Chapter 7, Problem 42A

To determine

The value of the attractive force existing between T and S.

Answer to Problem 42A

  $5.83\times {10}^{-10}\text{N}$

Explanation of Solution

Given:

The mass of T is $m_1=70.0\text{kg}$ .

The mass of S is $m_2=50.0\text{kg}$ .

The distance between T and Sis $r=20.0\text{}\text{m}$

Formula used:

Gravitational force:

  $F=\frac{G{m}_1{m}_2}{r^2}$

Where, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the masses.

Calculation:

The value of gravitational constant is:

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Then, the attractive force between T and S is,

  $\begin{array}{c}F=\frac{G{m}_1{m}_2}{r^2}\\ =\frac{6.67\times {10}^{-11}\times 70\times 50\text{}}{{\left(20\right)}^2}\\ =5.83\times {10}^{-10}\text{N}\end{array}$

Conclusion:

Thus, the attractive force between T and S is $5.83\times {10}^{-10}\text{N}$ .