Question

1. newton’s law of gravitation states that any two bodies attract each other with a force that is inversely proportional to the square of the distance between them. the force of attraction between two bodies is 450×10^(-12) n when they are 5 cm apart. what is the force on the bodies when they are 23 cm apart? express the answer in engineering notation.

Explanation:

Step1: Write the inverse - square law formula

According to Newton's law of gravitation, $F_1d_1^{2}=F_2d_2^{2}$, where $F_1$ and $d_1$ are the initial force and distance, and $F_2$ and $d_2$ are the final force and distance.

Step2: Identify the given values

We have $F_1 = 450\times10^{- 12}\text{ N}$, $d_1 = 5\text{ cm}=0.05\text{ m}$, and $d_2 = 23\text{ cm}=0.23\text{ m}$.

Step3: Solve for $F_2$

From $F_1d_1^{2}=F_2d_2^{2}$, we can express $F_2$ as $F_2=\frac{F_1d_1^{2}}{d_2^{2}}$.
Substitute the values:
\[

$$\begin{align*} F_2&=\frac{450\times 10^{-12}\times(0.05)^{2}}{(0.23)^{2}}\\ &=\frac{450\times10^{-12}\times0.0025}{0.0529}\\ &=\frac{1.125\times10^{-12}}{0.0529}\\ &\approx21.3\times10^{-12}\text{ N}\\ & = 2.13\times10^{-11}\text{ N} \end{align*}$$

\]

Answer:

$2.13\times 10^{-11}\text{ N}$