Question

What is the force of attraction between a $50kg$woman and a $70kg$man sitting $1.0m$apart?

Short answer

The force of attraction between the man and woman is$2.33\times {10}^{-7}N.$

Step-by-step solution

Given information.

Mass of man $\left(m_1\right)$ = $70kg$, mass of woman $\left(m_2\right)$ = $50kg$, distance between them (r) = $1m$.

Calculation.

We know, gravitational force is given by : $F=\frac{G{m}_1{m}_2}{r^2}.$

Therefore, the gravitational force of attraction between the man and woman is :

$$\begin{gathered} F=\frac{G{m}_1{m}_2}{r^2}. \\ =\frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)(50\mathrm{kg})(70\mathrm{kg})}{(1.0\mathrm{m}{)}^2}. \\ =2.33\times {10}^{-7}\mathrm{N}. \end{gathered}$$

Final answer.

The gravitational force between the man and woman is$2.33\times {10}^{-7}\mathrm{N}.$