is reduced to one - third of its original mass, $m_{1f}=\frac{1}{3}m_{1i}$ and $m_{2f}=\frac{1}{3}m_{2i}$. Let the new distance be $r_f$. We want $F_f = G\frac{m_{1f}m_{2f}}{r_{f}^2}=2000\ N$. Substitute $m_{1f}$ and $m_{2f}$ into the formula for $F_f$: $F_f=G\frac{(\frac{1}{3}m_{1i})(\frac{1}{3}m_{2i})}{r_{f}^2}=G\frac{\frac{1}{9}m_{1i}m_{2i}}{r_{f}^2}$. Since $F_f = F_1=G\frac{m_{1i}m_{2i}}{r_{i}^2}$, we set the two expressions equal: $G\frac{\frac{1}{9}m_{1i}m_{2i}}{r_{f}^2}=G\frac{m_{1i}m_{2i}}{r_{i}^2}$. We can cancel out $G$, $m_{1i}$, and $m_{2i}$ from both sides, getting $\frac{\frac{1}{9}}{r_{f}^2}=\frac{1}{r_{i}^2}$. Cross - multiply to obtain $r_{i}^2 = 9r_{f}^2$, then take the square root of both sides (assuming positive distances) and we get $r_{i}=3r_{f}$, or $r_{f}=\frac{1}{3}r_{i}$. So the distance must be one - third the original distance.

Answer:

The distance must be one - third the original distance.