if the distance between two objects is decreased to $\frac{1}{10}$ of the original distance, how will it…

Explanation:

Step1: Recall gravitational - force formula

The gravitational force between two objects is given by $F = G\frac{m_1m_2}{r^2}$, where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them.

Step2: Let the original distance be $r_1$ and new distance be $r_2$

Given that $r_2=\frac{1}{10}r_1$. Let the original force be $F_1 = G\frac{m_1m_2}{r_1^2}$ and the new force be $F_2 = G\frac{m_1m_2}{r_2^2}$.

Step3: Substitute $r_2$ into the formula for $F_2$

Substitute $r_2=\frac{1}{10}r_1$ into $F_2$: $F_2 = G\frac{m_1m_2}{(\frac{1}{10}r_1)^2}=G\frac{m_1m_2}{\frac{1}{100}r_1^2}=100\times G\frac{m_1m_2}{r_1^2}$. Since $F_1 = G\frac{m_1m_2}{r_1^2}$, we have $F_2 = 100F_1$.

Answer:

A. The new force will be 100 times more than the original.