The separation distance between two 2.0 kg masses is decreased by two-thirds. How is the gravitational force between them affected?

[tex]\begin{gathered} F_1=G\times\frac{2\text{ kg}\times2\text{ kg}}{r^2} \\ \\ F_1=G\times\frac{4\text{ kg}^2}{r^2}..........(1) \end{gathered}[/tex]

Now, the distance between the mass is reduced by two-thirds. Thus, the new distance between them will be "R" which is given as:

[tex]R=r-\frac{2}{3}r=r(1-\frac{2}{3})=\frac{r}{3}[/tex]

Now, the gravitational force between two masses with their distance of separation reduced by two-thirds is given as:

[tex]F_2=G\frac{m_1\times m_2}{R^2}[/tex]

Substituting the values in the above equation, we get:

[tex]\begin{gathered} F_2=G\times\frac{2\text{ kg}\times2\text{ kg}}{(\frac{r}{3})^2} \\ \\ F_2=G\times\frac{4\text{ kg}^2}{\frac{r^2}{9}} \\ \\ \begin{equation*} F_2=G\times\frac{9\times4\text{ kg}^2}{r^2} \end{equation*} \\ \\ F_2=9\times(G\times\frac{4\text{ kg}^2}{r^2})..........(2) \end{gathered}[/tex]

Substituting equation (1) in equation (2), we get:

[tex]\begin{gathered} F_2=9\times(G\times\frac{4\text{ kg}^2}{r^2}) \\ \\ F_2=9F_1 \end{gathered}[/tex]

From the above equation, we observe that the new gravitational force F2 between two masses when their distance of separation is reduced by two-thirds will be nine times that of the original value of gravitational force F1.

Final answer:

The new gravitational force F2 between two masses when their distance of separation is reduced by two-thirds will be nine times that of the original value of gravitational force F1.